Absolute continuity equivalence

Question

Show that a function $f:[a,b] \rightarrow \mathbb{R}$ is absolutely continuous if and only if there exists a sequence of Lipschitz functions $(f_n)$ such that $V([a,b], f-f_n)$ converges to $0$.

Answer 1

With some help from Daniel Fischer (see above), I have written up the following solution.

First assume $f$ is absolutley continuous. Since $f$ is absolutely continuous on $[a,b]$, we know that $f'$ exists almost everywhere on $[a,b]$ and $f(x)= f(a) + \int_a^x f' .$ Because $f'$ is integrable it can be approximated in the sup-norm by continuous functions, which can in turn be approximated by piece-wise linear functions in the sup-norm. Since $[a,b]$ has finite measure approximation in the sup-norm implies approximation in the $L^1-$ norm. So let $(g_n)$ be a sequence of piece-wise linear functions that converge to $f'$ in the sup-norm and hence the $L^1-$ norm. Then set $$f_n(x)= f(a)+ \int_a^xg_n,$$ which can be rewritten as $f_n(x)=f(a)+M_n+ Q_n(x)$, where $Q_n$ is some quadratic and $M_n$ is a constant.

Thus, $(f_n)$ converges to $f$ and since the $f_n$ are continuously diferentiable on $[a,b]$ they are Lipshcitz.

Finally, $$V([a,b], f-f_n)=\int_a^b|(f-f_n)'|= \int_a^b|f'-g_n|.$$ And we know this goes to zero by construction.

Now assume we have a sequence of Lipschitz functions satisfying the above condition. First note that for any finite disjoint collection of open intervals $\{(a_k,b_k) \}_{k=1}^m$ and any $n$, $$\sum_{k=1}^m |f(b_k)-f(a_k)| \leq \sum_{k=1}^m |(f-f_n)(b_k)-(f-f_n)(a_k)|+ \sum_{k=1}^m |f_n(b_k)-f_n(a_k)|.$$ Thus, $$\sum_{k=1}^m |f(b_k)-f(a_k)| \leq V([a,b], f-f_n) + \sum_{k=1}^m |f_n(b_k)-f_n(a_k)|.$$

Now let $\epsilon >0$ be given. Take $n$ such that $V([a,b], f-f_n)<\epsilon/2$, and since $f_n$ is Lipschitz it is absolutely continuous so we can find $\delta$ such that $\sum_{k=1}^m (b_k - a_k)< \delta$ implies $$\sum_{k=1}^m |f_n(b_k)-f_n(a_k)|< \epsilon/2.$$ It follows that $f$ is absolutely continuous on $[a,b]$.