Contents / Real Analysis / Lebesgue Integration

Chapter 6

Lebesgue Integration

Why the Riemann integral fails under pointwise limits; outer measure, Caratheodory measurability and measurable functions; the integral built from simple functions; the Monotone Convergence Theorem, Fatou's lemma proved from it and Dominated Convergence proved from Fatou; agreement with the Riemann integral; differentiation under the integral sign; and the counterexamples showing domination is necessary.

Introduction

The Riemann integral, built in Riemann Integration, is a complete success at what it was designed for: it integrates every continuous function on a closed bounded interval, it obeys the Fundamental Theorem of Calculus, and it computes the areas that calculus asks about. Its failure is of a different kind, and it is a failure of closure. The class of Riemann-integrable functions is not closed under pointwise limits. Take a sequence of perfectly well-behaved Riemann-integrable functions, let them converge pointwise to a limit, and the limit may not be Riemann integrable at all — so the question "does $\int f_{n} \to \int f$ ? " cannot even be asked, because the right-hand side does not exist.

That is fatal for analysis, because analysis is the subject of limits. A Fourier series is a limit. A solution of a differential equation obtained by iteration is a limit. An expectation in probability is an integral, and the laws of large numbers are statements about limits of such integrals. A theory of integration that loses its objects when you take a limit cannot support any of this.

Lebesgue's repair is a change of the primitive operation. Riemann slices the domain into small intervals and asks that the function be nearly constant on each — which is a demand about the function's continuity. Lebesgue slices the range into horizontal bands and asks how big the set of inputs landing in each band is — which is a demand only that those sets have a size. The demand on the function is thereby weakened enormously, and the price is that we must first say what "how big" means for a set that is not an interval. That is measure theory, and the first two sections build exactly as much of it as the integral needs.

The payoff is three convergence theorems — monotone convergence, Fatou's lemma, dominated convergence — which together say that under hypotheses one can actually check, the limit may be moved inside the integral. They are proved here in a chain: monotone convergence from the definition, Fatou from monotone convergence, dominated convergence from Fatou. The chapter closes with differentiation under the integral sign, which is dominated convergence applied to difference quotients, and with the counterexamples that show each hypothesis is doing real work.

Throughout, $m$ denotes Lebesgue measure on $R$ , $m^{*}$ outer measure, $χ_{E}$ the characteristic function of $E$ , and "a.e." means "except on a set of measure zero". Facts about suprema and infima come from The Real Number System; facts about pointwise and uniform convergence come from Sequences and Series; Riemann sums and Darboux criteria come from Riemann Integration.

6.1Why the Riemann integral is not enough

Before building anything, it is worth seeing the failure concretely, because the shape of the counterexample dictates the shape of the repair.

Definition 6.1 (The Dirichlet function). The Dirichlet function $D : [0, 1] \to R$ is

D (x) = {1, 0, x \in Q, x \in / Q .

Proposition 6.2 (Dirichlet's function is not Riemann integrable). $D$ is bounded on $[0, 1]$ but not Riemann integrable there.

Proof. Let $P$ be any partition of $[0, 1]$ into subintervals $[x_{i - 1}, x_{i}]$ of positive length. Every such subinterval contains both a rational and an irrational, since $Q$ and its complement are each dense in $R$ — density of $Q$ is the theorem Density of the rationals of The Real Number System, and density of the irrationals follows by adding $2 / k$ to a rational in a suitably shrunken subinterval. Hence

M_{i} = [x_{i - 1}, x_{i}] sup D = 1, m_{i} = [x_{i - 1}, x_{i}] in f D = 0

for every $i$ . The upper and lower Darboux sums are therefore $U (D, P) = \sum_{i} 1 \cdot (x_{i} - x_{i - 1}) = 1$ and $L (D, P) = 0$ for every partition $P$ , so the upper integral is $1$ , the lower integral is $0$ , and they do not agree.∎

The proposition alone is only mildly annoying: $D$ is an artificial function. What makes it fatal is that $D$ is a pointwise limit of functions that are as tame as one could wish.

Theorem 6.3 (Riemann integrability is not closed under pointwise limits). There is a sequence $(f_{n})$ of Riemann-integrable functions on $[0, 1]$ , each with $\int_{0}^{1} f_{n} = 0$ , converging pointwise on $[0, 1]$ to a function that is not Riemann integrable.

Proof. Enumerate the rationals of $[0, 1]$ as $q_{1}, q_{2}, q_{3}, \dots$ ; this is possible because $Q \cap [0, 1]$ is countable. Define

f_{n} = χ_{{q_{1}, \dots, q_{n}}}, that is, f_{n} (x) = {1, 0, x \in {q_{1}, \dots, q_{n}}, otherwise.

Each $f_{n}$ differs from the zero function at exactly $n$ points, so it is Riemann integrable with $\int_{0}^{1} f_{n} = 0$ : given $ε > 0$ , cover the $n$ offending points by $n$ subintervals of total length less than $ε$ , and any partition refining that cover has $U (f_{n}, P) < ε$ while $L (f_{n}, P) = 0$ .

The sequence increases, $f_{n} \leq f_{n + 1}$ , and it converges pointwise to $D$ : if $x$ is irrational then $f_{n} (x) = 0$ for all $n$ ; if $x$ is rational then $x = q_{N}$ for some $N$ , and $f_{n} (x) = 1$ for all $n \geq N$ . By the proposition Dirichlet's function is not Riemann integrable, the limit is not Riemann integrable.∎

So a monotone, uniformly bounded sequence of Riemann-integrable functions, with all integrals equal to $0$ , can converge pointwise to something outside the theory entirely. Note where the trouble is not: the integrals $\int f_{n} = 0$ converge perfectly well. The defect is that the limit function has no integral to converge to.

Intuition. Riemann's method surveys a landscape by walking left to right and measuring the height at each step, so it needs the height to change slowly as you walk. Lebesgue's method surveys the same landscape by asking, for each altitude band, how much land lies at that altitude — and then adding up altitude times area. The second question never cared whether neighbouring points had similar heights, which is exactly why $D$ , whose neighbouring points have wildly different heights, is easy for it: the set at height $1$ is $Q$ , a set of total length $0$ , so it contributes nothing.

Figure 6.4. Riemann partitions the domain into vertical strips and needs the function nearly constant on each; Lebesgue partitions the range into horizontal bands and needs only the preimage of each band to have a measure.

Exercises for §6.1 →

6.2Outer measure and measurable sets

The repair needs a notion of the size of an arbitrary subset of $R$ . The construction is in two steps: define an outer size for every set by economical covering, then restrict attention to the sets on which that outer size behaves additively.

Definition 6.5 (Lebesgue outer measure). For $E \subseteq R$ , the outer measure of $E$ is

m^{*} (E) = in f {k = 1 \sum \infty ℓ (I_{k}) : E \subseteq k = 1 ⋃ \infty I_{k}, I_{k} open intervals},

where $ℓ ((a, b)) = b - a$ . The infimum is over all countable covers of $E$ by open intervals.

Definition 6.6 (Null set). $E$ is a null set, or has measure zero, if $m^{*} (E) = 0$ : for every $ε > 0$ there is a countable cover of $E$ by open intervals of total length less than $ε$ . A property holds almost everywhere (a.e.) if the set where it fails is null.

Proposition 6.7 (Basic properties of outer measure). Outer measure satisfies: (i) $m^{*} (\emptyset) = 0$ ; (ii) monotonicity, $A \subseteq B ⟹ m^{*} (A) \leq m^{*} (B)$ ; (iii) countable subadditivity,

m^{*} (k = 1 ⋃ \infty A_{k}) \leq k = 1 \sum \infty m^{*} (A_{k});

and (iv) $m^{*} (I) = ℓ (I)$ for every interval $I$ .

Proof. (i) and (ii) are immediate: the empty set is covered by an interval of any length, and any cover of $B$ covers $A$ , so the infimum for $A$ is over a larger collection.

(iii) Fix $ε > 0$ . For each $k$ choose a cover ${I_{k, j}}_{j}$ of $A_{k}$ with $\sum_{j} ℓ (I_{k, j}) \leq m^{*} (A_{k}) + ε 2^{- k}$ , which is possible by the definition of the infimum (if some $m^{*} (A_{k}) = \infty$ the inequality is trivial). The doubly-indexed family ${I_{k, j}}_{k, j}$ is countable and covers $⋃_{k} A_{k}$ , so

m^{*} (k ⋃ A_{k}) \leq k \sum j \sum ℓ (I_{k, j}) \leq k \sum m^{*} (A_{k}) + ε k \sum 2^{- k} = k \sum m^{*} (A_{k}) + ε .

As $ε > 0$ was arbitrary, (iii) follows.

