A problem about convergence in finite measure space 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!
 A: 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.
