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.
This is more or less the continuity of the Lebesgue integral. Since $g \in L^2(0,1) \subset L^1(0,1)$, you have $$\int_E g \, dx \to 0$$ as $\mu(E) \to 0$, where $\mu$ is the Lebesgue measure of $E$.
Now, take $\varepsilon > 0$ arbitrary and pick $\delta > 0$, such that $\int_E g \,dx \le \varepsilon$ for all measurable $E$ with $\mu(E) \le \delta$. This immediately implies the absolute continuity of $f$.