Theorems for the Lebesgue Integral University （勒贝格积分定理的大学）.pdf

Theorems for the Lebesgue Integral Dung Le1 We now prove some convergence theorems for Lebesgue’s integral. The main question is this. Let {fn } be a sequence of Lebesgue integrable functions on E and assume that fn converges a.e. to f on E . Is f Lebesgue integrable and lim E fn = E f ? Let’s look at the following examples. Define the following functions on [0, 1]. 1 fn (x) = χ[1/n,1] (x) and gn (x) = nχ[0,1/n] (x). x We easily see that fn → f with f (x) = 1/x if x 0 and f (0) = 0. But f is not Lebesgue integrable (why?)! Meanwhile, gn → g with g ≡ 0. But [0,1] g = 0 = 1 = lim [0,1] gn . We see that appropriate hypotheses need to be assumed to answer our question. 1 Convergence theorems We start with the monotone convergence theorem. Theorem 1.1 If fn is a sequence of nonnegative measurable functions defined on E and fn f on E then lim E fn = E f . Proof: Obviously, E f ≥ lim E fn (why?). To prove the opposite inequality, for each integer m choose an increasing sequence {sm,n } of simple functions so that sm,n fm as n → ∞. We then define Sn (x) = sup{sk,n (x) : 1 ≤ k ≤ n}, for n ≥ 1. Because sk,n ≤ sk,n+1 , we have (check it!) Sn ≤ Sn+1 and sm,n ≤ Sn ≤ fn for 1 ≤ m ≤ n. Letting n → ∞ in the last inequality, we get fm ≤ lim Sn ≤ f n→∞ for each m ∈ N . Let m → ∞ in the above inequality, we see that Sn f . Therefore lim E Sn = E f. 1Department of Applied Mathematics, University of Texas at San Antonio, 6900 North L