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.
 A: Here is a way of proving this. (Of course, you should also  try to learn more about Helly's selection theorem, as pointed out by Brian.)
A 2 lines proof giving no detail could read as follows: a function is a.e. equal to a BV function if and only if its derivative in the distribution sense is a measure, and in this case the BV norm of the function is equal to the norm of this measure. From this, the result follows immediately.
Of course, this is not very self-contained... However, the details that follow are really nothing more than that.
The key point is the following lemma.
Lemma. Denote by $\mathcal E$ the linear space of all functions $\phi\in\mathcal C^1([a,b])$ such that $\phi(a)=0=\phi(b)$. For an integrable function $f:[a,b]\to\mathbb R$, the following are equivalent.
(1) $f$ is almost everywhere equal to a BV function $\widetilde f$;
(2) there is a constant $C$ such that $$\left\vert\int_a^b f(t) \phi'(t)\, dt\right\vert\leq C\, \Vert\phi\Vert_\infty\qquad\hbox{ for all $\phi\in\mathcal E$};$$
(3) there is a signed Borel measure $\mu$ on $[a,b]$ such that $$\int_a^b \phi'(t)f(t)\, dt=\int_{[a,b]} \phi\, d\mu\qquad\hbox{for all $\phi\in\mathcal E$}.$$
Moreover, if $\widetilde f$ is as in (1), then one can take $C:=V_a^b(\widetilde f)$ in (2).
Proof of $(1)\implies (2)$. Assume that $f=\widetilde f$ a.e., where $\widetilde f$ is BV. Let also $\phi\in\mathcal E$. Then, $\widetilde f$ being the difference of two monotonic functions and $\phi$ being continuous and BV, $\widetilde f$ is Riemann-Stieltjes integrable with respect to $d\phi$ (this is not difficult, and proved for example in Rudin's Principles of mathematical analysis; you can do it as an exercise, if you remember well the proof that monotonic functions are Riemann-integrable). It follows that $\int_a^b f(t) \phi'(t)\, dt$ is the limit of the Riemann-Stieltjes sums
$$S:=\sum_{j=0}^{N-1} \widetilde f(t_j) \bigl( \phi(t_{j+1})-\phi(t_j)\bigr)$$
as the mesh of the subdivision $a=t_0<\cdots <t_N=b$ goes to $0$. Now, since $\phi(t_0)=0=\phi(t_N)$, your sum $S$ can be written as follows:
$$S=-\sum_{j=1}^{N-1} \phi(t_{j})\bigl( \widetilde f(t_{j})-\widetilde f(t_{j-1})\bigr)\, ;$$
and since $\widetilde f$ is BV, we get
$$\vert S\vert\leq V_a^b(\widetilde f)\times \Vert \phi\Vert_\infty\, .$$
Passing to the limit, this gives
$$\left\vert\int_a^b f(t) \phi'(t)\, dt\right\vert=\left\vert\int_a^b \widetilde f(t) \phi'(t)\, dt\right\vert \leq C\, \Vert\phi\Vert_\infty\, ,$$
with $C:=V_a^b(\widetilde f)$.
Proof of $(2)\implies (3)$. Assume that $C$ is as in (2). This means that the linear functional $\phi\mapsto \int_a^b f(t)\phi'(t)\, dt$ is continuous on $(\mathcal E,\Vert\,\cdot\,\Vert_\infty)$. By the Hahn-Banach theorem, this linear functional may be extended to a bounded linear functional on the space $\mathcal C([a,b])$, with norm at most $C$; and by the Riesz representation theorem, this gives (3).
Proof of $(3)\implies (1)$. Assume that (3) holds true. Define a function $\widetilde f:[a,b]\to \mathbb R$ by
$$\widetilde f(t):= \mu\bigl((a,t]\bigr)\, \,,$$
This is a BV function, because for any subdivision $a=t_0<\cdots <t_N=b$, we have
$$\sum_{j=0}^{N-1} \vert \widetilde f(t_{j+1}-\widetilde f(t_j)\vert=\sum_{j=0}^{N-1} \left\vert \mu\bigl((t_j,t_{j+1}]\bigr)\right\vert\leq \Vert \mu\Vert\, .$$
Moreover, if $\phi\in\mathcal E$ then one can "integrate" by parts to get that
$$\int_{[a,b]} \phi\, d\mu=\int_a^b \widetilde f(t)\phi'(t)\, dt\, .$$
Indeed, since $\phi(a)=0$ we have $\phi(x)=\int_a^x\phi'(t)\, dt=\int_{[a,x)} \phi'(t)\, dt$ for all $x\in[a,b]$, so that (by Fubini's theorem)
$$\int_{[a,b]} \phi\, d\mu=\int_{[a,b]}\left(\int_{[a,x)} \phi'(t)\, dt\right) d\mu(x)=\int_a^b\phi'(t) \mu\bigl((t,b]\bigr)\, dt\, ,$$
and hence
$$\int_{[a,b]} \phi\, d\mu=\int_a^b \phi'(t) \bigl(\widetilde f(b)-\widetilde f(t)\bigr)\, dt=\int_a^b \phi'(t)\widetilde f(t)\, dt$$
because $\int_a^b \phi'(t)\, dt=\phi(b)-\phi(a)=0$.
Since (after all) we are assuming (3), we conclude that
$$\int_a^b \phi'(t)f(t)\, dt=\int_a^b\phi'(t)\widetilde f(t)\, dt\qquad\hbox{for all $\phi\in\mathcal E$}\, .$$
From this, one can then show that the function $f-\widetilde f$ is constant almost everywhere. This is not trivial, but quite classical (in sophisticated terms : if a function has derivative $0$ in the distribution sense, then it is a constant almost everywhere); and since this post is already long enough, I do not give a detailed proof. So $f$ is almost everywhere equal to a BV function $\widetilde f+cte$.
Let us now (finally) prove what is asked in the question. Sinc $V_a^b(f_n)\leq C$ for all $n$, we get by the Lemma
$$\left\vert \int_a^b \phi'(t)f_n(t)\, dt\right\vert\leq C\, \Vert\phi\Vert_\infty\qquad\hbox{for all $\phi\in\mathcal E$}\, .$$
Since $f_n\to f$ in $L^1$, it follows that
$$\left\vert \int_a^b \phi'(t)f(t)\, dt\right\vert\leq C\, \Vert\phi\Vert_\infty\qquad\hbox{for all $\phi\in\mathcal E$}\, ;$$
and hence $f$ is equal a.e. to a BV function, again by the Lemma.
A: Nice problem which opens up a few interesting doors. Our anonymous (and prolific) friend Etienne immediately associated the
phrase "equivalent to a function of bounded variation" with the world of distributional derivatives, thus providing an elegant
and informative discussion.
For me "uniformly of bounded variation" brings to mind, just as quickly, Helly.  Helly was 28 when he proved this selection theorem in 1912, not the only thing for which he is famous.  These were, no doubt, happy times.  To grow up in Vienna, study at Göttingen in the presence of Hilbert and Courant!  Alas, not much else afterwards is happy.  He enlisted in the Austrian army in 1914, was wounded and captured by the Russians where he spent the rest of the war and even later in Siberia.  He returned to an increasingly anti-intellectual and anti-semitic Vienna.  When the Nazis took over in 1938 it would have been clear what to do, and he did with the help of Einstein.  He got a low level teaching position${}^1$ in New Jersey and died in 1943.
But I digress.  Here is another very classical attack at this problem.  We assume that $f_n\to f$ in $L_1[a,b]$ and that
there is a uniform bound on the variation of the functions $f_n$.  If we write $F_n(x)=\int_a^x f_n(t)\,dt$ and
$F(x)=\int_a^x f(t)\,dt$ for $a\leq x \leq b$, then $F_n\to F$ uniformly.  The question now arises, how would we know whether $F$
is the indefinite integral of a function $g$ of bounded variation on $[a,b]$?  If so then $f$ and $g$ are equivalent. (These integrals can all 
be interpreted as Riemann integrals so the problem is not at any really deep level.)
Fortunately we do not have any real work to do (or I would abandon the field to someone more energetic).  There is a theorem of 
Riesz${}^2$ (yes the same Riesz that assisted in the solution given by Etienne).

