Showing $\mathcal{H}$ is a hilbert space.

Showing $\mathcal{H}$ is a hilbert space.

So this is an early exercise in Conway's A Course In Functional Analysis.
I'm trying to get to grips with this upto open mapping and closed graph to
see if I want to do any more functional analysis. Analysis isn't really my
thing but knowing lots of math is empowering and yadda yadda. Anyway the
problem:
Let $\mathcal{H}=\{f:[0,1]\rightarrow \mathbb{F} \,:\, f\text{ is
absolutely continuous, }f(0)=0, f'\in\mathcal{L}^2(0,1)\}$ (where
$\mathbb{F}$ is either $\mathbb{R}$ or $\mathbb{C}$).
Define $\langle f,g\rangle=\int_0^1 f'(t)\overline{g'(t)}\, dt$
Show $\mathcal{H}$ is a Hilbert space.
So I'm happy that $\langle \cdot,\cdot\rangle$ defines an inner product, I
just want to show completeness in the metric induced by the norm. So my
first stab was given a cauchy sequence $\{f_n\}$ to define
$$g(x)=\lim_{n\rightarrow\infty} f_n'(x)\text{ (in $\mathcal{L}^2$'s
norm)}$$ This defines $g$ almost everywhere as the sequence $f'(n)$ is
Cauchy in $\mathcal{L}^2(0,1)$ (this is direct from the definition of
$\mathcal{H}$'s norm). Then I want to define $f(x)=\int_0^x g(t)dt$.
Now I am struggling to show that $f$ is absolutely continuous (it's been a
while since real analysis, I could be missing something simple), I'm happy
that FTC deals with $f(0)=0$ and $f'\in\mathcal{L}^2(0,1)$ once I've shown
this.