The limit of a sequence of uniform bounded variation functions in $L_1$ is almost sure a bounded variation function Let  $\{f_n\} $be a sequence of functions on $[a,b] $ that $\sup V^b_a (f_n) \le C$,
if $f_n \rightarrow f $ in $L_1$ ,Prove that $f $ equals to a bounded variation function almost every where. 
I really have no idea to solve this, is there any hint ? 
Thanks a lot.

I wouldn't care much to do this question without relying on Helly's selection theorem but, then, I'm pretty lazy and am quite satisfied with proofs that take a single paragraph.

Theorem (Helly)  Let $f_n:[a,b]\to\mathbb{R}$ be a sequence of
  functions of bounded variation with
(i)  $V(f_n,[a,b])\leq C$ for all $n$  (i.e., uniformly bounded
  variation) and
(ii)  For at least one point $x_0\in [a,b]$, $\{f_n(x_0)\}$ is
  bounded.
Then there is a subsequence $f_{n_k}$ that converges pointwise to a
  function $g$ that has bounded variation on $[a,b]$. (Naturally since
  $V(g,[a,b])\leq C$.)

There problem solved!  Well no, it is never quite that easy.  We don't know about condition (ii) so you will have to pass to a subsequence of the original sequence using the other hypotheses in order to apply the theorem.  But Helly does all the work, we just tidy up the details.  (We do remember, however, that $L_1$ convergence implies convergence in measure and convergence in measure implies a a.e. convergent subsequence.)
