Basic Lebesgue Integration1

John Nachbar Washington University October 23, 2017

1 Introduction.

These are very brief notes on integration. Some proofs are omitted. Throughout most of these these notes, functions are real-valued with domain [0, 1]. Recall that a property (such as continuity) holds almost everywhere (a.e.) iﬀ it holds except, perhaps, on a set of measure zero.

2 Measurable Functions.

Let B(R) denote the Borel σ-algebra on R (the smallest σ algebra on R containing all open intervals). If I write just B, with no argument, then I mean the Borel σ-algebra on [0, 1]. Recall that M is the set of measurable sets on [0, 1]; B ⊂ M.

−1 Definition 1. f : [0, 1] → R is measurable iff for any B ∈ B(R), f (B) ∈ M. Remark 1. Note the asymmetry in the σ algebras used. In particular, if f takes values in [0, 1], then the definition of measurability for f uses a weaker σ algebra for the target space (B) than for the range (M). This definition turns out to be “correct” in the sense that it exactly characterizes what functions are Lebesgue integrable: Theorem5 below states that a (bounded) function is Lebesgue integrable iff is measurable.

Theorem 1. f : [0, 1] → R is measurable iﬀ any one of the following conditions hold: for any y ∈ R, 1. f −1((−∞, y)) ∈ M,

2. f −1((−∞, y]) ∈ M,

3. f −1((y, ∞)) ∈ M,

4. f −1([y, ∞)) ∈ M. 1cbna. This work is licensed under the Creative Commons Attribution-NonCommercial- ShareAlike 4.0 License.

1 Proof. Omitted but not diﬃcult. One must show that these conditions (a) are −1 equivalent, (b) imply that f (A) is measurable for any open interval A ⊆ R, and −1 (c) imply that the set of sets A ⊆ R such that f (A) is measurable constitutes a σ −1 algebra. (b) and (c) imply that the set of sets A ⊆ R such that f (A) is measurable is a σ algebra that contains the Borel σ algebra.

−1 Remark 2. In particular, if f is measurable then for any y ∈ R, f (y) is measurable. The converse, however, is not true. For example, suppose that V is a Vitali set and let f(x) = x for x ∈ V and f(x) = −e−x otherwise. Then f −1([0, ∞)) = V , which is not measurable, violating condition (4). But f is 1-1, hence f −1(y) is measurable for any y. The deﬁnition of measurability resembles that of continuity. Since any open set in R is in B(R), and any open set in [0, 1] is in B, any continuous function is measurable. But a measurable function need not be continuous. Recall that if A ⊆ [0, 1], then the indicator function 1A is deﬁned by ( 1 if x ∈ A 1A(x) = 0 if x ∈ Ac.

Example 1. If A = [0, 1] ∩ Q then f = 1A is the Dirichlet function. f is measurable but not continuous at any point. Theorem 2. Let f, g be measurable real-valued functions with domain [0, 1] and let a ∈ R. Then af, f + g, and fg are measurable. Proof. Omitted but not diﬃcult.

Theorem 3. Let f, g be real-valued functions with domain [0, 1]. If f = g a.e. and f is measurable then g is measurable. −1 Proof. By Theorem1, it suﬃces to check that, for any y ∈ R, g ((y, ∞)) is measurable. Let E be the set of points for which f 6= g. By assumption, E has measure zero. By a result in the Measure Theory notes, since E has measure zero, E is measurable and hence Ec is also measurable. g−1((y, ∞)) = [f −1((y, ∞)) ∩ Ec] ∪ [g−1((y, ∞)) ∩ E]. Since E has measure zero, g−1((y, ∞)) ∩ E also has measure zero, and hence is measurable. Since f is measurable, f −1((y, ∞)) is measurable, hence f −1((y, ∞)) ∩ Ec is measurable, which implies the result.

3 Simple Functions.

Recall that A = {A1,...,AK } ⊆ P([0, 1]) is a (ﬁnite) partition of [0, 1] iﬀ A is S pairwise disjoint (Ai ∩ Aj = ∅ for all Ai,Aj ∈ A) and k Ak = [0, 1].

