Vtyjkyuk

I'm trying to solve the following problem:

$f_n:[a,b] to mathbbR$ sequence of absolutely continuous functions converging pointwise to $f: [a,b] to mathbbR$. Say if $f$ is of bounded variation and/or absolutely continuous if:

a) $exists C : int_a^b |f_n'(x)|dx leq C ; forall n$

b) $exists g in L^1((a,b)): |f_n'| leq g text a.e. in (a,b) ; forall n$

My attempt:

Consider $a = x_0 < x_1 < cdots < x_N = b$.

Since $f_n$ are absolutely continuous, we have:
$$
f_n(x_j)-f_n(x_j-1) = int_x_j-1^x_j f_n'(t)dt
$$
$$
|f(x_j) - f(x_j-1)| = lim_n to infty |f_n(x_j)-f_n(x_j-1)| leq lim_n to infty int_x_j-1^x_j |f_n'(t)|dt \
implies sum_j=1^N|f(x_j) - f(x_j-1)| leq lim_n to infty sum_j=1^N int_x_j-1^x_j |f_n'(t)|dt = lim_n to infty int_a^b |f_n'(t)|dt
$$
Then, both $(a)$ and $(b)$ imply $f$ of bounded variation, since:
$$
(a) implies int_a^b |f_n'(t)|dt leq C \
(b) implies int_a^b |f_n'(t)|dt leq int_a^b g(t)dt < infty
$$
Is this right? However, I do not know how to proceed for the absolute continuity of $f$.
Can someone help? Thank you

Hints: For (a), consider $f_n(x)=x^n$ on $[0,1]$. For (b), you need to use very similar arguments as above, just with intervals whose length sum $|I|$ is small, and remember that the Lebesgue integral is absolutely continuous, i.e. for all $epsilon>0$ you can find $delta>0$ such that $|I|<delta$ implies $int_I g(t)dt<epsilon$.

A little bit of measure theory easily gives absolute continuity of $f$. We have $sum |f_n(y_i)-f_n(x_i)|=|int _x_i^y_i f_n'(t) dt| leq int _x_i^y_i g(t) dt$. Now use the fact that given $epsilon >0$ there exist $delta >0$ such that $int_A g(t), dt <epsilon$ whenever $m(A) <delta$ where $m$ is Lebesgue measure. Take $A$ to be union of the intervals $(x_i,y_i)$. You will get uniform absolute continuity of the $f_n$'s from this and letting $n to infty$ completes the proof.