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I want to show that if $f_n \rightarrow f$ a.e. on $A, \lambda(A) < \infty$ then $\lim_{n\rightarrow \infty} \int_A f_n(x) dx = \int_A f(x) dx$.

I've tried to prove this but my proof is sitting uncomfortably with me and I think it's missing things. In addition the question also has the definition of an equiintegrable sequence of functions in it, which I haven't used. If anyone could point out what I've done wrong or what I'm missing I'd be really grateful! Thanks in advance!

What I've done: Using Egorov's theorem since $\lambda(A)<\infty$ and $f_n \rightarrow f$ on $A$: $\forall \epsilon <0,\ \exists B\subset A:\ \lambda(A\backslash B)< \epsilon$ and $f_n\rightarrow f$ uniformly on $B$.

Because of this uniform convergence I can say that $|f_n -f| < \delta$ for all $\delta>0$ as from some $N$ large enough. Since $A= A\backslash B \cup B$ I have:

$\int_A |f_n -f|dx = \int_{A\backslash B}|f_n -f|dx + \int_B |f_n -f|dx \leq \delta \cdot ( \lambda(B) + \lambda(A\backslash B)) = \delta \cdot \lambda(A) \rightarrow 0$ and hence $\lim_{n\rightarrow \infty} \int_A f_n(x) dx = \int_A f(x) dx$.

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The result as stated is false. Let $A=[0,1]$ and $$ f_n(x)=\begin{cases} n &\text{if }0\le x\le1/n,\\ 0 & \text{if }1/n< x\le1. \end{cases}$$ Then $\int_Af_n=1$ for all $n$, but $\lim_{n\to\infty}f_n(x)=0$ for $0<x\le1$.

You certainly have to use the equiintegrability condition.

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  • $\begingroup$ Ok, do you have a hint as to how to start the proof? I'm a bit lost at the moment $\endgroup$ – Longeyes Nov 27 '13 at 18:39
  • $\begingroup$ What other convergence theorems do you know? Try to learn from their proofs. $\endgroup$ – Julián Aguirre Nov 27 '13 at 18:52

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