2 Deﬁnition 2. f : [0, 1] → R is a simple function iﬀ there are real numbers a1, . . . , aK (not necessarily distinct) and measurable sets A1,...,AK ∈ M such that A = {A1,...,AK } is a partition of [0, 1] and X f(x) = akIAk (x). k

If the Ak are intervals then a simple function is called a step function.

Example 2. Let A1 = [0, 1/4), a1 = 4/3, A2 = [1/4, 1], a2 = 8/9. The associated f is a step function that takes the value 4/3 on [0, 1/4) and 8/9 on [1/4, 1]. The Dirichlet function (Example1) is a simple function that is not a step function.

Deﬁnition 3. Let f : [0, 1] → R be a simple function deﬁned by X f(x) = aiIAk (x). k

Then the (Lebesgue) integral of f, written R f, is given by

Z X f = akλ(Ak). k Example 3. For the step function in Example2, Z f = (4/3 × 1/4) + (8/9 × 3/4) = 1/3 + 2/3 = 1.

Example 4. If f is the Dirichlet function, as in Example1, then Z f = 1 × 0 + 0 × 1 = 0.

The goal now is to extend the definition of integration to more general functions. To do this, I first make a detour to discuss Riemann integration, the integration of first year calculus.

4 Riemann Integration.

The following treatment of Riemann integration is somewhat non-standard but can (easily) be shown to be equivalent to the standard version.

3 For a bounded function f : [0, 1] → R, the upper Riemann integral, written U R(f), is defined to be, Z U R(f) = inf g, g≥f where the infimum is taken over all step functions g : [0, 1] → R for which g(x) ≥ f(x) for all x. This construction makes sense because f is bounded: there is an M such that |f(x)| ≤ M for all x, hence the constant function g(x) = M is a step function for which g ≥ f, hence the set of step functions g such that g ≥ f is not empty; moreover, for all step functions g such that g ≥ f, g(x) ≥ −M for all x, hence R R g ≥ −M, hence infg≥f g exists by the least upper bound property. Similarly, the lower Riemann integral, written LR(f), is defined to be, Z LR(f) = sup g, g≤f where the supremum is taken over all step functions g : [0, 1] → R for which g(x) ≤ f(x) for all x. It is easy to show that U R(f) ≥ LR(f). f is Riemann integrable iff the two are equal. If f is Riemann integrable, then the Rieman integral, written R f(x)dx, is defined to be, Z f(x)d(x) = LR(f).

It is easy to see that if f is a step function then f is Riemann integrable with Z Z f(x) d(x) = f.

But even f is a simple function, if f is not a step function, then f may not be Riemann integrable. Example 5. If f is the Dirichlet function, as in Example1, then U R(f) = 1 and R L (f) = 0, hence f is not Riemann integrable. The following theorem, due to Lebesgue, completely characterizes what functions are Riemann integrable.

Theorem 4. Let f : [0, 1] → R be bounded. f is Riemann integrable iff it is continuous a.e. Proof. Omitted and somewhat involved. Here is a sketch. ⇒. Note first that if f is continuous, then it is Riemann integrable. To see this, recall that any continuous function is uniformly continuous on a compact set. That is, for any ε > 0, there is a δ > 0 such that f varies by less than ε on any interval of length less than δ. Since [0, 1] is compact, it can be partitioned into a finite number of intervals of length less than δ, which implies that the difference between the upper

