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Chapter 4. the Dominated Convergence Theorem and Applica- Tions Contents

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Chapter 4. the Dominated Convergence Theorem and Applica- Tions Contents Chapter 4. The dominated convergence theorem and applica- tions The Monotone Covergence theorem is one of a number of key theorems alllowing one to ex- change limits and [Lebesgue] integrals (or derivatives and integrals, as derivatives are also a sort of limit). Fatou’s lemma and the dominated convergence theorem are other theorems in this vein, where monotonicity is not required but something else is needed in its place. In Fatou’s lemma we get only an inequality for lim inf’s and non-negative integrands, while in the dominated con- vergence theorem we can manage integrands that change sign but we need a ‘dominating’ inte- grable function as well as existence of pointwise limits of the sequence of inetgrands. Contents 4.1 Fatou’s Lemma . .1 4.2 Almost everywhere . .2 4.3 Dominated convergence theorem . .4 4.4 Applications of the dominated convergence theorem . .5 4.5 What’s missing? . .8 4.1 Fatou’s Lemma This deals with non-negative functions only but we get away from monotone sequences. Theorem 4.1.1 (Fatou’s Lemma). Let fn : R ! [0; 1] be (nonnegative) Lebesgue measurable functions. Then Z Z lim inf fn dµ ≥ lim inf fn dµ n!1 n!1 R R Proof. Let gn(x) = infk≥n fk(x) so that what we mean by lim infn!1 fn is the function with value at x 2 R given by lim inf fn (x) = lim inf fn(x) = lim inf fk(x) = lim gn(x) n!1 n!1 n!1 k≥n n!1 1 Notice that gn(x) = infk≥n fk(x) ≤ infk≥n+1 fk(x) = gn+1(x) so that the sequence (gn(x))n=1 is monotone increasing for each x and so the Monotone convergence theorem says that Z Z Z lim gn dx = lim gn dµ = lim inf fn dµ n!1 n!1 n!1 R R R But also gn(x) ≤ fk(x) for each k ≥ n and so Z Z gn dµ ≤ fk dµ (k ≥ n) R R 2 2017–18 Mathematics MA2224 or Z Z gn dµ ≤ inf fk dµ k≥n R R Hence Z Z Z Z lim inf fn dµ = lim inf fk dµ ≥ lim gn dµ = lim inf fn dµ n!1 n!1 k≥n n!1 n!1 R R R R Example 4.1.2. Fatou’s lemma is not true with ‘equals’. R For instance take, fn = χ[n;2n] and notice that fn dµ = n ! 1 as n ! 1 but for each R x 2 R, limn!1 fn(x) = 0. So Z Z Z lim inf fn dµ = 1 > lim inf fn dµ = 0 dµ = 0 n!1 n!1 R R R This also shows that the Monotone Convergence Theorem is not true without ‘Monotone’. 4.2 Almost everywhere Definition 4.2.1. We say that a property about real numbers x holds almost everywhere (with respect to Lebesgue measure µ) if the set of x where it fails to be true has µ measure 0. Proposition 4.2.2. If f : R ! [−∞; 1] is integrable, then f(x) 2 R holds almost everywhere (or, equivalently, jf(x)j < 1 almost everywhere). Proof. Let E = fx : jf(x)j = 1g. What we want to do is show that µ(E) = 0. R We know jfj dµ < 1. So, for any n 2 , the simple function nχE satisfies nχE(x) ≤ R N jf(x)j always, and so has Z Z nχE dµ = nµ(E) ≤ jfj dµ < 1: R R But this can’t be true for all n 2 N unless µ(E) = 0. Proposition 4.2.3. If f : R ! [−∞; 1] is measurable, then f satisfies Z jfj dµ = 0 R if and only if f(x) = 0 almost everywhere. R Proof. Suppose jfj dµ = 0 first. Let En = fx 2 : jf(x)j ≥ 1=ng. Then R R 1 χ ≤ jfj n En Lebesgue integral 3 and so Z 1 1 Z χEn dµ = µ(En) ≤ jfj dµ = 0: R n n R Thus µ(En) = 0 for each n. But E1 ⊆ E2 ⊆ · · · and 1 [ En = fx 2 R : f(x) 6= 0g: n=1 So 1 ! [ µ(fx 2 : f(x) 6= 0g) = µ En = lim µ(En) = 0: R n!1 n=1 Conversely, suppose now that µ(fx 2 R : f(x) 6= 0g) = 0. We know jfj is a non-negative 1 measurable function and so there is a monotone increasing sequence (fn)n=1 of measurable simple functions that converges pointwise to jfj. From 0 ≤ fn(x) ≤ jf(x)j we can see that fx 2 R : fn(x) 6= 