$\{f_n\}$ absolutely continuous converging pointwise to $f$. Is $f$ of bounded variation and/or absolutely continuous?

Question

I'm trying to solve the following problem:
$\{f_n\}:[a,b] \to \mathbb{R}$ sequence of absolutely continuous functions converging pointwise to $f: [a,b] \to \mathbb{R}$. 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

Answer 1

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 $\int_I g(t)dt<\epsilon$.

Answer 2

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.