4 and lower integrals is less than ε. Since ε can be made arbitrarily small, this implies that the difference between the upper and lower integrals is, in fact, zero. Suppose next that f is continuous a.e. and that the set of discontinuities is closed (it need not be) and hence, as a subset of [0, 1], compact. Since the set of discontinuities is a compact set of measure zero, it can be covered by a finite number of intervals of total length less than, say, γ > 0. One can, therefore, partition [0, 1] in such a way that (a) all of the discontinuities are covered by partition elements of total length less than γ and (b) on each of the other partition elements, where the function is continuous, the function varies by less than ε > 0 (as in the preceding argument); the total length of these other partition elements is at least 1 − γ but no more than 1. The difference between the upper and lower integrals is then less than 2Mγ + ε (where M is the bound on f). Since both γ and ε can be made arbitrarily small, this implies that the difference between the upper and lower integrals is, in fact, zero. Unfortunately, the set of discontinuities need not be closed, hence it need not be compact. One can repair the proof by instead looking at “ε discontinuities”: f is ε discontinuous at x iff for any open interval containing x, there are points a, b in the interval such that |f(a) − f(b)| ≥ ε. It is not hard to show that, for any ε > 0, the set of ε discontinuities is closed, hence compact as a subset of [0, 1]. The proof then proceeds along the same lines as above. ⇐. Conversely, suppose that f is Riemann integrable. For any ε > 0, since the difference between the upper and lower integrals must be zero, it must be possible to partition [0,1] in such a way that the ε discontinuities (defined above) are covered by intervals of arbitrarily small total length, which implies that the set of ε discontinuities has measure zero. Since this holds for any ε > 0, the set of discontinuities has measure zero.

5 Lebesgue Integration

For a bounded function f : [0, 1] → R, define the upper Lebesgue integral, written U L(f), to be, Z U L(f) = inf g, g≥f where the infimum is taken over all simple functions g : [0, 1] → R for which g(x) ≥ f(x) for all x. Define the lower Lebesgue integral, written LL(f), to be, Z LL(f) = sup g, g≤f

where the supremum is taken over all simple functions g : [0, 1] → R for which g(x) ≤ f(x) for all x. Again, it is easy to see that LL(f) ≤ U L(f).

5 L Definition 4. A bounded function f : [0, 1] → R is Lebesgue integrable iff L (f) = U L(f). If a function is Lebesgue integrable, then the Lebesgue integral, written R f, is defined to be Z f = LL(f).

Example 6. The Dirichlet function (Example1) is Lebesgue, but not Riemann, integrable. The Dirichlet example raises the question of what functions are Lebesgue integrable. The answer is that, for bounded functions with domain [0, 1], measurability is both suﬃcient and necessary for Lebesgue integrability.

Theorem 5. If f : [0, 1] → R is bounded then it is Lebesgue integrable iﬀ it is measurable.

Proof. Omitted. Here is a sketch of a proof. ⇒. Fix ε > 0. Since f is bounded, its range lies in some interval [−M,M]. Partition [−M,M] into subintervals each of length less than ε. Since f is measurable, the preimage of a partition element is measurable. Construct a simple function gL by, for example, if (a, b] is an element of the partition of [−M,M], then gL(x) = a −1 for x ∈ f ((a, b]). Similarly, deﬁne a simple function gU by, for example, gU (x) = b −1 for x ∈ f ((a, b]). By construction, gU ≥ f ≥ gL and Z Z

gU − gL < ε.

Since ε was arbitrary, if follows that f is integrable. ⇐. This direction is more involved, but the idea is that if f is integrable then there is a sequence of simple functions that converges to f a.e., which (by an argument involving Theorem3) implies that f is measurable.

Theorem 6. If a bounded function f : [0, 1] → R is Riemann integrable then it is Lebesgue integrable, hence measurable, and Z Z f(x) dx = f.

Proof. Almost immediate, since step functions are simple functions, hence LR(f) ≤ L L R L (f) ≤ U (f) ≤ U (f).

Remark 3. Theorems4,5,6 jointly imply that a bounded function that is continuous a.e. is measurable. In fact, one can show that any function that is continuous a.e., even if not bounded, is measurable. For later reference, I note the following facts about integration.

6 Theorem 7. Let f : [0, 1] → R, and fˆ : [0, 1] → R be bounded, measurable functions and let a, b ∈ R. 1. R (af + bfˆ) = a R f + b R fˆ.

