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Basic Lebesgue Integration1 John Nachbar Washington University October 23, 2017 Basic Lebesgue Integration1 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 2 B(R), f (B) 2 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 2 R, 1. f −1((−∞; y)) 2 M, 2. f −1((−∞; y]) 2 M, 3. f −1((y; 1)) 2 M, 4. f −1([y; 1)) 2 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 2 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 2 V and f(x) = −e−x otherwise. Then f −1([0; 1)) = 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 2 A 1A(x) = 0 if x 2 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 2 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 2 R, g ((y; 1)) 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; 1)) = [f −1((y; 1)) \ Ec] [ [g−1((y; 1)) \ E]. Since E has measure zero, g−1((y; 1)) \ E also has measure zero, and hence is measurable. Since f is mea- surable, f −1((y; 1)) is measurable, hence f −1((y; 1)) \ Ec is measurable, which implies the result. 3 Simple Functions. Recall that A = fA1;:::;AK g ⊆ P([0; 1]) is a (finite) partition of [0; 1] iff A is S pairwise disjoint (Ai \ Aj = ; for all Ai;Aj 2 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 2 M such that A = fA1;:::;AK g 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 func- tion. 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 jf(x)j ≤ 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 " > 0, there is a δ > 0 such that f varies by less than " 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 ". Since " 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 " > 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 + " (where M is the bound on f). Since both γ and " 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.
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