Assume that $(X,M,\mu)$ is a $\sigma$-finite space. Suppose that $|f_n|\leq g\in L^+$ and $f_n\rightarrow f$ in measure. Show that $\int f=\lim_{n\rightarrow\infty}\int f_n$.

I tried taking a subsequence of $f_n$, call it $f_{n_j}$ and $f_{n_j}\rightarrow f$ almost everywhere. Also, $f_{n_j}\leq g\in L^+$. So by Dominated Convergence Theorem, $f=\lim_{j\rightarrow\infty}\int f_{n_j}$ And then I say that by $\lim_{n\rightarrow\infty}\int f_n=\lim_{j\rightarrow\infty}\int f_{n_j}$, I get the equality.

However, I received a comment on my solution saying that the final statement only holds if $\lim_{n\rightarrow\infty}\int f_n$ exists. Can anybody provide a clearer explanation why this is the case and how I can fix my proof?

  • $\begingroup$ I knew it's a duplicate question regarding the exact statement of the question. But I thought I was being specific enough in terms of where I was having problems, to the extent that I think it is a different question altogether. Thanks anyway. :) $\endgroup$ – Haikal Yeo Apr 11 '13 at 15:12

I will add something to make Thomas's answer clearer

Consider this sequence

$(\int f_n)$. Now take any subsequence $(\int f_{n_k})$ Now apply your argument to $(\int f_{n_k})$, there exists a sub-subseqeuence $\int f_{n_{k_j}}$ converging $\int f$.

Now what you have is a sequence of real numbers such that for any subsequence, there is a sub-subsequence which converges to a limit.

This means, the sequence must converges to that limit, i.e.$(\int f_n)$ converges to $\int f$.


If the limit of a sequence $(x_n)_n$ exists, then all subsequence converge to the same limit. But without knowing that $(x_n)_n$ converges, the object $\lim_{n \to \infty} {x_n}$ is not defined.

Hint for fixing the proof: Let $\left(\lim_{n \to \infty} {\int f_n}\right)_n$ have the limit points $x,y \in \overline{\mathbb{R}}$. Use your argument to conclude that $x=y$.


Not the answer you're looking for? Browse other questions tagged or ask your own question.