(iv) For a bounded closed interval $[a, b]$ : the cover ${(a - ε, b + ε)}$ gives $m^{*} ([a, b]) \leq b - a + 2 ε$ for every $ε$ , hence $m^{*} ([a, b]) \leq b - a$ . For the reverse, let ${I_{k}}$ cover $[a, b]$ . By compactness of $[a, b]$ — the Heine–Borel theorem of Metric Spaces — finitely many $I_{1}, \dots, I_{N}$ already cover it, and an induction on $N$ over a chain of overlapping intervals gives $\sum_{k = 1}^{N} ℓ (I_{k}) \geq b - a$ . Hence every cover has total length at least $b - a$ , so $m^{*} ([a, b]) \geq b - a$ . The other interval types differ from $[a, b]$ by at most two points, which are null, so (iii) and (ii) force the same value.∎

Corollary 6.8 (Countable sets are null). Every countable set $E = {x_{1}, x_{2}, \dots} \subseteq R$ has $m^{*} (E) = 0$ .

Proof. Each singleton satisfies $m^{*} ({x_{k}}) \leq ℓ ((x_{k} - ε, x_{k} + ε)) = 2 ε$ for every $ε > 0$ , so $m^{*} ({x_{k}}) = 0$ . Countable subadditivity from the proposition Basic properties of outer measure gives $m^{*} (E) \leq \sum_{k} 0 = 0$ .∎

In particular $m^{*} (Q) = 0$ , which is already the whole explanation of why the Dirichlet function will turn out to have integral $0$ .

Outer measure is defined for every set, which sounds like a triumph but is in fact the problem: it is not additive on disjoint sets in general. Vitali's construction produces disjoint sets $V_{1}, V_{2}, \dots$ whose union is contained in $[0, 3]$ but with $\sum_{k} m^{*} (V_{k})$ either $0$ or $\infty$ , neither compatible with a finite total. The fix is to discard the offenders.

Definition 6.9 (Measurable set (Carathéodory)). A set $E \subseteq R$ is Lebesgue measurable if for every $A \subseteq R$ ,

m^{*} (A) = m^{*} (A \cap E) + m^{*} (A \cap E^{c}) .

For measurable $E$ we write $m (E) = m^{*} (E)$ and call it the Lebesgue measure of $E$ . The collection of measurable sets is denoted $M$ .

The condition says $E$ splits every test set additively, not merely itself. By subadditivity the inequality $\leq$ always holds, so only $\geq$ needs checking in practice.

Theorem 6.10 (The measurable sets form a σ-algebra and $m$ is countably additive). $M$ contains $\emptyset$ and $R$ , is closed under complements and under countable unions and intersections, and contains every interval and every null set. Moreover, if $E_{1}, E_{2}, \dots \in M$ are pairwise disjoint then

m (k = 1 ⋃ \infty E_{k}) = k = 1 \sum \infty m (E_{k}) .

Proof. Closure under complement is immediate, the Carathéodory condition being symmetric in $E$ and $E^{c}$ . For null $E$ : $m^{*} (A \cap E) \leq m^{*} (E) = 0$ and $m^{*} (A \cap E^{c}) \leq m^{*} (A)$ by monotonicity, so the two terms sum to at most $m^{*} (A)$ , and $\geq$ is subadditivity.

For finite unions, let $E, F \in M$ and test with $A$ . Applying the condition for $E$ , then for $F$ with test set $A \cap E^{c}$ :

m^{*} (A) = m^{*} (A \cap E) + m^{*} (A \cap E^{c} \cap F) + m^{*} (A \cap E^{c} \cap F^{c}) .

Since $A \cap (E \cup F) = (A \cap E) \cup (A \cap E^{c} \cap F)$ , subadditivity bounds $m^{*} (A \cap (E \cup F))$ by the sum of the first two terms, so $m^{*} (A) \geq m^{*} (A \cap (E \cup F)) + m^{*} (A \cap (E \cup F)^{c})$ , which is the required inequality. Hence $M$ is an algebra.

For disjoint $E_{1}, \dots, E_{N} \in M$ and any $A$ , an induction using the condition for $E_{N}$ with test set $A \cap ⋃_{k \leq N} E_{k}$ gives

m^{*} (A \cap k = 1 ⋃ N E_{k}) = k = 1 \sum N m^{*} (A \cap E_{k}) . (*)

Now let $E = ⋃_{k = 1}^{\infty} E_{k}$ with the $E_{k}$ disjoint and measurable, and put $F_{N} = ⋃_{k \leq N} E_{k} \in M$ . For any $A$ ,

m^{*} (A) = m^{*} (A \cap F_{N}) + m^{*} (A \cap F_{N}^{c}) \geq k = 1 \sum N m^{*} (A \cap E_{k}) + m^{*} (A \cap E^{c}),

using $(*)$ and $F_{N}^{c} \supseteq E^{c}$ . Letting $N \to \infty$ and then applying countable subadditivity to $A \cap E = ⋃_{k} (A \cap E_{k})$ ,

m^{*} (A) \geq k = 1 \sum \infty m^{*} (A \cap E_{k}) + m^{*} (A \cap E^{c}) \geq m^{*} (A \cap E) + m^{*} (A \cap E^{c}),

so $E \in M$ ; a general countable union is disjointified by $E_{k}^{'} = E_{k} ∖ ⋃_{j < k} E_{j}$ , each measurable since $M$ is an algebra. Taking $A = E$ in the displayed chain turns both inequalities into equalities and yields countable additivity. Intersections follow from unions by De Morgan. Finally $(a, \infty) \in M$ by a direct check on covers, and the intervals generate from these by complement and intersection.∎

Example 6.11 (The Cantor set is uncountable and null). Let $C \subseteq [0, 1]$ be the Cantor set, obtained by removing the open middle third of $[0, 1]$ , then the open middle thirds of the two remaining intervals, and so on. Compute $m (C)$ .

Solution.

At stage $n$ the set $C_{n}$ retained is a union of $2^{n}$ closed intervals each of length $3^{- n}$ , and $C = ⋂_{n} C_{n}$ .
Each $C_{n}$ is a finite union of intervals, hence measurable, and $m (C_{n}) = 2^{n} \cdot 3^{- n} = (2/3)^{n}$ by finite additivity from the theorem The measurable sets form a σ-algebra and $m$ is countably additive.
$C \subseteq C_{n}$ for every $n$ , so monotonicity gives $m^{*} (C) \leq (2/3)^{n}$ for every $n$ . Since $(2/3)^{n} \to 0$ , $m^{*} (C) = 0$ and $C$ is null, hence measurable.
Alternatively, sum the removed pieces: at stage $n$ one removes $2^{n - 1}$ intervals of length $3^{- n}$ , total $\sum_{n \geq 1} 2^{n - 1} 3^{- n} = \frac{1}{3} \sum_{n \geq 0} (2/3)^{n} = \frac{1}{3} \cdot 3 = 1$ , leaving measure $1 - 1 = 0$ .
Yet $C$ is uncountable: the map sending a ${0, 2}$ -valued ternary expansion $\sum a_{k} 3^{- k}$ to the binary number $\sum (a_{k} /2) 2^{- k}$ is a surjection $C \to [0, 1]$ .
Sanity check on the moral: "null" is a statement about length, not about cardinality. The corollary Countable sets are null gives one direction only — countable forces null, null does not force countable.

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Example 6.12 (A fat Cantor set: nowhere dense with positive measure). Repeat the Cantor construction but at stage $n$ remove from each of the $2^{n - 1}$ surviving intervals a centred open interval of length $4^{- n}$ . Find the measure of the resulting set $F$ .

Solution.

At stage $n$ there are $2^{n - 1}$ intervals, and from each one removes length $4^{- n}$ , so the total removed at stage $n$ is $2^{n - 1} 4^{- n}$ .
Summing over $n$ : $n = 1 \sum \infty 2^{n - 1} 4^{- n} = \frac{1}{2} n = 1 \sum \infty (\frac{2}{4})^{n} = \frac{1}{2} \cdot \frac{1/2}{1 - 1/2} = \frac{1}{2} .$
By countable additivity, $m (F) = 1 - \frac{1}{2} = \frac{1}{2}$ .
$F$ contains no interval — every surviving interval at stage $n$ has length at most $2^{- n} \to 0$ — so $F$ is nowhere dense, and its complement in $[0, 1]$ is dense and open.
Sanity check: a nowhere dense set with measure $\frac{1}{2}$ shows "small in the topological sense" and "small in the measure sense" are genuinely different notions. This is exactly why $χ_{F}$ is a bounded function whose discontinuity set is $F$ itself, so by the theorem Lebesgue's criterion for Riemann integrability below it is not Riemann integrable.

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Pitfall. Null does not mean empty, finite, countable, or closed under any topological operation. The Cantor set is null, uncountable, closed and perfect; a fat Cantor set is nowhere dense with measure $\frac{1}{2}$ ; $Q$ is null and dense. The only thing $m (E) = 0$ asserts is that $E$ can be hidden inside open intervals of arbitrarily small total length.

Exercises for §6.2 →

6.3Measurable functions

With a notion of size for sets, the class of admissible functions is determined by the slicing picture: we need every "above height $a$ " set to be measurable.

Definition 6.13 (Measurable function). An extended-real-valued function $f$ on a measurable set $E$ is measurable if

{x \in E : f (x) > a} \in M for every a \in R .

