Bounded sequence of functions has subsequence convergent a.e.? 
Does bounded sequence of functions have subsequence convergent a.e.?

There is a equivalent question :

Does weak convergent sequence of functions have subsequence convergent a.e.?

These two are equivalent because of Banach-Steinhaus theorem.
Before I ask here, I thought Bolzano-Weierstrass theorem which is every bounded sequence has convergent subsequence. And my question comes up from 'this theorem can be applied to sequence of functions?'.
If the space of functions should be special (like reflexive or compact or housdorff), please mention about it.
I didn't study functional analysis but attended courses : Partial differential equation and real analysis for graduate student.
 A: Not necessarily. For each $x\in [0,1)$ let $B(x)=(x_n)_{n\in \mathbb N}$ be the sequence of digits for the representation of $x$ in base $2,$ with $\{n:x_n=0\}$ being infinite.  For $n\in \mathbb N$ let $f_n(x)=x_n.$ Let $g:\mathbb N\to \mathbb N$ be strictly increasing. A convergent binary sequence is eventually constant . So if $(f_{g(n)}(x))_n=(x_{g(n)})_n$  converges then $$x\in \cup_{m\in \mathbb N}(A_m\cup B_m)$$ $$ \text {where } A_m=\{x: \forall n\geq m\;(x_{g(n)}=0)\}$$ $$ \text  { and } B_m=\{x:\forall n\geq m\;(x_{g(n)}=1)\}.$$ We show that  $\mu(A_m)=\mu(B_m)=0$  for each $m,$ where $\mu$ is Lebesgue  measure, so  the set of $x$ for which $(f_{g(n)}(x))_n$ converges is Lebesgue-null. 
(i). Fix $m$. Let $S(n)=\{g(m+j):0\leq j\leq n\}.$ Let $2^{S(n)}$ denote  the set of all functions from $S(n)$ to $\{0,1\}.$
(ii). For $h\in 2^{S(n)}$ let $$R(h)=\{x\in [0,1): \forall i\in S(n)\;(x_i=h(i)\}$$ and  let $V(h)=\sum_{i\in S(n)}2^{-i}h(i).$ 
Let $h_0\in 2^{S(n)}$ where $h_0(j)=0$ for all $j\in S(n).$ 
(iii). For each $h\in 2^{S(n)}$ we have $R(h)=\{x+V(h): x\in R(h_0)\},$ so $$\mu (R(h))=\mu (R(h_0)).$$  Now if $h,h'$ are distinct members of $2^{S(n)}$ then $R(h)\cap R(h')=\emptyset.$ Also $\cup \{R(h):h\in 2^{S(n)}\}=[0,1).$ And the number of members of $2^{S(n)}$ is $2^n.$ 
(iv).Putting all the parts of (iii) together we have $$1=\mu ([0,1))=\sum_{h\in 2^{S(n)}}\mu (R(h))=2^n\mu (R(h_0)).$$ Since $A_m\subset R(h_0)$  we have $1\geq 2^n \mu (A_m)$ for every $n.$ Therfore $\mu (A_m)=0.$ 
The proof that $\mu (B_m)=0$ is similar.
Note: I neglected to show that $R(h_0)$ is a Lebesgue-measurable set. I will leave this as an exercise because it's 8 A.M. and I've been up all night. Happy New Year. 
A: There is no subsequence of $\sin (nx)$ that converges a.e. In fact, every subsequence $\sin (n_kx)$ diverges a.e. For a proof of this, see Pointwise almost everywhere convergent subsequence of $\{\sin (nx)\}$
