Monday, 30 January 2017

real analysis - About an exercise in Rudin's book

In the book of Walter Rudin Real And Complex Analysis page 31, exercise number 10 said:
Suppose μ(X)<, {fn} a sequence of bounded complexes measurables functions on X , and fnf uniformly on X. Prove that limnXfndμ=Xfdμ



And the answer is the following



Let ϵ>0. Since fnf uniformly, therefore there exists n0N
such that
|fn(x)f(x)|<ϵn>n0


Therefore |f(x)|<|fn0(x)|+ϵ. Also |fn(x)|<|f(x)|+ϵ. Combining both
equations, we get
|fn(x)|<|fn0|+2ϵn>n0

Define g(x)=max(|f1(x)|,···,|fn01(x)|,|fn0(x)|+2ϵ), then fn(x)g(x)
for all n. Also g is bounded. Since μ(X)<, therefore gL1(μ). Now

apply DCT to get
limnXfndμ=Xfdμ



What didn't I understand is why the condition fnf uniformly on X is necessary? I proceed like the following



For every xX we have
|fn(x)|h(x)=maxi(|fi(x)|)


such as every fi is bounded so h is, and in the other hand we have μ(X)<, therefore hL1(μ), then we can apply DCT.



Where I am wrong please? are there any counter examples?

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