Proposition 6.14 (Equivalent forms of measurability). For $f$ on a measurable $E$ , the following are equivalent: ${f > a}$ measurable for all $a$ ; ${f \geq a}$ measurable for all $a$ ; ${f < a}$ measurable for all $a$ ; ${f \leq a}$ measurable for all $a$ .

Proof. ${f \geq a} = ⋂_{n \geq 1} {f > a - 1/ n}$ is a countable intersection of measurable sets, so the first condition gives the second. ${f < a} = E ∖ {f \geq a}$ gives the third from the second by complementation, ${f \leq a} = ⋂_{n} {f < a + 1/ n}$ gives the fourth from the third, and ${f > a} = E ∖ {f \leq a}$ closes the cycle. Each step uses only that $M$ is a σ-algebra, by the theorem The measurable sets form a σ-algebra and $m$ is countably additive.∎

Theorem 6.15 (Measurability is preserved by limits). If $f_{1}, f_{2}, \dots$ are measurable on $E$ , then $sup_{n} f_{n}$ , $in f_{n} f_{n}$ , $lim sup_{n} f_{n}$ and $lim inf_{n} f_{n}$ are measurable. In particular a pointwise limit of measurable functions is measurable.

Proof. For the supremum, $x$ satisfies $sup_{n} f_{n} (x) > a$ exactly when $f_{n} (x) > a$ for at least one $n$ , so

{n sup f_{n} > a} = n = 1 ⋃ \infty {f_{n} > a},

a countable union of measurable sets. For the infimum, ${in f_{n} f_{n} < a} = ⋃_{n} {f_{n} < a}$ , measurable by the proposition Equivalent forms of measurability. Then

n lim sup f_{n} = k \geq 1 in f n \geq k sup f_{n}, n lim inf f_{n} = k \geq 1 sup n \geq k in f f_{n},

each an iterated sup/inf of measurable functions, hence measurable. If $f_{n} \to f$ pointwise then $f = lim sup_{n} f_{n}$ .∎

This theorem is the whole point. Compare it with the theorem Riemann integrability is not closed under pointwise limits: the class of measurable functions is closed under exactly the operation that destroyed the Riemann class. Continuous functions are measurable, since ${f > a} = f^{- 1} ((a, \infty))$ is open; monotone functions are measurable, since ${f > a}$ is an interval; and everything obtainable from these by countable limiting operations stays inside.

Example 6.16 (Measurability of the Dirichlet function). Show that $D = χ_{Q}$ is measurable on $R$ .

Solution.

Test ${D > a}$ for each real $a$ . If $a \geq 1$ the set is $\emptyset$ ; if $0 \leq a < 1$ it is $Q$ ; if $a < 0$ it is $R$ .
$\emptyset$ and $R$ are measurable, and $Q$ is null by the corollary Countable sets are null, hence measurable by the theorem The measurable sets form a σ-algebra and $m$ is countably additive.
Every super-level set is measurable, so $D$ is measurable.
Sanity check: measurability is a far weaker demand than Riemann integrability. $D$ fails the latter by the proposition Dirichlet's function is not Riemann integrable, yet passes the former with three lines of checking.

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Note. Measurability of $f$ does not require $f$ to be finite: the extended values $\pm \infty$ are allowed, and one checks ${f = \infty} = ⋂_{n} {f > n}$ is measurable automatically. This matters because $sup_{n} f_{n}$ may genuinely be infinite somewhere, and the monotone convergence theorem below is stated so that an infinite limit is permitted rather than excluded.

Exercises for §6.3 →

6.4Simple functions and the integral

The integral is now built in three stages, each forced by the previous one.

Definition 6.17 (Simple function). A measurable function $φ$ is simple if it takes only finitely many values. Writing its distinct nonzero values as $c_{1}, \dots, c_{N}$ and $E_{k} = φ^{- 1} ({c_{k}})$ , the canonical form is

φ = k = 1 \sum N c_{k} χ_{E_{k}},

with the $E_{k}$ measurable and pairwise disjoint.

Definition 6.18 (Integral of a nonnegative simple function). For a nonnegative simple $φ = \sum_{k = 1}^{N} c_{k} χ_{E_{k}}$ in canonical form and measurable $A$ ,

\int_{A} φ d m = k = 1 \sum N c_{k} m (E_{k} \cap A),

with the convention $0 \cdot \infty = 0$ .

Definition 6.19 (Integral of a nonnegative measurable function). For measurable $f \geq 0$ on $A$ ,

\int_{A} f d m = sup {\int_{A} φ d m : φ simple, 0 \leq φ \leq f on A} \in [0, \infty] .

Definition 6.20 (Integrable function and the general integral). For measurable $f$ , write $f^{+} = max (f, 0)$ and $f^{-} = max (- f, 0)$ , so $f = f^{+} - f^{-}$ and $∣ f ∣ = f^{+} + f^{-}$ . If $\int_{A} ∣ f ∣ d m < \infty$ we call $f$ integrable on $A$ and set

\int_{A} f d m = \int_{A} f^{+} d m - \int_{A} f^{-} d m .

The stages are forced in this order because each supplies what the next needs. Stage one is a definition rather than a theorem only because "height times measure" is what we decided the integral should mean. Stage two is a supremum rather than a limit because no canonical approximating sequence has been singled out yet; the next theorem supplies one and shows the supremum is achieved along it. Stage three splits a signed function because the supremum in stage two is meaningless without a sign.

Figure 6.21. A nonnegative simple function is a finite stack of blocks; its integral is the sum of height times the measure of the base of each block.

Theorem 6.22 (Simple approximation theorem). Let $f \geq 0$ be measurable on $E$ . There is an increasing sequence of nonnegative simple functions $φ_{1} \leq φ_{2} \leq \dots$ with $φ_{n} \to f$ pointwise on $E$ . The convergence is uniform on any set where $f$ is bounded.

Proof. For each $n$ , cut the range $[0, n)$ into $n 2^{n}$ bands of height $2^{- n}$ and define

φ_{n} (x) = ⎩ ⎨ ⎧ \frac{k - 1}{2 ^{n}}, n, if \frac{k - 1}{2 ^{n}} \leq f (x) < \frac{k}{2 ^{n}}, k = 1, \dots, n 2^{n}, if f (x) \geq n .

Each level set of $φ_{n}$ is of the form ${a \leq f < b}$ or ${f \geq n}$ , measurable by the proposition Equivalent forms of measurability, and $φ_{n}$ takes at most $n 2^{n} + 1$ values, so it is simple; clearly $0 \leq φ_{n} \leq f$ .

Monotonicity: refining from level $n$ to level $n + 1$ halves each band, so the value assigned to $x$ either stays put or increases by $2^{- (n + 1)}$ ; and where $f (x) \geq n$ , the new value is at least $n$ . Hence $φ_{n} \leq φ_{n + 1}$ .

Convergence: if $f (x) < \infty$ , then for all $n > f (x)$ we have $0 \leq f (x) - φ_{n} (x) < 2^{- n} \to 0$ . If $f (x) = \infty$ then $φ_{n} (x) = n \to \infty = f (x)$ . Where $f \leq M$ , every $n > M$ gives the uniform bound $∥ f - φ_{n} ∥_{\infty} \leq 2^{- n}$ .∎

This is the "range-slicing" picture made into an algorithm, and it is what the theorem Monotone Convergence Theorem below will be applied to constantly: to prove something about $\int f$ for general $f \geq 0$ , prove it for simple functions and pass to the limit along $φ_{n}$ .

Proposition 6.23 (Linearity and monotonicity of the integral). For integrable $f, g$ and $α, β \in R$ , $α f + β g$ is integrable with

\int_{A} (α f + β g) d m = α \int_{A} f d m + β \int_{A} g d m,

and if $f \leq g$ a.e. then $\int_{A} f d m \leq \int_{A} g d m$ . Moreover $\int_{A} f d m \leq \int_{A} ∣ f ∣ d m$ .

Proof. For nonnegative simple $φ, ψ$ , refine to the common partition generated by the level sets of both; on each piece of that finite partition $φ + ψ$ is constant, and finite additivity of $m$ gives $\int (φ + ψ) = \int φ + \int ψ$ , while $\int (α φ) = α \int φ$ for $α \geq 0$ is immediate from the definition. For nonnegative measurable $f, g$ , take simple sequences $φ_{n} ↑ f$ and $ψ_{n} ↑ g$ from the theorem Simple approximation theorem; then $φ_{n} + ψ_{n} ↑ f + g$ , and the theorem Monotone Convergence Theorem transfers additivity to the limit. For signed integrable $f, g$ , write $h = f + g$ and note $h^{+} - h^{-} = f^{+} - f^{-} + g^{+} - g^{-}$ , so $h^{+} + f^{-} + g^{-} = h^{-} + f^{+} + g^{+}$ ; integrating this identity between nonnegative functions and rearranging the (finite) terms gives additivity. Negative scalars are handled by $(- f)^{\pm} = f^{\mp}$ .