0g ⊆ fx 2 R : f(x) 6= 0g and so µ(fx 2 R : fn(x) 6= 0g) ≤ µ(fx 2 R : f(x) 6= 0g) = 0. Being a simple function fn has a largest value yn (which is finite) and so if we put En = fx 2 R : fn(x) 6= 0g we have Z Z Z fn ≤ ynχEn ) fn dµ ≤ ynχEn dµ = yn χEn dµ = ynµ(En) = 0: R R R From the Monotone Convergence Theorem Z Z Z jfj dµ = lim fn dµ = lim fn dµ = lim 0 = 0: n!1 n!1 n!1 R R R The above result is one way of saying that integration ‘ignores’ what happens to the integrand on any chosen set of measure 0. Here is a result that says that in way that is often used. Proposition 4.2.4. Let f : R ! [−∞; 1] be an integrable function and g : R ! [−∞; 1] a Lebesgue measurable function with f(x) = g(x) almost everywhere. Then g must also be integrable and R g dµ = R f dµ. R R Proof. Let E = fx 2 R : f(x) 6= g(x)g (think of E as standing for ‘exceptional’) and note that f(x) = g(x) almost everywhere means µ(E) = 0. Write f = (1 − χE)f + χEf. Note that both (1 − χE)f and χEf are integrable because they are measurable and satisfy j(1 − χE)fj ≤ jfj and jχEfj ≤ jfj. Also Z Z χEf dµ ≤ jχEfj dµ = 0 R R R as χEf = 0 almost everywhere. Similarly χEg dµ = 0. R 4 2017–18 Mathematics MA2224 So Z Z f dµ = ((1 − χE)f + χEf) dµ R ZR Z = (1 − χE)f dµ + χEf dµ ZR R = (1 − χE)f dµ + 0 ZR = (1 − χE)g dµ R R R The same calculation (with jfj in place of f) shows (1 − χE)jgj dµ = jfj dµ < 1, so R R that (1 − χE)g must be integrable. Thus g = (1 − χE)g + χEg is also integrable (because R jχEgj dµ = 0 and so χEg is integrable and g is then the sum of two integrable functions). R R R R R Thus we have g dµ = (1 − χE)g dµ + χEg dµ = f dµ + 0. R R R R Remark 4.2.5. It follows that we should be able to manage without allowing integrable functions to have the values ±∞. The idea is that, if f is integrable, it must be almost everywhere finite. If we change all the values where jf(x)j = 1 to 0 (say) we are only changing f on a set of x’s of measure zero. This is exactly changing f to the (1 − χE)f in the above proof. The changed function will be almost everywhere the same as the original f, but have finite values everywhere. So from the point of view of the integral of f, this change is not significant. However, it can be awkward to have to do this all the time, and it is better to allow f(x) = ±∞. 4.3 Dominated convergence theorem Theorem 4.3.1 (Lebesgue dominated convergence theorem). Suppose fn : R ! [−∞; 1] are (Lebesgue) measurable functions such that the pointwise limit f(x) = limn!1 fn(x) exists. Assume there is an integrable g : R ! [0; 1] with jfn(x)j ≤ g(x) for each x 2 R. Then f is integrable as is fn for each n, and Z Z Z lim fn dµ = lim fn dµ = f dµ n!1 n!1 R R R R R Proof. Since jfn(x)j ≤ g(x) and g is integrable, jfnj dµ ≤ g dµ < 1. So fn is integrable. R R We know f is measurable (as a pointwise limit of measurable functions) and then, similarly, jf(x)j = limn!1 jfn(x)j ≤ g(x) implies that f is integrable too. The proof does not work properly if g(x) = 1 for some x. We know that g(x) < 1 almost everywhere. So we can take E = fx 2 R : g(x) = 1g and multiply g and each of the functions fn and f by 1 − χE to make sure all the functions have finite values. As we are changing them all only on the set E of measure 0, this change does not affect the integrals or the conclusions. We assume then all have finite values. Lebesgue integral 5 Let hn = g − fn, so that hn ≥ 0. By Fatou’s lemma Z Z Z lim inf (g − fn) dµ ≥ lim inf(g − fn) dµ = (g − f) dµ n!1 n!1 R R R and that gives Z Z Z Z Z Z lim inf g dµ − fn dµ = g dµ − lim sup fn dµ ≥ g dµ − f dµ n!1 R R R n!1 R R R or Z Z lim sup fn dµ ≤ f dµ (1) n!1 R R Repeat this Fatou’s lemma argument with g + fn rather than g − fn.
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