2. If fˆ ≥ f a.e. then R fˆ ≥ R f.

3. R |f| ≥ | R f|.

Proof.

1. Omitted but not diﬃcult.

2. Since f and fˆ are both bounded, there is an M so that for all x, |f(x)| < M and |fˆ(x)| < M. Let E be the set on which fˆ−f < 0. Then λ(E) = 0. Deﬁne the simple function g by g(x) = 0 if x ∈ Ec and g(x) = −2M for x ∈ E. Then g ≤ fˆ − f and R g = 0, which implies R (fˆ − f) ≥ 0, which (by the previous result) implies R fˆ ≥ R f.

3. Since |f| ≥ f and |f| ≥ −f, it follows that R |f| ≥ R f and R |f| ≥ R (−f) = − R f, from which the result follows.

Finally, if f : [0, 1] → R is bounded and measurable then for any measurable A ⊆ [0, 1], write Z f A for the Lebesgue integral over the set A; that is, consider simple functions with domain A, rather than all of [0, 1], and modify the deﬁnition of Lebesgue integral accordingly.

Theorem 8. Let f : [0, 1] → R be bounded and measurable. Then for any disjoint measurable sets A, B ⊆ [0, 1], Z Z Z f = f + f. A∪B A B Proof. Almost immediate, since any simple function defined on A ∪ B can be split into two simple functions, one defined on A and the other defined on B, and conversely, any two simple functions, one defined on A and the other defined on B, can be merged into a single simple function defined on A ∪ B. R In particular, note that if A has measure zero then for any bounded f, A f = 0. It follows that if f, g are bounded measurable functions with f = g a.e. then R f = R g.

7 6 Sequences of Functions.

In this section, convergence is pointwise: ft → f iﬀ ft(x) → f(x) for every x ∈ [0, 1]. The ﬁrst theorem says that the limit of a sequence of measurable functions is measurable.

Theorem 9. Let (ft) be a sequence of measurable functions from [0, 1] to R. If ft → f then f is measurable.

Proof. By Theorem1, it suﬃces to show that for any y ∈ R, the set

A = {x ∈ [0, 1] : f(x) > y} is measurable. Note that, for any given x, f(x) > y iﬀ there is a k ∈ {1, 2, 3,... } and a T such that ft(x) > y + 1/k for all t > T . For each k, let, [ \ Ak = {x ∈ [0, 1] : ft(x) > y + 1/k}. T t>T

Ak is the set of x for which there is a T such that ft(x) > y + 1/k for all t > T . Ak S is measurable (since the ft are measurable). Finally, A = k Ak, and is therefore measurable.

The next theorem says that if a sequence of measurable functions converges pointwise then it converges almost uniformly.

Theorem 10. Let (ft) be a sequence of measurable functions. If ft → f, then for any δ > 0 and any ε > 0, there is a measurable set A ⊆ [0, 1] with λ(A) < δ such that for every x ∈ Ac, |ft(x) − f(x)| < ε. Proof. For each t ∈ {1, 2,... }, let

At = {x ∈ [0, 1] : |fs(x) − f(x)| ≥ ε for at least one s ≥ t}.

At is the set of x for which, at time t, the sequence (ft(x)) has not yet converged to within ε of f(x). Each At is measurable since it is the union, for s ≥ t, of the sets {x ∈ [0, 1] : |fs(x) − f(x)| ≥ ε}, and these sets are measurable since |fs − f| is measurable (by Theorem2, Theorem9, and the fact, which is easy to show, that a continuous transformation of a measurable function is measurable). For all t, At+1 ⊆ At, hence for any T ,

T \ At = AT . t=1

8 T∞ Moreover, since ft(x) → f(x) for each x, t=1 At = ∅. By a theorem on nested sequences of measurable sets that appears in the Measure Theory notes, this implies that, lim λ(At) = 0.

It follows that there is a t such that λ(At) < δ. Set A = At and the proof follows.

