Show that H$(I)$ is a closed subspace of $L^2(I)$

Question

EDIT: Original statement is not true, added condition.

Let $I$ be the unit interval, define $H(I) = \{f\in AC(I)$ and $f'\in L^2(I)\}$. I want to show that $H(I)$ a closed subspace of $L^2(I)$. Suppose there is a limit point $f\in L^2(I)$ such that $\int |f_n-f|^2d\mu\to 0$ for $\{f_n\}\subset$ H$(I)$, I want to show that $f$ is in H$(I)$.

Let $\{[a_k,b_k]\}$ be a finite family of disjoint intervals contained in $I$, I observe $$\sum_k |f(b_k)-f(a_k)| \le \sum_k |f(b_k)-f_n(b_k)|+\sum_k |f_n(b_k)-f_n(a_k)|+\sum_k |f_n(a_k)-f(a_k)|$$ where the middle term is easily bounded using AC of $f_n$, but I don't know how to bound first and last term. Also, how do I show that $f'\in L^2(I)$?

Answer 1

Unfortunately this is not true, as the absolutely continuous functions constitute a dense subset of $L^2(I)$.

If, even the set $P(I)$ of trigonometric polynomials of the form $$ \sum_{|k|\le N}a_k\mathrm{e}^{2k\pi i} $$ for some $N\in\mathbb N$, where $a_j$ constants, are dense in $L^2(I)$, and it is a subset of the set of absolutely continuous functions.