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 fn→f uniformly on X. Prove that limn→∞∫Xfndμ=∫Xfdμ
And the answer is the following
Let ϵ>0. Since fn→f uniformly, therefore there exists n0∈N
such that
|fn(x)−f(x)|<ϵ∀n>n0
Therefore |f(x)|<|fn0(x)|+ϵ. Also |fn(x)|<|f(x)|+ϵ. Combining both
equations, we get
|fn(x)|<|fn0|+2ϵ∀n>n0
Define g(x)=max(|f1(x)|,···,|fn0−1(x)|,|fn0(x)|+2ϵ), then fn(x)≤g(x)
for all n. Also g is bounded. Since μ(X)<∞, therefore g∈L1(μ). Now
apply DCT to get
limn→∞∫Xfndμ=∫Xfdμ
What didn't I understand is why the condition fn→f uniformly on X is necessary? I proceed like the following
For every x∈X 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 h∈L1(μ), then we can apply DCT.
Where I am wrong please? are there any counter examples?
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