Remark 4. With only a slight complication to the proof, Theorem 10 can be strength- ened to hold whenever ft → f a.e. Example 7. Let f(x) = xt. Then f converges pointwise to f given by ( 0 if x ∈ [0, 1) ft(x) = 1 if x = 1.

2 Note that ft is continuous while f is not. ft does not converge uniformly to f. Nevertheless, for any ε > 0 and any δ one can ﬁnd a T such that, setting A = [1−δ, 1], |ft(x) − f(x)| < ε for all t > T and all x∈ / [1 − γ, 1]. The next theorem, the main result of this section, has many applications. Among other things, it implies that if a (uniformly bounded) sequence of random variables converges pointwise then the limit of the means is the mean of the limit. This need not be true if we instead use Riemann integration.

Theorem 11 (Bounded Convergence Theorem). Let (ft) be a sequence of measurable functions that is uniformly bounded: there is an M > 0 such that |ft(x)| ≤ M for every t and every x. If ft → f then f is bounded and integrable, with Z Z lim ft = f.

Proof. To see that f is bounded, note that, since ft(x) → f(x), for any x and any γ > 0 there is a t such that |ft(x) − f(x)| < γ, which implies |f(x)| < |ft(x)| + γ ≤ M + γ. Since γ was arbitrary, |f(x)| ≤ M. Fix γ > 0. I will show that there is a T such that for t > T , Z Z

ft − f < γ, which implies the result.

2The set of continuous real-valued functions with domain [0, 1] is not complete under pointwise convergence, much as the space `∞ was not complete under pointwise convergence (as discussed in ω the notes on Completeness and Compactness in R ).

ft − f = (ft − f)

Z ≤ |ft − f| Z Z = |ft − f| + |ft − f| A Ac < (2M)δ + ε = γ.

Example 8. Since A = Q ∩ [0, 1] is countable, I can represent it as A = {a1, a2,... }. For each t, let ( 1 if x ∈ {a1, . . . , at} ft(x) = 0 otherwise

For every t, ft is Riemann integrable, with Z ft(x) dx = 0.

This sequence of ft converges to the Dirichlet function (Example1), call this f, and indeed, for all t, Z Z ft = 0 = f.

But f is not Riemann integrable, so the conclusion of Theorem 11 is false if we use Riemann rather than Lebesgue integration.

Example 9. Recall Example7. Even though each ft is continuous but f is not, Theorem 11 holds, and indeed, Z Z lim ft = 0 = f.

10 7 Extensions.

Very briefly, Lebesgue integration can be extended to unbounded functions, although there are subtleties, one of which is that a function is said to be integrable iff it has a finite integral (an issue that does not arise with unbounded functions). The generalization of the Bounded Convergence Theorem to (possibly) unbounded functions is the Lebesgue Convergence Theorem, which says that if ft is a sequence of integrable (but not necessarily bounded) functions converging to f a.e. and if, in addition, there is an integrable (but not necessarily bounded) function g such that |ft| ≤ g for all t, then f is integrable and Z Z lim ft = f.

Example 10. For each t, deﬁne ft by, ( t if x ∈ (0, 1/t], ft(x) = 0 if x ∈ {0} ∪ (1/t, 1].

For every t, ft is integrable (indeed, Riemann integrable), with Z ft = t(1/t) = 1.

But ft converges to the zero function, f(x) = 0 for all x ∈ [0, 1], and Z f = 0.

Hence the conclusion of the Lebesgue Convergence Theorem fails in this example. The problem here is that ft is not uniformly bounded by an integrable function. In particular, ft is uniformly bounded by g deﬁned by, ( 1/x if x ∈ (0, 1], g(x) = 0 if x = 0, but g is not integrable because its integral is inﬁnite. (Note that g is measurable, since it is continuous except at one point; the fact that g is not integrable does not contradict Theorem5 since g is not bounded.) The theory can also be extended to functions with domains other than [0, 1]. If the domain is contained in a bounded interval then the theory is almost unchanged. If the domain is unbounded then are additional considerations, but I won’t pursue this.