Monotonicity: if $f \leq g$ a.e. then every simple $φ$ with $0 \leq φ \leq f$ satisfies $φ \leq g$ off a null set, and altering a function on a null set does not change the integral, since the corresponding term $c m (\emptyset -up-to-null)$ contributes $0$ . So the supremum defining $\int f$ is over a subfamily of that defining $\int g$ . The final bound follows from $- ∣ f ∣ \leq f \leq ∣ f ∣$ .∎

Corollary 6.24 (Vanishing on null sets). If $m (E) = 0$ then $\int_{E} f d m = 0$ for every measurable $f$ . Consequently, if $f = g$ a.e. then $\int_{A} f d m = \int_{A} g d m$ .

Proof. Any simple $0 \leq φ \leq ∣ f ∣$ has $\int_{E} φ = \sum_{k} c_{k} m (E_{k} \cap E) = 0$ , since each $m (E_{k} \cap E) \leq m (E) = 0$ . Taking the supremum, $\int_{E} ∣ f ∣ d m = 0$ , so $\int_{E} f = 0$ . If $f = g$ a.e., let $N$ be the null set where they differ and split $A = (A ∖ N) \cup N$ ; the integrals agree on $A ∖ N$ and both vanish on $N$ .∎

Example 6.25 (Integrating a simple function). Compute $\int_{R} φ d m$ for $φ = 2 χ_{[0, 1)} + 5 χ_{[1, 3)} + 4 χ_{[3, 3.5)}$ .

Solution.

The function is already in canonical form: distinct values $2, 5, 4$ on disjoint sets.
Measures of the bricks: $m ([0, 1)) = 1$ , $m ([1, 3)) = 2$ , $m ([3, 3.5)) = 0.5$ .
By the definition Integral of a nonnegative simple function, $\int φ = 2 (1) + 5 (2) + 4 (0.5) = 2 + 10 + 2 = 14$ .
Sanity check: $φ$ is a step function, so its Riemann integral over $[0, 3.5]$ is the same sum of rectangle areas, $14$ — as the theorem Riemann and Lebesgue agree below requires.

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Example 6.26 (A simple function on a null set). Compute $\int_{[0, 1]} (7 χ_{Q} + 3 χ_{C}) d m$ , where $C$ is the Cantor set.

Solution.

Both $Q \cap [0, 1]$ and $C$ are null: the first by the corollary Countable sets are null, the second by the example The Cantor set is uncountable and null.
The function is simple, taking values in ${0, 3, 7, 10}$ on measurable sets.
Every set carrying a nonzero value is contained in the null set $(Q \cap [0, 1]) \cup C$ , whose measure is $0$ by subadditivity.
By the corollary Vanishing on null sets, the integral is $0$ .
Sanity check: the function is unbounded in neither value nor support, and is $0$ a.e.; the Lebesgue integral cannot see anything else. It is not Riemann integrable, being discontinuous at every point of $[0, 1]$ .

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Exercises for §6.4 →

6.5The Monotone Convergence Theorem

We now have an integral. The question that motivated the whole construction is whether it commutes with pointwise limits, and the first answer is the best one available.

Theorem 6.27 (Monotone Convergence Theorem). Let $f_{n}$ be measurable on $E$ with $0 \leq f_{1} \leq f_{2} \leq \dots$ and $f_{n} \to f$ pointwise. Then $f$ is measurable and

n \to \infty lim \int_{E} f_{n} d m = \int_{E} f d m,

both sides being allowed the value $+ \infty$ .

Proof. $f = lim f_{n} = sup_{n} f_{n}$ is measurable by the theorem Measurability is preserved by limits. Since $f_{n} \leq f_{n + 1} \leq f$ , monotonicity of the integral gives an increasing sequence $\int f_{n}$ , bounded above by $\int f$ ; let $L = lim_{n} \int_{E} f_{n} \leq \int_{E} f$ . Only $L \geq \int_{E} f$ remains.

Fix a simple $φ$ with $0 \leq φ \leq f$ , and fix $c \in (0, 1)$ . Put

E_{n} = {x \in E : f_{n} (x) \geq c φ (x)} .

Each $E_{n}$ is measurable, $E_{n} \subseteq E_{n + 1}$ because $f_{n}$ increases, and $⋃_{n} E_{n} = E$ : if $φ (x) = 0$ then $x \in E_{1}$ ; if $φ (x) > 0$ then $c φ (x) < φ (x) \leq f (x)$ , so $f_{n} (x)$ , increasing to $f (x)$ , eventually exceeds $c φ (x)$ . Now

\int_{E} f_{n} d m \geq \int_{E_{n}} f_{n} d m \geq c \int_{E_{n}} φ d m = c k \sum c_{k} m (A_{k} \cap E_{n}),

writing $φ = \sum_{k} c_{k} χ_{A_{k}}$ in canonical form. For each $k$ the sets $A_{k} \cap E_{n}$ increase to $A_{k} \cap E$ , so $m (A_{k} \cap E_{n}) \to m (A_{k} \cap E)$ by countable additivity — write the increasing union as a disjoint union of the successive differences and apply the theorem The measurable sets form a σ-algebra and $m$ is countably additive. Letting $n \to \infty$ in the display, the sum being finite,

L \geq c k \sum c_{k} m (A_{k} \cap E) = c \int_{E} φ d m .

Let $c ↑ 1$ : $L \geq \int_{E} φ d m$ . Taking the supremum over all simple $φ \leq f$ , the definition Integral of a nonnegative measurable function gives $L \geq \int_{E} f d m$ .∎

Note what the theorem does not require: no bound on the $f_{n}$ , no integrability of $f$ , no uniformity. Monotonicity alone is enough, because mass cannot escape from an increasing sequence — it can only be added. When $\int f = \infty$ the conclusion is that $\int f_{n} \to \infty$ too, which is a genuine statement, not a vacuous one.

Corollary 6.28 (Term-by-term integration of nonnegative series). If $g_{k} \geq 0$ are measurable on $E$ , then

\int_{E} k = 1 \sum \infty g_{k} d m = k = 1 \sum \infty \int_{E} g_{k} d m .

Proof. The partial sums $s_{n} = \sum_{k \leq n} g_{k}$ are measurable, nonnegative and increasing, with $s_{n} \to \sum_{k} g_{k}$ pointwise. By linearity, $\int s_{n} = \sum_{k \leq n} \int g_{k}$ , and the theorem Monotone Convergence Theorem lets $n \to \infty$ on both sides.∎

Example 6.29 (Monotone convergence in action). Evaluate $n \to \infty lim \int_{0}^{1} \frac{n x ^{n - 1}}{2} d x$ and then $n \to \infty lim \int_{0}^{\infty} \frac{x}{1 + x ^{n}} d m$ on $(0, 1)$ , deciding in each case whether monotone convergence applies.

Solution.

First integral, directly: $\int_{0}^{1} \frac{n}{2} x^{n - 1} d x = \frac{n}{2} \cdot \frac{1}{n} = \frac{1}{2}$ for every $n$ , so the limit is $\frac{1}{2}$ . The pointwise limit of the integrand is $0$ for $x \in (0, 1)$ , so here the limit of the integrals is not the integral of the limit — and indeed the sequence $n x^{n - 1} /2$ is not monotone in $n$ at any fixed $x$ , so the theorem never applied.
Second, on $(0, 1)$ the functions $h_{n} (x) = x / (1 + x^{n})$ increase with $n$ , since $x^{n}$ decreases for $0 < x < 1$ and the denominator therefore shrinks.
Pointwise, $x^{n} \to 0$ on $(0, 1)$ , so $h_{n} (x) \to x$ .
The theorem Monotone Convergence Theorem applies: $lim_{n} \int_{0}^{1} h_{n} d m = \int_{0}^{1} x d m = \frac{1}{2}$ .
Sanity check on the contrast: both limits equal $\frac{1}{2}$ , but only the second is obtained by moving the limit inside. The first is a warning that computing $lim \int$ and $\int lim$ and finding them equal does not mean any theorem licensed the exchange.

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Example 6.30 (Summing a series under the integral sign). Compute $\int_{0}^{1} k = 1 \sum \infty \frac{x ^{k}}{2 ^{k}} d m$ .

Solution.

Each term $g_{k} (x) = x^{k} / 2^{k}$ is nonnegative and measurable on $[0, 1]$ , so the corollary Term-by-term integration of nonnegative series applies with no further hypotheses.
$\int_{0}^{1} \frac{x ^{k}}{2 ^{k}} d x = \frac{1}{2 ^{k} ( k + 1 )}$ .
Hence the answer is $k = 1 \sum \infty \frac{1}{2 ^{k} ( k + 1 )}$ . Multiply and divide by $2$ : $\sum_{k \geq 1} \frac{2}{2 ^{k + 1} ( k + 1 )} = 2 \sum_{j \geq 2} \frac{( 1/2 ) ^{j}}{j} = 2 (- ln (1 - \frac{1}{2}) - \frac{1}{2}) = 2 ln 2 - 1.$
Sanity check: $2 ln 2 - 1 \approx 0.3863$ . Directly, the sum inside is the geometric series $\frac{x /2}{1 - x /2} = \frac{x}{2 - x}$ , and $\int_{0}^{1} \frac{x}{2 - x} d x = \int_{0}^{1} (\frac{2}{2 - x} - 1) d x = 2 ln 2 - 1$ . The two agree.

