# absolutely continuous and g satisfies a Lipschitz condition [on hold]

1. If $f$ is absolutely continuous and $g$ satisfies a Lipschitz condition, then $g \circ f$ is absolutely continuous.

2. If $f$ is absolutely continuous and strictly increasing and $g$ is absolutely continuous, then $g \circ f$ is absolutely continuous.

3. There exist absolutely continuous functions $f$ and $g$ defined on $[0,1]$ such that $g\circ f$ is not absolutely continuous.

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## put on hold as too broad by Mice Elf, avid19, Solid Snake, John Ma, Claude Leibovici24 secs ago

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The solutions to both (a) and (b) follow fairly directly from the definitions of absolutely continuous and Lipschitz continuous. Here's the solution to the first one:

(a) Since $g$ is Lipschitz continuous an interval $[a,b]$, there is some $M>0$ such that $|g(x)-g(y)|<M|x-y|$ for any $x, y \in [a,b]$. Now since $f$ is absolutely continuous on some other interval $[c,d]$, there is some $\delta >0$ such that, given $\epsilon > 0$,

$$\displaystyle \sum_{k=1}^n |f(x_k)-f(y_k)|<\epsilon/M$$

whenever $\{[x_k, y_k] \mid k=1, \ldots, n \}$ is a (clearly, finite) collection of mutually disjoint subintervals of $[c,d]$ such that $\displaystyle \sum_{k=1}^n (y_k-x_k)<\delta$. Now it is straightforward to see that

$$\displaystyle \sum_{k=1}^n |(g \circ f)(x_k)-(g \circ f)(y_k)|= \sum_{k=1}^n|g(f(x_k))-g(f(y_k))|$$

$$\leq \sum_{k=1}^n M|f(x_k)-f(y_k)|<M(\epsilon / M) = \epsilon$$

Thus $g \circ f$ is absolutely continuous on $[c,d]$.

Is it clearer what to do for part (b)? For part (c) I suggest looking at this previous question.

EDIT: If you need further explanation, please let me know.

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