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Suppose we have a measure space $(X,M,\mu)$, where $\mu(X)<+\infty$. Let ${f_n}$ be a sequence of non-negative functions, $f_n\in L_1,\ \forall n$ and $f_n$ converges to some $f$ pointwise. $f$ is not necessarily in $L_1$. Suppose now $\underset{X}{\int} f_nd\mu$ also converges to $\underset{X}{\int}fd\mu$, as $n\rightarrow \infty$. Is it true that $\underset{E}{\int} f_nd\mu\rightarrow\underset{E}{\int}fd\mu,\ \forall E\in M$? If not, can you give a counter-example?

I've encountered a similar problem, where I have the condition $f\in L_1(X,M,\mu)$ but don't have the condition $\mu(X)<+\infty$. In that case it is pretty easy to prove this problem. But here I don't have this condition, and $\underset{X}{\int}fd\mu = \infty$ can indeed happen, so how can I prove the problem now? I tried using Egoroff's theorem, but I failed to get it to work. Any suggestions? Thanks!

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    @JonasMeyer I have edited the question. Hopefully it is clearer. But I get your point. If $f$ is not in $L_1$, then $\int |f_n - f|\rightarrow 0$ will fail. I mainly focused my attention on the first part of the problem, so I didn't think of this issue. But is it possible that the first part is still true? I feel it's true, but can't prove it.2012-12-14

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For a counterexample to the first claim where $\|f\|_{L^1} = \infty$ consider the space $[0,1]$ with the Lebesgue measure, $f(x) = \frac{1}{x}$ and $f_n(x) = \frac{1}{x} \chi_{[\frac{1}{n},1]} + n \chi_{[1-\frac{1}{n},1)}$ so $\int f_n \to \int f = \infty$ clearly we have pointwise convergence $f_n \to f$ everywhere and $\int_0^1 |f_n - f| = \infty$ for all $n$. Moreover, $\int_{[\frac{1}{2},1]} f_n - f = 1$ for any $n \geq 2$ so the second claim fails as well.

You do not need the hypothesis that $\mu(X) < \infty$ for the rest though if $\int f < \infty$. The triangle inequality gives for positive numbers $a$ and $b$ $0 \leq |a - b| + a - b \leq 2a$

First, let's apply the dominated convergence theorem to the inequality

$0 \leq |f - f_n| + f - f_n \leq 2f$

Since $\int f_n \to \int f$ this gives that $\|f - f_n \|_{L^1} \to 0$

Notice that for $E \subset X$ measurable this implies that $\int_E |f - f_n| \to 0$

since we have (the integrand being positive) $\int_E |f - f_n| \leq \int_X |f - f_n|$

$L^1$ convergence implies convergence of norms, so we're done.

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    I see. Thanks again! Your answer is very helpful!2012-12-14