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Exercises for §6.5 →

6.6Fatou's lemma

Monotonicity is a strong hypothesis. Dropping it entirely still leaves one inequality, and that inequality is the engine for everything else.

Lemma 6.31 (Fatou's lemma). Let $f_{n} \geq 0$ be measurable on $E$ . Then

\int_{E} n \to \infty lim inf f_{n} d m \leq n \to \infty lim inf \int_{E} f_{n} d m .

Proof. Put $g_{k} = in f_{n \geq k} f_{n}$ . Each $g_{k}$ is measurable by the theorem Measurability is preserved by limits, nonnegative, and the sequence increases with $k$ because the infimum is over a shrinking family. By definition $g_{k} ↑ lim inf_{n} f_{n}$ pointwise.

The theorem Monotone Convergence Theorem applies to $(g_{k})$ and gives

\int_{E} n lim inf f_{n} d m = k \to \infty lim \int_{E} g_{k} d m .

For every $n \geq k$ we have $g_{k} \leq f_{n}$ , so monotonicity of the integral gives $\int_{E} g_{k} \leq \int_{E} f_{n}$ for all $n \geq k$ , hence

\int_{E} g_{k} d m \leq n \geq k in f \int_{E} f_{n} d m .

Letting $k \to \infty$ , the left side tends to $\int_{E} lim inf_{n} f_{n}$ and the right side tends to $lim inf_{n} \int_{E} f_{n}$ by definition of the lower limit.∎

The inequality is genuinely one-directional, and the direction has a physical reading: mass can leak away in the limit, so the limit function may have less integral than the sequence did, but mass cannot be manufactured, so the limit function can never have more.

Example 6.32 (Fatou's inequality is strict). Exhibit nonnegative measurable $f_{n}$ on $R$ for which Fatou's inequality is strict.

Solution.

Take $f_{n} = χ_{[n, n + 1]}$ .
Pointwise: fix $x$ ; for all $n > x$ we have $f_{n} (x) = 0$ , so $lim inf_{n} f_{n} = 0$ everywhere and the left side of Fatou is $\int 0 = 0$ .
Integrals: $\int_{R} f_{n} d m = m ([n, n + 1]) = 1$ for every $n$ , so $lim inf_{n} \int f_{n} = 1$ .
The inequality reads $0 \leq 1$ : strict, by exactly the amount of mass that walked off to $+ \infty$ .
Sanity check with the spike $g_{n} = n χ_{(0, 1/ n)}$ : again $lim inf g_{n} = 0$ but $\int g_{n} = 1$ , so $0 \leq 1$ . Horizontal escape and vertical escape are the two ways a sequence loses mass, and Fatou tolerates both.

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Exercises for §6.6 →

6.7The Dominated Convergence Theorem

Fatou is an inequality, and an inequality is not an interchange. To get equality one needs a hypothesis that forbids mass from escaping, and that hypothesis is a single integrable envelope.

Theorem 6.33 (Dominated Convergence Theorem). Let $f_{n}$ be measurable on $E$ with $f_{n} \to f$ a.e., and suppose there is an integrable $g$ with $∣ f_{n} ∣ \leq g$ a.e., for every $n$ . Then $f$ is integrable and

n \to \infty lim \int_{E} f_{n} d m = \int_{E} f d m .

Moreover $\int_{E} ∣ f_{n} - f ∣ d m \to 0$ , that is, $f_{n} \to f$ in $L^{1}$ .

Proof. Passing to the limit in $∣ f_{n} ∣ \leq g$ gives $∣ f ∣ \leq g$ a.e., so $\int_{E} ∣ f ∣ \leq \int_{E} g < \infty$ and $f$ is integrable; altering everything on the union of the countably many null sets involved changes no integral, by the corollary Vanishing on null sets, so we may assume the hypotheses hold everywhere.

Apply the lemma Fatou's lemma to the nonnegative sequence $g + f_{n}$ :

\int_{E} g + \int_{E} f = \int_{E} (g + f) \leq n lim inf \int_{E} (g + f_{n}) = \int_{E} g + n lim inf \int_{E} f_{n},

using linearity and $lim inf_{n} (g + f_{n}) = g + f$ . Since $\int_{E} g$ is finite, it cancels, leaving

\int_{E} f d m \leq n lim inf \int_{E} f_{n} d m .

Now apply Fatou to the nonnegative sequence $g - f_{n}$ :

\int_{E} g - \int_{E} f \leq n lim inf \int_{E} (g - f_{n}) = \int_{E} g - n lim sup \int_{E} f_{n},

because $lim inf (- a_{n}) = - lim sup (a_{n})$ . Cancelling the finite $\int_{E} g$ and negating reverses the inequality to

n lim sup \int_{E} f_{n} d m \leq \int_{E} f d m .

Combining, $lim sup_{n} \int f_{n} \leq \int f \leq lim inf_{n} \int f_{n}$ , which forces all three to coincide. For the $L^{1}$ statement, apply the same argument to $h_{n} = ∣ f_{n} - f ∣$ , which satisfies $h_{n} \to 0$ a.e. and $h_{n} \leq 2 g$ with $2 g$ integrable; the conclusion is $\int_{E} h_{n} \to \int_{E} 0 = 0$ .∎

Three points about the hypotheses deserve emphasis. The dominator must be a single $g$ working for all $n$ simultaneously — a bound $∣ f_{n} ∣ \leq g_{n}$ that drifts with $n$ is no bound at all. The dominator must be integrable, which on an infinite-measure space rules out constants. And the finiteness of $\int g$ is what licenses the cancellations in the proof; with $\int g = \infty$ the two displays would read $\infty \leq \infty$ and say nothing.

Example 6.34 (A DCT computation). Compute $n \to \infty lim \int_{0}^{\infty} \frac{e ^{- x}}{1 + n x ^{2}} d m$ .

Solution.

Pointwise limit: for $x > 0$ the denominator $\to \infty$ , so the integrand $\to 0$ . At $x = 0$ the value is $1$ for every $n$ , but ${0}$ is null, so $f_{n} \to 0$ a.e.
Dominator: $1 + n x^{2} \geq 1$ gives $0 \leq f_{n} (x) \leq e^{- x} =: g (x)$ for all $n \geq 1$ , and $g$ does not depend on $n$ .
Integrability of $g$ : $\int_{0}^{\infty} e^{- x} d x = 1 < \infty$ .
The theorem Dominated Convergence Theorem gives $lim_{n} \int_{0}^{\infty} f_{n} d m = \int_{0}^{\infty} 0 d m = 0$ .
Sanity check: the areas are squeezed under a fixed envelope of area $1$ and collapse pointwise to $0$ , so they must collapse to $0$ .

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Example 6.35 (Bounded convergence on a finite-measure space). On $[0, 1]$ , compute $lim_{n} \int_{0}^{1} x^{n} d m$ using domination, and confirm directly.

Solution.

Pointwise, $x^{n} \to 0$ for $0 \leq x < 1$ and $1^{n} = 1$ ; since ${1}$ is null, $f_{n} \to 0$ a.e.
Dominator: $∣ x^{n} ∣ \leq 1$ on $[0, 1]$ , and the constant $1$ is integrable there because $\int_{0}^{1} 1 d m = m ([0, 1]) = 1 < \infty$ .
By the theorem Dominated Convergence Theorem, the limit is $\int_{0}^{1} 0 d m = 0$ .
Direct confirmation: $\int_{0}^{1} x^{n} d x = \frac{1}{n + 1} \to 0$ .
Sanity check: the constant dominator worked only because $m ([0, 1]) < \infty$ . On $[0, \infty)$ the constant $1$ has infinite integral and dominates nothing usefully.

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Example 6.36 (The domination hypothesis is necessary: the moving bump). Let $f_{n} = χ_{[n, n + 1]}$ on $R$ . Show that the conclusion of dominated convergence fails, and identify precisely which hypothesis is absent.

Solution.

Pointwise: for each fixed $x$ , $f_{n} (x) = 0$ once $n > x$ , so $f_{n} \to 0$ everywhere — the convergence hypothesis holds in its strongest form.
Integrals: $\int_{R} f_{n} d m = 1$ for every $n$ , so $lim_{n} \int f_{n} = 1$ .
Integral of the limit: $\int_{R} 0 d m = 0 \neq = 1$ . The conclusion fails.
Which hypothesis is missing: any $g$ with $g \geq f_{n}$ for every $n$ must satisfy $g \geq 1$ on each of the disjoint intervals $[n, n + 1]$ , $n = 1, 2, \dots$ . By countable additivity, $\int_{R} g d m \geq \sum_{n = 1}^{\infty} m ([n, n + 1]) = \sum_{n = 1}^{\infty} 1 = \infty$ . No integrable dominator exists.
Sanity check against Fatou: the lemma Fatou's lemma still holds here and reads $0 \leq 1$ — it never promised equality, and it is exactly the gap between $0$ and $1$ that domination is designed to close.

