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Let $\{f_n\}$ be a sequence of measurable functions on $R$ with Lesbegue measure and $f$ be a measurable function.

I have to show that $\{f_n\}$ converges to $f$ in measure if and only if any subsequence of $\{f_n\}$ has a subsequence that converges to $f$ almost everywhere. It was not difficult to show the "only if" part. But I am stuck at the "if" part. Could anyone help me with this?

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Suppose $\{f_n\}$ converges to $f$ in measure. Let $\{f_{n_k}\}$ be a subsequence of $\{f_n\}$. Choose $n_{k_1} < n_{k_2} < n_{k_3} < \cdots$ such that $m(|f_{n_{k_j}} - f| > 2^{-j}) < 2^{-j}$ for all $j \in \Bbb N$. Let $A_j = (|f_{n_{k_j}} - f| > 2^{-j})$ and $A = \limsup A_j$. Since $\sum_j m(A_j) < \infty$,

$$m(A) = \lim_{j\to \infty} m\left(\bigcup_{k\ge j} A_k\right) \le \lim_{j\to \infty} \sum_{k = j}^\infty m(A_k) = 0.$$

Given $x\notin A$, there exists a positive integer $N$ such that $x\notin A_j$ for all $j\ge N$, i.e., $|f_{n_{k_j}}(x) - f(x)| \le 2^{-j}$ for all $j \ge N$. Thus $f_{n_{k_j}}(x) \to f(x)$. As $x$ was arbitrary, $f_{n_{k_j}} \to f$ a.e..

Now suppose $\{f_n\}$ does not converge to $f$ in measure. Decompose $\Bbb R$ as the countable union of open intervals $I_k$. Then there exists an index $\ell$, positive numbers $\epsilon$ and $\alpha$, and a sequence of indices $n_1 < n_2 < n_3 < \cdots$ such that $m(\{x\in I_\ell :|f_{n_k}(x) - f(x)| > \epsilon\}) \ge \alpha$ for all $k \in \Bbb N$. Let $\{f_{n_{k_j}}\}$ be any subsequence of $\{f_{n_k}\}$. If $X$ is the set of points $x\in I_\ell$ such that $\lim\limits_{j\to \infty} f_{n_{k_j}}(x) \neq f(x)$, then $X = \cup_{m\ge 1} (\limsup\limits_{j\to \infty} X^j_m)$, where $X_m^j = \{x\in I_\ell :|f_{n_{k_j}} - f| > \frac{1}{m}\}$. Let $N$ be a positive integer such that $\frac{1}{N} < \epsilon$. Then

$$ m(X) \ge m(\limsup\limits_{j\to \infty} X_j^N) \ge \limsup_{j\to \infty}\, m(X_j^N) \ge \alpha.$$

It follows that $\{f_{n_{k_j}}\}$ does not converge to $f$ pointwise almost everywhere.

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  • $\begingroup$ This proves the "only if" part. Wasn't OP stuck on the "if" part? $\endgroup$ – grand_chat Apr 9 '15 at 15:43
  • $\begingroup$ Could you explain more about the last part of your answer? Why is the measure of limsup X_(N,j) larger than α? $\endgroup$ – Keith Apr 27 '15 at 13:17
  • $\begingroup$ But, in oder for the inequality to hold, the union of X_(N, j) must be of finite measure. $\endgroup$ – Keith Apr 27 '15 at 14:09
  • $\begingroup$ Thank you for your explanantion. I was stuck at that part. $\endgroup$ – Keith Apr 27 '15 at 14:10
  • $\begingroup$ Wait! does the assumption that the measure of X is zero implies that the union of X_(N, j) is of finite measure? I think that doesn't have to be true. $\endgroup$ – Keith Apr 27 '15 at 14:32

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