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I have a question regarding the final step in my proof of the following exercise:

Exercise 2.34 Suppose $|f_n|\leq g \in L^1$ and $f_n \to f$ in measure. Show $\int f = \lim_{n\to\infty}\int f_n$.

Since $f_n\to f$ in measure, there exists a subsequence ${f_{n_j}}$ such that $f_{n_j}\to f$ a.e.. Now since $\forall j\in \mathbb{N}$ $|f_{n_j}|<g$ by the dominated convergence theorem we have that

$$\int f = \int \lim_{j\to\infty}f_{n_j}=\lim_{j\to\infty}\int f_{n_j}$$

Now all I need to do is show that $\lim_{j\to\infty}\int f_{n_j}=\lim_{n\to\infty}\int f_{n}$ however I cannot see why this is true.

What is the reasoning which allows those two limits to be the same?

random0620
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  • When you say converges in measure, do you mean local or global convergence in measure? – Mark Saving Aug 04 '21 at 21:46
  • @MarkSaving The definition of "converges in measure" given by Folland is for every $\epsilon >0$ $\mu({x:|f_n(x)-f(x)|>\epsilon})\to 0$ as $n \to \infty$ which I believe is local convergence in measure. – random0620 Aug 04 '21 at 21:48

1 Answers1

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This is a general fact (at least in, say, metric spaces): If you have a sequence $(f_n)$ and $f$ such that every subsequence has a further subsequence $(f_{n_{j_k}})$ converging to $f$, then the original sequence converges to $f$. This is just the negation of convergence (once you unpack the definitions). The key is that the limit is the same no matter the subsequence.

In your case, you have a sequence of real numbers $\left(\int f_n\right)_n$.

Jose27
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  • A sequence with a convergent subsequence doesn't necessarily have a limit. It's only if it does have a limit that you can make such a claim. – Mark Saving Aug 04 '21 at 21:49
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    @MarkSaving: I'm not claiming that; I'm assuming that every subsequence has a further convergent subsequence, and that the limit is independent of which subsequence you pick. – Jose27 Aug 04 '21 at 21:51
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    as a side note for OP: your argument can be reworded slightly to show that the sequence of real numbers $a_n:=|f_n-f|_{L^1}$ is such that every subsequence has a further subsequence which converges to $0$. Thus, by the lemma mentioned here, $a_n\to 0$, i.e $f_n\to f$ in $L^1$. – peek-a-boo Aug 04 '21 at 22:08
  • Thanks! Do you know of a text which has this theorem? It didn't click that this was true before seeing this here. – random0620 Aug 04 '21 at 22:31
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    For the case of real numbers I think Bartle's "Introduction to real analysis" has it. I'd be surprised if baby Rudin didn't. I'm pretty sure it'd be stated as an exercise though, in both of those. – Jose27 Aug 04 '21 at 22:43