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Example 6.37 (The domination hypothesis is necessary: the tall spike). Let $f_{n} = n χ_{(0, 1/ n)}$ on $R$ . Repeat the analysis.

Solution.

Pointwise: if $x \leq 0$ then $f_{n} (x) = 0$ always; if $x > 0$ then $n > 1/ x$ forces $x \in / (0, 1/ n)$ , so $f_{n} (x) = 0$ eventually. Thus $f_{n} \to 0$ everywhere.
$\int f_{n} = n \cdot m ((0, 1/ n)) = n \cdot \frac{1}{n} = 1$ for every $n$ , so $lim_{n} \int f_{n} = 1 \neq = 0 = \int lim_{n} f_{n}$ .
Missing hypothesis: a dominator must satisfy $g (x) \geq n$ whenever $0 < x < 1/ n$ , that is $g (x) \geq ⌊ 1/ x ⌋$ , so $g (x) \geq 1/ x - 1$ for $0 < x < 1$ . But $\int_{0}^{1} (1/ x - 1) d m = \infty$ , since $\int_{ε}^{1} d x / x = - ln ε \to \infty$ .
Sanity check on the contrast with the moving bump: there the mass escaped horizontally to $+ \infty$ ; here it escapes vertically at the origin. Both are ruled out by a single integrable envelope, which is a finite-area ceiling in space as well as in height.

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Figure 6.38. Two ways a sequence loses its mass: the bump walks off to the right, the spike grows narrow and tall. Both have integral 1 for every n and both converge pointwise to 0.

Exercises for §6.7 →

6.8Fatou and monotone convergence as the engine of the proof

The three theorems are not independent facts to be memorised in parallel; they are a chain, and knowing which link you are standing on tells you what you are allowed to do next.

Summary.

Monotone convergence is proved directly from the definition of the integral as a supremum over simple functions, using only countable additivity of $m$ .
Fatou's lemma is proved from monotone convergence by replacing $f_{n}$ with $g_{k} = in f_{n \geq k} f_{n}$ , which is increasing by construction.
Dominated convergence is proved from Fatou applied twice, to $g + f_{n}$ and to $g - f_{n}$ ; the finiteness of $\int g$ is what lets the common term cancel.

Each link costs a hypothesis and buys a conclusion. Monotone convergence demands an ordering and gives equality for free, even when the answer is $\infty$ . Fatou demands nothing beyond nonnegativity and gives only an inequality. Dominated convergence buys back equality by paying for an envelope.

Proposition 6.39 (Monotone convergence from Fatou). The Monotone Convergence Theorem can also be deduced from Fatou's lemma together with monotonicity of the integral.

Proof. Let $0 \leq f_{n} ↑ f$ . Since $f_{n} \leq f$ for every $n$ , monotonicity of the integral gives $\int f_{n} \leq \int f$ , hence $lim sup_{n} \int f_{n} \leq \int f$ . The lemma Fatou's lemma gives the other direction: $lim inf_{n} f_{n} = f$ pointwise, so $\int f \leq lim inf_{n} \int f_{n}$ . Combining, $\int f \leq lim inf_{n} \int f_{n} \leq lim sup_{n} \int f_{n} \leq \int f$ , so all are equal.∎

The proposition is not circular — it merely records that once Fatou is available by any route, monotone convergence is two lines. In the development above, Fatou was derived from monotone convergence, so the logical order is fixed; some books reverse it, proving Fatou first from the definition.

Example 6.40 (Running the DCT squeeze by hand). Show $n \to \infty lim \int_{0}^{\infty} \frac{e ^{- x} sin ( n x )}{n} d m = 0$ by imitating the proof of the theorem Dominated Convergence Theorem rather than quoting it.

Solution.

Set $f_{n} (x) = e^{- x} sin (n x) / n$ and $g (x) = e^{- x}$ . Since $∣ sin (n x) ∣ \leq 1$ and $n \geq 1$ , $∣ f_{n} ∣ \leq e^{- x} = g$ , and $\int_{0}^{\infty} g d m = 1 < \infty$ .
Pointwise, $∣ f_{n} (x) ∣ \leq e^{- x} / n \to 0$ , so $f_{n} \to 0$ everywhere.
Fatou on $g + f_{n} \geq 0$ : $\int g + 0 = \int (g + lim f_{n}) \leq lim inf_{n} \int (g + f_{n}) = \int g + lim inf_{n} \int f_{n}$ , and cancelling the finite $\int g = 1$ gives $0 \leq lim inf_{n} \int f_{n}$ .
Fatou on $g - f_{n} \geq 0$ gives, after the same cancellation, $lim sup_{n} \int f_{n} \leq 0$ .
Therefore $0 \leq lim inf_{n} \int f_{n} \leq lim sup_{n} \int f_{n} \leq 0$ , so the limit exists and equals $0$ .
Sanity check by direct computation: $\int_{0}^{\infty} e^{- x} sin (n x) d x = \frac{n}{1 + n ^{2}}$ , so $\int f_{n} = \frac{1}{1 + n ^{2}} \to 0$ . The squeeze gave the right answer without needing that formula.

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Exercises for §6.8 →

6.9Riemann and Lebesgue agree where both exist

A new integral that disagreed with the old one on the functions calculus cares about would be useless. It does not.

Theorem 6.41 (Lebesgue's criterion for Riemann integrability). A bounded function $f$ on $[a, b]$ is Riemann integrable if and only if its set of discontinuities is null.

Proof. Let $ω (x)$ denote the oscillation of $f$ at $x$ , the infimum of $sup_{I} f - in f_{I} f$ over open intervals $I ∋ x$ ; then $f$ is continuous at $x$ exactly when $ω (x) = 0$ , and the discontinuity set is $⋃_{k \geq 1} D_{k}$ with $D_{k} = {ω \geq 1/ k}$ , each $D_{k}$ closed.

( $\Rightarrow$ ) Suppose $f$ is Riemann integrable and fix $k$ . Given $ε > 0$ choose a partition with $U (f, P) - L (f, P) < ε / k$ . Every open subinterval of $P$ meeting $D_{k}$ contributes at least $(1/ k) \times$ its length to that difference, so those subintervals have total length below $ε$ ; the finitely many partition points are null. Hence $m^{*} (D_{k}) \leq ε$ for every $ε$ , so $D_{k}$ is null, and $⋃_{k} D_{k}$ is null by subadditivity.

( $\Leftarrow$ ) Suppose the discontinuity set is null and $∣ f ∣ \leq M$ . Fix $ε > 0$ and $k$ with $1/ k < ε$ . Since $D_{k}$ is null and closed, hence compact, it is covered by finitely many open intervals of total length under $ε$ . On the compact complement of that cover, $ω < 1/ k$ at every point, so by compactness there is $δ > 0$ such that every subinterval of length $< δ$ meeting it has oscillation below $1/ k$ . Take a partition of mesh $< δ$ refining the cover's endpoints. Subintervals inside the cover contribute at most $2 Mε$ to $U - L$ ; the rest contribute at most $(1/ k) (b - a) < ε (b - a)$ . So $U (f, P) - L (f, P) < (2 M + b - a) ε$ , and $f$ is Riemann integrable by the Riemann criterion of Riemann Integration.∎

Theorem 6.42 (Riemann and Lebesgue agree). If $f$ is Riemann integrable on $[a, b]$ , then $f$ is Lebesgue measurable and integrable there, and

\int_{a}^{b} f (x) d x = \int_{[a, b]} f d m .

Proof. Take partitions $P_{n}$ of mesh tending to $0$ , each refining the last, and let $ℓ_{n}$ and $u_{n}$ be the step functions equal to $in f f$ and $sup f$ on each subinterval of $P_{n}$ . Then $ℓ_{n} \leq f \leq u_{n}$ , the $ℓ_{n}$ increase, the $u_{n}$ decrease, and the Lebesgue integrals of these step functions are exactly the Darboux sums:

\int_{[a, b]} ℓ_{n} d m = L (f, P_{n}), \int_{[a, b]} u_{n} d m = U (f, P_{n}),

by the definition Integral of a nonnegative simple function applied after adding the constant $M$ to make everything nonnegative.

Set $ℓ = lim_{n} ℓ_{n}$ and $u = lim_{n} u_{n}$ , both measurable by the theorem Measurability is preserved by limits, with $ℓ \leq f \leq u$ . All functions are bounded by $M$ on the finite-measure set $[a, b]$ , so the constant $M$ is an integrable dominator and the theorem Dominated Convergence Theorem gives

\int_{[a, b]} ℓ d m = n lim L (f, P_{n}) = \int_{a}^{b} f, \int_{[a, b]} u d m = n lim U (f, P_{n}) = \int_{a}^{b} f,

the last equalities because $f$ is Riemann integrable. Hence $\int_{[a, b]} (u - ℓ) d m = 0$ with $u - ℓ \geq 0$ , which forces $u = ℓ$ a.e.: otherwise ${u - ℓ > 1/ k}$ would have positive measure for some $k$ and contribute a positive amount to the integral. Squeezed between two functions equal a.e., $f$ agrees with $ℓ$ a.e., so $f$ is measurable (differing from a measurable function on a null set) and, by the corollary Vanishing on null sets, $\int_{[a, b]} f d m = \int_{[a, b]} ℓ d m = \int_{a}^{b} f$ .∎

The converse fails, and the theorem Riemann integrability is not closed under pointwise limits already told us it must: $D$ is Lebesgue integrable with $\int_{[0, 1]} D d m = 1 \cdot m (Q \cap [0, 1]) = 0$ , and not Riemann integrable.

