Lecture 11: The Lebesgue Integral of a Nonnegative Function and Convergence Theorems

March 30, 2021 Last time, we defined the Lebesgue integral for simple functions: for any simple function φ written in the canonical R P P way nj=1 aj χAj for disjoint sets Aj , we have E φ = nj=1 aj m(Aj ), and we proved some properties about this integral R R last time (we have linearity of the integral, if f (x) ≤ g(x) for all x, then f ≤ g, and so on). Today, we’ll define the integral for general nonnegative measurable functions, and much like Riemann sums give better and better approximations for Riemann integrals as the rectangles become thinner, we can think of Lebesgue integrals as being the result of a similar limiting procedure. We saw last time already that for a nonnegative measurable function f , we can always find a sequence of simple functions that increase pointwise to f . So it makes sense to try to define the Lebesgue integral as the limit of the integrals of the simple functions, but then we run into issues where the final integral may depend on the specific sequence of simple functions that we chose. Definition 108 Let f ∈ L+ (E). Then the Lebesgue integral of f is Z  Z + f = sup φ : φ ∈ L (E) simple, φ ≤ f . E E Proposition 109 Let E ⊂ R be a set with m(E) = 0. Then for all f ∈ L+ (E), we have R E f = 0. In other words, it’s only interesting to take integrals over functions of positive measure. (And this is sort of like how Riemann integrals over a point are always zero.) Proof. Working from the definition, start with our function f ∈ L+ (E). If φ is a simple function in the canonical form Pn Pn j=1 aj χ(Aj ) with φ ≤ f , then m(Aj ) ≤ m(A) = 0, so in the sum j=1 aj m(Aj ), all terms must be zero. So we R always have E φ = 0, and the supremum over all simple functions φ is also zero, as desired. We can also verify a bunch of results that were true of the Lebesgue integral for simple functions: Proposition 110 If φ ∈ L+ (E) is a simple function, then the two definitions of R E f (for simple functions and general nonnegative + measurable functions) agree with each other. If f , g ∈ L (E), c ∈ [0, ∞) is a nonnegative real number, and R R R R R R f ≤ g, then we have E cf = c E f and E f ≤ E g. Finally, if f ∈ L+ (E) and F ⊂ E, then F f ≤ E f . (The proof will be left for our homework, but the idea is that taking supremums shouldn’t change our inequalities.) We can actually relax the second statement here to an “almost-everywhere” statement as well: Proposition 111 If f , g ∈ L+ (E), and f ≤ g almost everywhere on E, then R E f ≤ R E g. Proof. Define the set F = {x ∈ E : f (x) ≤ g(x)}; this is a measurable set because g − f is measurable, so the inverse image of [0, ∞] is measurable (with some small details about how functions behave at ∞, but we’re dealing with that on our homework). By assumption, m(F c ) = 0, and thus by Proposition 109 and Proposition 110, Z Z Z Z Z Z Z Z f = f + f = f ≤ g= g+ g= g, E Fc F F F Fc F E as desired. R f = E g. We may notice that we’re missing R R R the linearity that we had for simple functions: we haven’t mentioned that E f + E g = E (f + g). To prove that, In particular, if we know that f = g almost everywhere on E, then R E we’ll need one of the big three results in Lebesgue integration: Theorem 112 (Monotone Convergence Theorem) If {fn } is a sequence of nonnegative measurable functions (in L+ (E)) such that f1 ≤ f2 ≤ · · · pointwise on E, and fn → f pointwise on E for some f (which will be in L+ (E) because the pointwise limit of measurable functions is measurable), then Z Z lim n→∞ fn = f. E E Notice that the assumption of pointwise convergence here is much weaker than the uniform convergence we usually need to assume for Riemann integration. Proof. Since f1 ≤ f2 ≤ · · · , we know that R E f1 ≤ R E f2 ≤ · · · . Thus, R E fn is a nonnegative increasing sequence of R E fn exists in [0, ∞]. Furthermore, because limn→∞ fn (x) = f (x) R for all x, we know that fn ≤ f for all n, which means that E f (which is also some number in [0, ∞]) must satisfy Z Z Z Z fn ≤ f =⇒ lim fn ≤ f. nonnegative numbers, meaning that the limit limn→∞ E n→∞ E R R E E R It suffices to prove the reverse inequality (that E f ≤ E limn→∞ E fn ), and we can show this by showing that R R R E φ ≤ E limn→∞ E fn for every simple function φ ≤ f (the point being that eventually fn will be larger than φ). P We’ll first take some ε ∈ (0, 1) as “breathing room.” If φ = m j=1 aj χAj is an arbitrary simple function with φ ≤ f , then we can define the set En = {x ∈ E : fn (x) ≥ (1 − ε)φ(x)} . Since (1 − ε)φ(x) < f (x) for all x (we have strict equality now that ε is positive), and limn→∞ fn (x) = f (x), every x must lie in some En . Therefore, we have ∞ [ En = E. n=1 Furthermore, because f1 ≤ f2 ≤ · · · , we know that E1 ⊂ E2 ⊂ · · · (the sets En are increasing by inclusion). So now notice that Z Z fn ≥ E Z fn ≥ En Z (1 − ε)φ = (1 − ε) En φ = (1 − ε) En m X aj m(Aj ∩ En ) j=1 (because the inequality holds on En , and the Aj ∩ En . are measurable and disjoint). And now, because En increases to E, and therefore E1 ∩ Aj ⊂ E2 ∩ Aj ⊂ · · · increases to Aj , continuity of Lebesgue measure tells us that as n → ∞, m(Aj ∩ En ) → m(Aj ). Therefore, we can take limits on both sides and find (because we have a finite sum on the right-hand side) that for all ε ∈ (0, 1), we have Z fn ≥ lim (1 − ε) lim n→∞ E n→∞ m X j=1 aj m(Aj ∩ En ) = (1 − ε) m X j=1 Z aj m(Aj ) = (1 − ε) φ. E Taking ε → 0 yields the desired inequality R E φ ≤ limn→∞ R E fn , and combining the two inequalities finishes the proof. With this result, we now have tools for evaluating Lebesgue integrals that aren’t just using the definition directly: Corollary 113 Let f ∈ L+ (E), and let {φn }n be a sequence of simple functions such that 0 ≤ φ1 ≤ φ2 ≤ · · · ≤ f , with φn → f R R pointwise. Then E f = limn→∞ E φn . In other words, we can take any pointwise increasing sequence of simple functions and compute the limit, instead of needing to compute the supremum explicitly. (And this follows because we can just plug in the φn s as fn s into the Monotone Convergence Theorem.) Corollary 114 If f , g ∈ L+ (E), then R E (f + g) = R E f + R E g. Proof. Let {φn }n and {ψn }n be two sequences of simple functions, such that 0 ≤ φ1 ≤ φ2 ≤ · · · ≤ f and φn → f pointwise, and similarly 0 ≤ ψ1 ≤ ψ2 ≤ · · · ≤ g and ψn → g pointwise. Then we have 0 ≤ φ1 + ψ1 ≤ φ2 + ψ2 ≤ · · · ≤ f + g, where φn + ψn → f + g pointwise, and each φi + ψi is a simple function (because it’s the sum of two simple functions). Then the Monotone Convergence Theorem tells us that Z Z Z Z (f + g) = lim (φn + ψn ) = lim φn + ψn E n→∞ n→∞ E E E by using linearity for simple functions, and then the Monotone Convergence Theorem again tells us that this is R R E f + E g, as desired. In fact, we have something stronger than finite additivity: Theorem 115 Let {fn }n be a sequence in L+ (E). Then Z X E fn = XZ n n fn . E (The left-hand side is defined here, because we’re summing a bunch of nonnegative real numbers pointwise, and we’re allowing ∞ as an output of the our functions.) Proof. By induction, Corollary 114 tells us that for each N, we have Z X N fn = N Z X E n=1 n=1 fn . E Now because 1 X n=1 fn ≤ 2 X n=1 fn ≤ · · · , PN and by definition of the infinite sum, we have pointwise convergence n=1 fn → P∞ n=1 fn as N → ∞, the Monotone Convergence Theorem tells us that Z X ∞ E n=1 fn = lim N→∞ Z X N fn = lim N→∞ E n=1 N Z X fn = E n=1 ∞ Z X n=1 fn , E as desired. (And again, this kind of result is not going to hold for Riemann integration, if for example we enumerate the