sequence of absolutely continuous functions, uniform convergence Let $\{f_n\}$ be absolutely continuous functions on $[0,1]$, $f_n \geq 0$ $\forall n \in \mathbb{N}$. 
Suppose that $\lim_{n \to \infty} \int^1_0 f_n dx = 0$. 
Moreover, suppose that $\forall$ $\epsilon > 0$ $\exists$ $\delta > 0$ such that $\int_{E} |f_n'(x)|dx$ < $\epsilon$ $\forall n \in \mathbb{N}$,  if $m(E) &lt \delta$. 
Prove that $f_n$ converges to $0$ uniformly on $[0,1]$.
Can someone help show this?
Thanking you in advance.
 A: Suppose that $f_n$ does not converge uniformly to $0$. Then, there is an $\epsilon>0$ so that for all $N$ there is an $n>N$ and an $x_n$ so that 
$$
f_n(x_n)>2\epsilon\tag{1}
$$
Find $\delta>0$ so that $|E|&lt\delta$ implies
$$
\int_E|f_n^\prime(x)|\,\mathrm{d}x&lt\epsilon\tag{2}
$$
Suppose that $n$ is such that $(1)$ is true. Let
$$
E_n=\left\{x\in[0,1]:|x-x_n|&lt\dfrac{\delta}{3}\right\}\tag{3}
$$
Note that $\dfrac\delta3\le|E_n|\le\dfrac{2\delta}{3}&lt\delta$ (since $x_n$ could be near the boundary of $[0,1]$).
Then $(2)$ says that for $x\in E_n$ we have
$$
|f_n(x)-f_n(x_n)|\le\int_{E_n}|f_n^\prime(t)|\,\mathrm{d}t&lt\epsilon\tag{4}
$$
The triangle inequality applied to $(1)$ and $(4)$ says that for $x\in E_n$, $f_n(x)>\epsilon$. Thus,
$$
\int_{E_n}f_n(x)\,\mathrm{d}x>\epsilon\frac{\delta}{3}\tag{5}
$$
Since there are arbitrarily large $n$ so that $(1)$ is true, $(5)$ contradicts the assumption that
$$
\lim_{n\to\infty}\int_0^1f_n(x)\,\mathrm{d}x=0\tag{6}
$$
A: Following Mike's comment, pass to a subsequence to get almost everywhere convergence to 0.  Thus there is a set $A$ of measure 1 with $f_n(x) \to 0$ for all $x \in A$.  In particular $A$ is dense.  Now the assumption about the $f'$ guarantees that the family $\{f_n\}$ is equicontinuous (since $|f(x)-f(y)| \le \int_x^y |f'(t)|\,dt$).  One can now follow the proof of the Arzela-Ascoli theorem to obtain the uniform convergence.