Example 6.43 (Thomae against Dirichlet). Both Thomae's function $T$ — $T (p / q) = 1/ q$ in lowest terms, $T = 0$ on irrationals — and the Dirichlet function vanish a.e. on $[0, 1]$ . Decide Riemann integrability of each and compute both integrals.

Solution.

$T$ is continuous at every irrational and discontinuous at every rational: near an irrational $x$ , all fractions with small denominator are bounded away, so $T$ is small nearby; at a rational $p / q$ , irrationals arbitrarily close carry value $0$ while $T (p / q) = 1/ q > 0$ .
So the discontinuity set of $T$ is $Q \cap [0, 1]$ , null by the corollary Countable sets are null. By the theorem Lebesgue's criterion for Riemann integrability, $T$ is Riemann integrable.
Its value: $T = 0$ a.e., so $\int_{[0, 1]} T d m = 0$ by the corollary Vanishing on null sets, and the theorem Riemann and Lebesgue agree makes the Riemann integral $0$ as well.
$D$ is discontinuous at every point, so its discontinuity set has measure $1$ ; Lebesgue's criterion fails and $D$ is not Riemann integrable. Its Lebesgue integral is nonetheless $0$ .
Sanity check on what distinguishes them: not their values (both are $0$ a.e.), not their integrals (both $0$ ), but the size of the discontinuity set — which is the only thing Riemann integrability can see and the only thing the Lebesgue integral ignores.

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Pitfall. "Improper Riemann integrable" and "Lebesgue integrable" are not the same. $\int_{0}^{\infty} \frac{s i n x}{x} d x$ converges as an improper Riemann integral to $π /2$ , but $\int_{0}^{\infty} \frac{s i n x}{x} d m = \infty$ , so $sin (x) / x$ is not Lebesgue integrable on $(0, \infty)$ . The Lebesgue integral is absolute by construction, since it is built from $f^{+}$ and $f^{-}$ separately; a conditionally convergent improper integral depends on the order of cancellation, which the Lebesgue theory refuses to consider.

Exercises for §6.9 →

6.10Differentiation under the integral sign

The most-used corollary of dominated convergence is not about sequences at all. It is about a parameter, and it is dominated convergence applied to difference quotients.

Theorem 6.44 (Differentiation under the integral sign). Let $I \subseteq R$ be an open interval and $f : E \times I \to R$ satisfy: (i) $x \mapsto f (x, t)$ is integrable on $E$ for each $t \in I$ ; (ii) $\partial f / \partial t$ exists for every $x \in E$ and $t \in I$ ; (iii) there is an integrable $g$ on $E$ with

\frac{\partial f}{\partial t} (x, t) \leq g (x) for all x \in E, t \in I .

Then $F (t) = \int_{E} f (x, t) d m (x)$ is differentiable on $I$ and

F^{'} (t) = \int_{E} \frac{\partial f}{\partial t} (x, t) d m (x) .

Proof. Fix $t \in I$ and any sequence $h_{n} \to 0$ with $h_{n} \neq = 0$ and $t + h_{n} \in I$ . Put

q_{n} (x) = \frac{f ( x , t + h _{n} ) - f ( x , t )}{h _{n}} .

Each $q_{n}$ is measurable, being a difference of integrable functions divided by a constant, and by linearity

\frac{F ( t + h _{n} ) - F ( t )}{h _{n}} = \int_{E} q_{n} d m .

Pointwise, $q_{n} (x) \to \partial f / \partial t (x, t)$ by hypothesis (ii). For the dominator, the Mean Value Theorem of Differentiation applied to $s \mapsto f (x, s)$ on the interval between $t$ and $t + h_{n}$ gives a point $ξ$ there with

q_{n} (x) = \frac{\partial f}{\partial t} (x, ξ), so ∣ q_{n} (x) ∣ \leq g (x)

by hypothesis (iii), for every $n$ and $x$ — one dominator, valid for all $n$ at once.

The theorem Dominated Convergence Theorem therefore gives

n \to \infty lim \frac{F ( t + h _{n} ) - F ( t )}{h _{n}} = \int_{E} \frac{\partial f}{\partial t} (x, t) d m (x) .

The limit is the same for every null sequence $(h_{n})$ , so the two-sided limit defining $F^{'} (t)$ exists and equals that integral.∎

The Mean Value Theorem step is where the hypothesis on the derivative rather than on $f$ itself earns its keep: it converts a bound on $\partial f / \partial t$ into a bound on every difference quotient, which is what dominated convergence needs.

Example 6.45 (Differentiating under the sign, verified two ways). Let $F (t) = \int_{0}^{1} e^{t x} d m (x)$ . Compute $F^{'} (t)$ by differentiating under the integral, and check against the closed form.

Solution.

Justify: restrict to $∣ t ∣ \leq M$ . Then $∣ \partial_{t} e^{t x} ∣ = ∣ x e^{t x} ∣ \leq e^{M}$ for $x \in [0, 1]$ , and the constant $e^{M}$ is integrable on $[0, 1]$ , which has finite measure. Hypothesis (iii) of the theorem Differentiation under the integral sign holds with $g \equiv e^{M}$ .
Differentiate inside: $F^{'} (t) = \int_{0}^{1} x e^{t x} d m (x)$ .
Evaluate by parts, for $t \neq = 0$ : $\int_{0}^{1} x e^{t x} d x = [\frac{x}{t} e^{t x}]_{0}^{1} - \frac{1}{t} \int_{0}^{1} e^{t x} d x = \frac{e ^{t}}{t} - \frac{e ^{t} - 1}{t ^{2}}$ .
Closed form: $F (t) = \frac{e ^{t} - 1}{t}$ for $t \neq = 0$ , so by the quotient rule $F^{'} (t) = \frac{t e ^{t} - ( e ^{t} - 1 )}{t ^{2}} = \frac{e ^{t}}{t} - \frac{e ^{t} - 1}{t ^{2}}$ . The two agree.
Sanity check at $t = 0$ : differentiating inside gives $F^{'} (0) = \int_{0}^{1} x d m = \frac{1}{2}$ , matching $lim_{t \to 0} (\frac{e ^{t}}{t} - \frac{e ^{t} - 1}{t ^{2}}) = \frac{1}{2}$ from the series $\frac{e ^{t} - 1}{t} = 1 + \frac{t}{2} + \frac{t ^{2}}{6} + \dots$ .

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Example 6.46 (A definite integral evaluated by a parameter). Evaluate $\int_{0}^{\infty} \frac{e ^{- x} - e ^{- 3 x}}{x} d m$ by introducing a parameter.

Solution.

Write $F (t) = \int_{0}^{\infty} \frac{e ^{- x} - e ^{- t x}}{x} d m$ for $t \geq 1$ ; the integrand extends continuously by $t - 1$ at $x = 0$ , and the target is $F (3)$ , with $F (1) = 0$ .
Formally $\partial_{t} \frac{e ^{- x} - e ^{- t x}}{x} = e^{- t x}$ . On a neighbourhood $t \in (a, b)$ with $1 < a < b$ , $∣ e^{- t x} ∣ \leq e^{- a x}$ , which is integrable with $\int_{0}^{\infty} e^{- a x} d m = 1/ a < \infty$ . Hypothesis (iii) holds.
So $F^{'} (t) = \int_{0}^{\infty} e^{- t x} d m = \frac{1}{t}$ .
Integrate: $F (3) = F (1) + \int_{1}^{3} \frac{d t}{t} = 0 + ln 3$ .
Sanity check numerically: $ln 3 \approx 1.0986$ , and this is the Frullani integral $\int_{0}^{\infty} \frac{e ^{- a x} - e ^{- b x}}{x} d x = ln (b / a)$ with $a = 1$ , $b = 3$ — the same value by an independent route.

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Pitfall. The dominator in hypothesis (iii) must bound $\partial f / \partial t$ uniformly in $t$ over a neighbourhood, not merely at the single point where you want the derivative. A bound that degenerates as $t$ approaches an endpoint gives no conclusion there. This is why the previous example restricted to $t \in (a, b)$ with $a > 1$ rather than working on all of $[1, \infty)$ at once: the bound $e^{- a x}$ needs a strictly positive $a$ .

Exercises for §6.10 →

6.11Where each theorem applies, and the theorems it generates

The three convergence theorems are rarely interchangeable in practice. Choosing correctly is a matter of reading which hypothesis you can actually verify.

