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Since $f_n \rightarrow f$ in measure then there exists a subsequence $f_{n_j}$ such that $f_{n_j} \rightarrow f$ a.e. and by the dominated convergence theorem $$\int f =\lim_{j\to\infty}\int f_{n_j}.$$ To finish the proof I need to show that $$\lim_{j\to\infty}\int f_{n_j} = \lim_{n\to\infty}\int f_{n}.$$

Folland 2.34 exercise involving dominated convergence theorem this question also got stuck at the same place I did but I don't understand the accepted answer which says "If you have a sequence $(f_n)$ and $f$ such that every subsequence has a further subseqeuence $(f_{n_{j_k}})$ converging to $f$, then the original sequence converges to $f$. In your case, you have a sequence of real numbers $\left(\int f_n\right)_n$".

How can I apply this to show that the two limits are the same?

Mathematics
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    It's essentially trying to show a property of a sequence of real numbers given by $\int f_n$, right? This seems relevant: https://math.stackexchange.com/questions/397978/every-subsequence-of-x-n-has-a-further-subsequence-which-converges-to-x-the – Trevor Norton Feb 11 '25 at 17:31
  • @TrevorNorton So we would have to show that every subsequence of $\left(\int f_n\right)_n$ converges to the same thing? – Mathematics Feb 11 '25 at 17:33
  • We want to show that every subsequence has a further subsequence which converges to the same limit. This has essentially already been done (a subsequence of $f_n$ will satisfy the same assumptions on the original sequence). – Trevor Norton Feb 11 '25 at 17:37
  • @TrevorNorton We proved that $\int f =\lim_{j\to\infty}\int f_{n_j}$ and since the sequence $(\int f_{n_j})_j$ converges any subsequence also converges to the same limit. But what if we chose a difference subsequence of $f_n$? How can we prove that a further subsequence of this subsequence must also converge to the same limit? – Mathematics Feb 11 '25 at 17:42
  • Any subsequence $f_{n_j}$ will converge to $f$ in measure and $|f_{n_j}| \leq g$, so the exact same argument can be applied. – Trevor Norton Feb 11 '25 at 17:45
  • @TrevorNorton Thanks that makes a lot sense now! – Mathematics Feb 11 '25 at 17:48

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The posting you are referring to can be explain as follows:

As $f_n$ converges (strongly) in measure (i.e. $\lim_{n\rightarrow\infty}\mu(|f_n-f|>\varepsilon)=0$ for all $\varepsilon>0$),

  • (a) Any subsequence $f_{n'}$ admits a subsequence $f_{n''}$ that converges to $f$ a.s.

  • (b) By Fatou's lemma, $f\in L_1$ (indeed, $\int_X|f|\leq\liminf_n\int |f|_n\leq\int g$).

  • (c) Dominated convergence implies that any subsequence $f_{n'}$ admits a subsequence $f_{n''}$ that converges to $f$ in $L_1$ (indeed, $|f_n|\leq g\in L_1$ for all $n$ and so, $|f_{n''}-f|\leq 2g$.)

Suppose that $f_n$ does not converge to $f$ in $L_1$. This means that there is $\varepsilon>0$ and subsequence $f_{n_k}$ of $f_n$ such that $$\|f_{n_k}-f\|_1\geq\varepsilon\qquad \forall k\tag{1}\label{one}$$ This contradicts (c), for no subsequence $f_{n''}$ of $f_{n_k}$ can converge in view of \eqref{one}.

Mittens
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