rationals and let fn be the function which is 1 at the first n rational numbers and 0 everywhere else.) Theorem 116 Let f ∈ L+ (E). Then R E f = 0 if and only if f = 0 almost everywhere on E. Proof. First of all, if f = 0 almost everywhere, then f ≤ 0 almost everywhere, meaning R E f ≤ R E 0 = 0, so the integral is indeed zero. For the other direction, define   1 , F = {x ∈ E : f (x) > 0}. Fn = x ∈ E : f (x) > n S 1 We know that F = ∞ n=1 Fn (because whenever f (x) > 0, we have f (x) > n for some large enough n), and we also have F1 ⊂ F2 ⊂ · · · ,. Now we can compute 1 0 ≤ m(Fn ) = n Z Fn 1 ≤ n Z Z f ≤ Fn f = 0, E which means that n1 m(Fn ) = 0 =⇒ m(Fn ) for all n, and thus by continuity of measure ! ∞ [ m(F ) = m Fn = lim m(Fn ) = 0, n→∞ n=1 as desired. We can now slightly relax the assumptions of the Monotone Convergence Theorem as well: Theorem 117 If {fn }n is a sequence in L+ (E) such that f1 (x) ≤ f2 (x) ≤ · · · for almost all x ∈ E and limn→∞ fn (x) = f (x), then R R E f = limn→∞ E fn . Proof. Let F be the set of x ∈ E where the two assumptions above hold. By assumption, m(E \F ) = 0, so f −χF f = 0 and fn − χF fn = 0 almost everywhere for all n. The Monotone Convergence Theorem then tells us that Z Z Z Z f = lim fn , f = f χF = E E n→∞ F F where the first equality holds because the two functions f , f χF are equal almost everywhere, and the third equality holds because {fn } satisfy the assumptions of the Monotone Convergence Theorem on F . We can then simplify this to Z = lim n→∞ Z fn = lim F n→∞ fn , E because E \ F has measure zero so any integral over the region has measure zero. In other words, sets of measure zero don’t affect our Lebesgue integral. We’re now ready for the second big convergence theorem – it’s equivalent to the Monotone Convergence Theorem, but it’s often a useful restatement: Theorem 118 (Fatou’s lemma) Let {fn }n be a sequence in L+ (E). Then Z Z lim inf fn (x) ≤ lim inf E n→∞ fn (x). n→∞ E (Recall that we define the liminf of a sequence via   lim inf an = sup inf ak , n→∞ n≥1 k≥n and then the liminf function is defined pointwise.) Proof. We know that   lim inf fn (x) = sup inf fk (x) , n→∞ n≥1 k≥n and the expression inside the brackets on the right is increasing in n (since we’re taking an infimum over a smaller set). So the supremum on the right-hand side is actually a limit of a pointwise increasing sequence of functions:   = lim inf fk (x) . n→∞ k≥n So now by the Monotone Convergence Theorem, we have  Z Z  lim inf fn = lim inf fk , n→∞ E k≥n E and now for all j ≥ n and for all x ∈ E, we know that inf k≥n fn (x) ≤ fj (x), so for all j ≥ n, we have a fixed bound Z Z inf fk ≤ fj , E k≥n E and thus we can take the infimum over all j on the right-hand side and still have a valid inequality: Z Z inf fk ≤ inf fj . E k≥n j≥n E So we’ve successfully “swapped the integral and infimum,” and plugging this into the Monotone Convergence Theorem equality above yields Z  Z lim inf fn = lim n→∞ E  inf fk E k≥n  ≤ lim Z inf n→∞ j≥n  Z fj = lim inf E fn , E as desired. We might be worried about the fact that our functions can take on infinite values, and this next result basically says that we don’t need to worry too much: Theorem 119 Let f ∈ L+ (E), and suppose that R E f < ∞. Then the set {x ∈ E : f (x) = ∞} is a set of measure zero. Proof. Define the set F = {x ∈ E : f (x) = ∞}. We know that for all n, we have nχF ≤ f χF , so integrating both sides yields Z nm(F ) ≤ E Therefore, for all n, m(F ) ≤ 1 n R E Z f χF ≤ f < ∞. E f , which goes to 0 as n → ∞, so we must have m(F ) = 0. Our next steps will be to define the set of all Lebesgue integrable functions, prove some more properties of the Lebesgue integral, and then starting looking into Lp spaces (the motivation for this theory of integration in the first place).