Method 6.47 (Choosing a convergence theorem).

Is the sequence nonnegative and increasing in $n$ at each point? Use the theorem Monotone Convergence Theorem — no bound is needed and the answer may be $\infty$ .
Is it a sum of nonnegative terms? Use the corollary Term-by-term integration of nonnegative series, which is monotone convergence in disguise.
Otherwise, hunt for a single integrable $g$ with $∣ f_{n} ∣ \leq g$ for all $n$ . If the measure space is finite and the $f_{n}$ are uniformly bounded, a constant will do.
If no dominator exists, you still have the lemma Fatou's lemma, an inequality — often enough to prove that a limit is finite, or that an integral vanishes.
If the conclusion fails, look for escaping mass: a bump translating to infinity, or a spike concentrating at a point.

Corollary 6.48 (Bounded Convergence Theorem). Let $m (E) < \infty$ , let $f_{n}$ be measurable on $E$ with $f_{n} \to f$ a.e., and suppose $∣ f_{n} ∣ \leq M$ a.e. for a constant $M$ . Then $\int_{E} f_{n} d m \to \int_{E} f d m$ .

Proof. The constant function $g \equiv M$ satisfies $\int_{E} g d m = M m (E) < \infty$ because $m (E)$ is finite, so $g$ is integrable and dominates every $f_{n}$ . Apply the theorem Dominated Convergence Theorem.∎

Corollary 6.49 (Countable additivity of the integral). If $E = ⨆_{k = 1}^{\infty} E_{k}$ is a disjoint union of measurable sets and $f$ is integrable on $E$ , then

\int_{E} f d m = k = 1 \sum \infty \int_{E_{k}} f d m,

the series converging absolutely.

Proof. For $f \geq 0$ apply the corollary Term-by-term integration of nonnegative series to $g_{k} = f χ_{E_{k}}$ , whose pointwise sum is $f χ_{E}$ . For general integrable $f$ , do this for $f^{+}$ and $f^{-}$ separately, both series having finite sums since $\int_{E} f^{\pm} \leq \int_{E} ∣ f ∣ < \infty$ , and subtract. Absolute convergence follows from applying the nonnegative case to $∣ f ∣$ .∎

Corollary 6.50 (Absolute continuity of the integral). If $f$ is integrable on $E$ , then for every $ε > 0$ there is $δ > 0$ such that $m (A) < δ$ and $A \subseteq E$ imply $\int_{A} ∣ f ∣ d m < ε$ .

Proof. Let $f_{n} = min (∣ f ∣, n)$ . These are nonnegative, increase to $∣ f ∣$ , so the theorem Monotone Convergence Theorem gives $\int_{E} f_{n} \to \int_{E} ∣ f ∣ < \infty$ ; choose $n$ with $\int_{E} (∣ f ∣ - f_{n}) d m < ε /2$ . Now take $δ = ε / (2 n)$ . For $m (A) < δ$ ,

\int_{A} ∣ f ∣ d m = \int_{A} (∣ f ∣ - f_{n}) d m + \int_{A} f_{n} d m \leq \frac{ε}{2} + n m (A) < \frac{ε}{2} + \frac{ε}{2} = ε .

∎

Example 6.51 (Choosing the right theorem). For each sequence on the indicated set, name the theorem that applies and give $lim_{n} \int f_{n}$ : (a) $f_{n} = χ_{[0, n]}$ on $[0, \infty)$ ; (b) $f_{n} (x) = \frac{s i n ( x / n )}{x ^{2} + 1}$ on $R$ ; (c) $f_{n} (x) = \frac{n}{1 + n ^{2} x ^{2}}$ on $[1, 2]$ .

Solution.

(a) The sequence is nonnegative and increasing to $χ_{[0, \infty)}$ . The theorem Monotone Convergence Theorem applies with no bound needed, and gives $lim_{n} \int f_{n} = \int_{[0, \infty)} 1 d m = \infty$ . Indeed $\int f_{n} = n \to \infty$ . Dominated convergence is unavailable — there is no integrable dominator — and it is not needed.
(b) $∣ f_{n} (x) ∣ \leq \frac{1}{x ^{2} + 1} =: g (x)$ , with $\int_{R} g d m = π < \infty$ , and $f_{n} (x) \to 0$ pointwise. The theorem Dominated Convergence Theorem gives $lim_{n} \int f_{n} = 0$ . The sequence is not monotone, so monotone convergence does not apply.
(c) On $[1, 2]$ , $f_{n} (x) = \frac{n}{1 + n ^{2} x ^{2}} \leq \frac{n}{n ^{2} x ^{2}} = \frac{1}{n x ^{2}} \leq 1$ , so the constant $1$ dominates on a set of finite measure, and $f_{n} \to 0$ pointwise. The corollary Bounded Convergence Theorem gives $lim_{n} \int_{1}^{2} f_{n} d m = 0$ .
Sanity check on (c) directly: $\int_{1}^{2} \frac{n d x}{1 + n ^{2} x ^{2}} = [arctan (n x)]_{1}^{2} = arctan (2 n) - arctan (n) \to \frac{π}{2} - \frac{π}{2} = 0$ .

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Example 6.52 (Fatou where no dominator exists). Suppose $f_{n} \geq 0$ are measurable on $E$ with $\int_{E} f_{n} d m \leq 5$ for every $n$ , and $f_{n} \to f$ pointwise. What can be concluded about $\int_{E} f d m$ ?

Solution.

No dominator is assumed, so the theorem Dominated Convergence Theorem is unavailable; no monotonicity is assumed, so the theorem Monotone Convergence Theorem is unavailable too.
The lemma Fatou's lemma needs only nonnegativity: $\int_{E} f d m = \int_{E} lim inf_{n} f_{n} d m \leq lim inf_{n} \int_{E} f_{n} d m \leq 5$ .
So $f$ is integrable with $\int_{E} f d m \leq 5$ . Equality need not hold: the moving bump has $\int f_{n} = 1 \leq 5$ for all $n$ and $\int f = 0$ .
Sanity check on how much was extracted from how little: from a uniform bound and pointwise convergence alone, we get integrability of the limit and a bound on its integral. That is exactly the service Fatou performs, and it is the reason it appears as a hypothesis-free safety net in the recipe above.

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Intuition. A useful mental ranking. Monotone convergence is the theorem for building up: you are assembling a function out of an increasing approximation and want the integral to assemble too. Dominated convergence is the theorem for staying put: nothing is growing, everything sits under one ceiling of finite area, and the only question is whether the limit passes through. Fatou is the theorem for when you know nothing: it gives you the one inequality that cannot fail, and half of an answer is often all a proof needs.

Exercises for §6.11 →

Common Mistakes

Using a dominator that depends on $n$ . The theorem Dominated Convergence Theorem requires a single integrable $g$ with $∣ f_{n} ∣ \leq g$ for every $n$ at once. " $∣ f_{n} ∣ \leq n e^{- x}$ , and each $n e^{- x}$ is integrable" establishes nothing.
Applying a constant dominator on an infinite-measure space. A constant is integrable over $[0, 1]$ and not over $R$ . The corollary Bounded Convergence Theorem carries the hypothesis $m (E) < \infty$ for exactly this reason; drop it and the moving bump $χ_{[n, n + 1]}$ , which is bounded by $1$ , becomes a counterexample.
Concluding that an interchange was legitimate because the two answers agreed. In the example Monotone convergence in action the sequence $n x^{n - 1} /2$ has $lim \int = \frac{1}{2}$ while $\int lim = 0$ ; the agreement in the other half of that example was earned by checking monotonicity.
Confusing null with small in other senses. The Cantor set is null and uncountable; a fat Cantor set is nowhere dense with measure $\frac{1}{2}$ ; $Q$ is null and dense. Only "coverable by intervals of arbitrarily small total length" is being asserted.
Assuming improper Riemann integrability implies Lebesgue integrability. $sin (x) / x$ on $(0, \infty)$ has a convergent improper integral and is not Lebesgue integrable, because $\int ∣ sin x / x ∣ d m = \infty$ and the Lebesgue integral is absolute.
Forgetting that monotone convergence permits an infinite answer. $\int f_{n} \to \infty$ is a valid conclusion, not a failure of the theorem; the theorem Monotone Convergence Theorem has no finiteness hypothesis and needs none.
Applying Fatou in the wrong direction. The inequality is $\int lim inf \leq lim inf \int$ . Mass can leak away, never appear from nowhere, so it is the limit function that can be deficient — writing it the other way around asserts that area is created in the limit.
Treating a.e. equality as everywhere equality when pointwise values matter. The integral cannot distinguish $f$ from $g$ when $f = g$ a.e., but continuity, differentiability and pointwise bounds at a specific point are not a.e. properties and are not preserved.

About the practice questions. They check that you can carry out this chapter's computations correctly, and each one is graded on a single answer. They are not proof exercises: working through them confirms the mechanics, not that you could prove the results yourself. For that, re-read the statements above and try to reconstruct their proofs with the page closed.

Practice this chapter