Convergence in measure and almost everywhere 
In a finite measure space, let $\{f_{n}\}$ be a sequence of measurable functions. Show that $f_{n}  \rightarrow f$ in measure if and only if every subsequence $\{f_{n_{k}}\}$ contains a subsequence $\{f_{n_{k_{j}}}\}$, that converges almost everywhere to f.

I can prove from converge in measure to converge almost everywhere, but I don't know how to write it down for the other direction.
 A: If not, then for some $\varepsilon>0$ and c>0, there exist a subsequence of $f_n(x)$, for example $f_{n_k}(x)$, which satisfies
$\forall n_k$,
$m\{x \in X : |f_{n_k}(x) - f(x)| > \varepsilon\}>c $ (*)
Since $f_{n_k}(x)$ has a subsequence $g_n(x)$, which converges almost everywhere to $f(x)$, and that the problem is discussed in a finite measure space, we can say $g_n(x)$ converges to $f(x)$ in measure. However, that contradicts the existence of $f_{n_k}(x)$. $g_n(x)$ is a subsequence of $f_{n_k}(x)$, but it doesn't follow (*).
A: I'll use the hint from Kolmogorov's book to suggest two possible proofs.
$\textbf{Hint}$. Let $\left \lbrace \delta_n \right \rbrace$ be a sequence of positive numbers such that 
\begin{align*}
\lim_{n\rightarrow \infty}\delta_n=0,
\end{align*}
and let $\left \lbrace \epsilon_n \right \rbrace$ be a sequence of positive numbers such that 
\begin{align*}
\sum_{n=1}^{\infty} \epsilon_n < \infty .
\end{align*}
Let $\left \lbrace n_k \right \rbrace$ be a sequence of positive integers such that $n_k>n_{k-1}$ and 
\begin{align*}
\mu \left \lbrace  x: \left| f_{n_k}(x) - f(x)  \right| \geq \delta_k\right\rbrace  < \epsilon_k \qquad (k=1,2,...).
\end{align*}
Moreover, let 
\begin{align*}
R_i = \bigcup_{k=i}^{\infty} \left  \lbrace  x: \left| f_{n_k}(x) - f(x)  \right| \geq \delta_k \right\rbrace, \qquad Q=\bigcap_{i=1}^{\infty}R_i.
\end{align*}
Then $\mu(R_i)\rightarrow\mu(Q)$  as $i\rightarrow \infty$, since $R_1\supset R_2 \supset \cdot\cdot\cdot$. On the other hand,
\begin{align*}
\mu(R_i) < \sum_{n=1}^{\infty} \epsilon_n,
\end{align*}
and hence $\mu(R_i)\rightarrow 0$, so that $\mu(Q)=0$.
Now show that $\left\lbrace f_{n_k} \right\rbrace$ converges to $f$ on $E-Q$.
$\textbf{Alternative 1}$
Let $B_k=\left\lbrace x: \left| f_{n_k}(x) - f(x)  \right| \geq \delta_k \right\rbrace$.
Note that
\begin{align*}
\forall x\in (E-Q) \implies x \notin Q \implies x \notin \bigcap_{i=1}^{\infty}R_i \implies x \notin R_i \, \text{ for some } i \implies x \notin B_k \, \text{ for } k \geq i  .
\end{align*}
In fact this implies that there is no $k\geq i$ for which  $\left| f_{n_k}(x) - f(x)  \right|\geq \delta_k$ holds in $E-Q$, since otherwise it would be contained in every $R_i$ and hence in $Q$.
So we have that in $E-Q$ there is no $k\geq i $ such that  $\left| f_{n_k}(x) - f(x)  \right| \geq \delta_k$ implying that the opposite is true, i.e., for all $k\geq i$,  $\left| f_{n_k}(x) - f(x)  \right| < \delta_k$, where $\delta_k \rightarrow 0$ as $k\rightarrow \infty$ (by its definition). So we have that $f_n(x)\rightarrow f(x), \, \forall x \in (E-Q)$, and finally that $f_n(x)\rightarrow f(x) \, a.e.$ since the set $Q$ has measure zero.
$\textbf{Alternative 2}$
Let $B_k=\left\lbrace x: \left| f_{n_k}(x) - f(x)  \right| \geq \delta_k \right\rbrace$.
We have 
\begin{align*}
(E-Q)  &= E-\bigcap_{i=1}^{\infty}R_i=\bigcup_{i=1}^{\infty}(E-R_i)=\bigcup_{i=1}^{\infty}R_i^{\complement} \\
&=\bigcup_{i=1}^{\infty}\left( \bigcup_{k=i}^{\infty}B_k\right)^\complement
\bigcup_{i=1}^{\infty}\bigcap_{k=i}^{\infty}B_k^{\complement}
\end{align*}
\begin{equation}
(E-Q)=\bigcup_{i=1}^{\infty}\bigcap_{k=i}^{\infty} \left\lbrace x: \left| f_{n_k}(x) - f(x)  \right| < \delta_k \right\rbrace.
\end{equation}
From the last equation, we can read that in this set there exists an integer $i$ such that for all $k\geq i$, for which $\left| f_{n_k}(x) - f(x)  \right| $ is less than $\delta_k$ which get arbitrarily small, implying the convergence of $f_{n_k}$ to $f$ on $(E-Q)$.
A: Assume $\{f_n\}$ converges pointwise a.e. to $f$. Fix $\varepsilon > 0$ and define:
$$
E_N = \{x \in X \mid \exists n > N : |f_n(x) - f(x)| > \varepsilon\}
$$
We have $E_1 \supset E_2 \supset \cdots$ and $\mu(E_1) \le \mu(X) < \infty$. Furthermore, since $\{f_n\}$ converges pointwise a.e. to $f$, we have $\mu\left(\bigcap_{N \in \mathbb{N}} E_N\right) = 0$.
Hence, $\lim_{N \to \infty} \mu(E_N) = 0$ and we can find an $N$ for which $\mu(E_N) < \varepsilon$.
It's clear that:
$$
\forall n > N : \{x \in X \mid |f_n(x) - f(x)| > \varepsilon\} \subset E_N
$$
Hence, $\{f_n\}$ converges to $f$ in measure.
A: "Convergence almost everywhere implies convergence in measure (for a finite measure)" follows almost trivially from the following lemma (e.g., lemma 2.5.4 of the Ash and Dade text):
Suppose $\mu$ is a finite measure. Then $f_n \to f$ pointwise $\mu$-a.e. if and only if
$$
\forall \epsilon > 0, \lim_{n \to \infty} \mu( \bigcup_{n \geq k} B_{k \epsilon}) = 0,
$$
where $B_{k\epsilon}$ is the set where the convergence is violated by the amount $\epsilon$ after index $k$,
$$
B_{k \epsilon} := \{x:  |f_k(x) - f(x)| > \epsilon \}.
$$
(we actually only need the $\implies$ direction).
Now to prove convergence in measure, fix any $\epsilon > 0$, then
$$
\lim_{n \to \infty} \mu(\{x: |f_n(x) - f(x)| > \epsilon \}) =  \lim_{n \to \infty} \mu(B_{n\epsilon}) \leq  \lim_{n \to \infty} \mu(\bigcup_{n \geq k} B_{k \epsilon}) = 0
$$
since $B_{n \epsilon} \subset  \bigcup_{n \geq k} B_{k \epsilon}$. That's it!