Theorem.(Riesz)  Let $F:[a,b]\to\mathbb{R}$ be a continuous function. 
  A necessary and sufficient condition that $$F(x) = C + \int_a^x
  g(t)\,dt$$ for some constant $C$ and some function $g$ of bounded
  variation on $[a,b]$ is that $F$ has bounded slope variation${}^3$ on
  $[a,b]$.

Now the solution of our problem is mostly done, with all the heavy lifting done by Riesz.  The uniform bounded variation condition on $\{f_n\}$ translates routinely to a uniform bounded
slope variation condition for  $\{F_n\}$ and then, because $F$ is a pointwise limit of these functions $F$ must have 
bounded
slope variation.

*Footnote*${}^1$: This was typical at the time.  Truly great mathematicians fleeing the Nazis at the time were reduced to taking whatever teaching positions they could.  My favorite example is Zygmund.  He was quite widely famous for his treatise on Trigonometric Series.  So for his first teaching position in the US the chairman thought that it was more than appropriate to assign him to teaching elementary trigonometry to the freshman students.

*Footnote*${}^2$: This is Frigyes Riesz, not to be confused with his younger brother Marcel.  The Riesz representation theorem and this theorem here
are his.  This is the citation for his theorem quoted here:

F. Riesz, Sur certain systèmes singulier d’equations intégrales,
  Annales de L’École Norm. Sup., Paris (3) 28 (1911), 33–62.


*Footnote*${}^3$:  A function $F$ has bounded slope variation
on $[a,b]$ if there is a number $K$ so that
$$
\sum_{k=1}^{n-1}\left|  \frac{F(x_{i+1})-F(x_{i})}{x_{i+1}-x_{i}} - 
 \frac{F(x_{i})-F(x_{i-1})}{x_i-x_{i-1}} 
\right| \leq K
$$
for all choices of $a=x_0<x_1<x_2< \dots < x_n=b$.
A: 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.)
