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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    $\begingroup$ This is not true in general if $\int_X fd\mu =\infty$. (It actually isn't clear exactly what the claim is if $\int_X fd\mu = \infty$, but in that case at least $f_n\to f$ in $L_1$ need not hold.) $\endgroup$ – Jonas Meyer Dec 14 '12 at 2:28
  • $\begingroup$ @JonasMeyer I think the claim is still clear if $\underset{X}{\int}fd\mu = \infty$, it simply means $\underset{X}{\int}f_nd\mu\rightarrow \infty$ as $n\rightarrow\infty$. Can you give a counterexample to show that $f_n\rightarrow f$in $L_1$ does not hold? And also, how to prove $\underset{E}{\int}f_nd\mu\rightarrow \underset{E}{\int}fd\mu$ in this case, if this holds? $\endgroup$ – qitao Dec 14 '12 at 3:26
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    $\begingroup$ qitao: $\int_X f_nd\mu\to \infty$ would be part of the hypothesis, but the claim is that the rest of the stuff follows. What does $f_n\to f$ in $L_1$ mean? It would typically mean that $f\in L_1$ and $\int|f_n-f|\to 0$. While the latter condition could make sense in general even if $f$ is not in $L_1$, it will definitely fail if $f$ is not in $L_1$ while each $f_n$ is in $L_1$, which seems to be part of the hypothesis. Even if the $f_n$s also have infinite integral there is no way this implies that $\int|f_n-f|\to 0$, as $\int|f_n-f|$ could be $\infty$ for all $n$. $\endgroup$ – Jonas Meyer Dec 14 '12 at 4:02
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    $\begingroup$ @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. $\endgroup$ – qitao Dec 14 '12 at 4:57

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.

  • $\begingroup$ Thanks for your answer. But I have to say I'm sorry for all the confusion. I've edited the problem again. As I said in the problem, with the condition $\int f < \infty$, I also know how to prove this problem. And with the help of @Jonas Meyer I see that if $\int f = \infty$, $\int_X |f_n - f|\rightarrow 0$ does not hold. But I still feel that $\int_E f_n \rightarrow \int_E f$ is true, even when $\int_X f = \infty$. Is it possible to prove this, with $\mu(X)<\infty$ but not $\int_X f <\infty$? $\endgroup$ – qitao Dec 14 '12 at 17:10
  • $\begingroup$ @qitao I fixed the counterexample to be a counterexample to that too. $\endgroup$ – Chris Janjigian Dec 14 '12 at 18:15
  • $\begingroup$ I see. Thanks again! Your answer is very helpful! $\endgroup$ – qitao Dec 14 '12 at 19:43

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