A sequence of absolutely continuous functions whose derivatives converge to $0$ a.e Here's the problem I'm having trouble with:

Let $f_{n}:[a,b] \to \mathbb{R}$ be a sequence of absolutely continuous functions such that $f_{n}(a)=0$, $||f_{n}'||_{2} \leq 1$ and $f_{n}' \to 0$ a.e. (in the standard Lebesgue measure):

*

*Prove that $||f_{n}'||_{1} \to 0$.

*Prove that $f_{n} \to 0$ a.e.


I know how to do part 2. once I do part 1: $|f_{n}(x)| = | \int_{a}^{x} f_{n}'(t)|dt \leq \int_{a}^{x}|f_{n}'(t)|dt \leq \int_{a}^{b} |f_{n}'(t)|dt \xrightarrow{n \to +\infty}0.$
I've tried solving one using the Holder inequality, as well as through the dominated convergence theorem (my attempt was, if $f_{n}$ is abs. cont, and $f'_{n}$ is bounded a.e, then $f_{n}$ is Lipschitz and has bounded derivative, so I can use the dominated convergence theorem. Unfortunately, $f_{n}'$ needn't be bounded, even if it tends to zero at a.e. point. How can I prove 1?
 A: As you may know, a.e. convergence does not imply $L^1$ convergence in general, with uniform integrability it will be the case.
We will use the following theorem :

A sequence of r.v. $(X_n)$ converges to $X$ in norm $L^1$ if it converges a.s. to $X$ and the sequence $(X_n)$ is uniformly integrable.

See here for more info and definitions.
Let $Y$ a uniform r.v. on $[a,b]$, $X=\phi(Y)$ with $\phi \equiv 0$ and $X_n=f'_n(Y)$.
$$E(|X_n|)=E(|f'_n(Y)|)=\frac 1 {b-a}\int_a^b |f'_n(y)|dy=\frac 1 {b-a} \|f'_n\|_1 \le \|f'_n\|_2.$$
The last inequality is by Holder's one by introducing $\chi_{[a,b]}$ the indicator function of the interval $[a,b]$.
Then :
$$E(|X_n|)\le 1.$$
Now let $ \varepsilon >0$ and $A$ a subset of $[a,b]$ s.t. $P(Y \in A) \le \varepsilon (b-a) $, then :
$$\frac 1 {b-a} \int |X_n| dy=\frac 1 {b-a}\int_A |f_n'(y)| dy =\frac 1 {b-a} \int_a^b |f_n'(y)| \chi_A dy \le \frac 1 {b-a} \|f'_n\|^2 \lambda(A) $$
Finally we get :
$$\frac 1 {b-a} \int |X_n| dy \le \frac 1 {b-a} \lambda(A) =\varepsilon$$
(We used Holder's inequality again and with $\lambda(.)$ the standard Lebesgue measure.)
Then the sequence $(X_n)$ is uniformly integrable and we can apply the theorem.
Then $E(|X_n|) = \frac 1 {b-a} \int_a^b |f_n'(y)| dy = \|f'_n\|_1 \to E(|X|)=0.$
PS : Note that the thorem is an "iff" if you replace a.s. convergence by convergence in probability. 
