# Proving $f$ is Absolutely Continuous

Let $\{f_{n}\}_{n}$ be a sequence of absolutely continuous function defined on $[0,1]$ such that $f_{n}(0)=0$ for all $n$. Assume that the sequence of derivatives $\{f_{n}^{`}\}_{n}$ is Cauchy in $L_{1}[0,1]$. Prove that the $\{f_{n}\}_{n}$ converges uniformly to a function $f$, and that $f$ is absolutely continuous in $[0,1]$.

Seriously, I have no idea where to start. This is one of the past qual question. Any help will be much appreciated.

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Absolute continuity is equivalent to showing $f$ can be written as the integral of its derivative. As well, $\int_0^xf_n'(t)dt=f_n(x)$ may be a useful observation. –  Alex R. Dec 23 '12 at 1:28

1) Note that, $f'_n$ is a Cauchy sequence in a complete space $(L_1[0,1],||.||_1)$,
2) Prove that $f_n$ is a uniformly Cauchy sequence.
3) $$|f_n(x)-f_m(x)|=\Bigg| \int_{0}^{x}f'_n(x)dx - \int_{0}^{x}f'_m(x)dx \Bigg|\leq \int_{0}^{x}|f'_n(x)-f'_m(x)|dx$$
$$\leq \int_{0}^{1}|f'_n(x)-f'_m(x)|dx\dots.$$