A relation between convergence in measure and pointwise convergence 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?
 A: 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.
