# Absolutely Continuous Function composed with an Exponential Function [duplicate]

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Absolutely continuous function question

If $f(x)$ is absolutely continuous on $[0,1]$ then $f(x)$ is bounded and $\exp\left(f(x)\right)$ is absolutely continuous.

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## marked as duplicate by JavaMan, Jonas Meyer, Byron Schmuland, Srivatsan, t.b.Dec 8 '11 at 7:32

A famous theorem states that $f(x)$ is absolutely continuous on $[0,1]$ if and only if $f'(x)$ exists almost everywhere in $[0,1]$, and

$f(x) = f(0) + \int_0^x f'(t) dt$

for all $x \in [0,1]$. Given that $f(x)$ is absolutely continuous on $[0,1]$, $f'(x)$ exists almost everywhere and so also $\exp(f(x))$ is also differentiable almost everywhere in $[0,1]$ by the chain rule.

Now notice that by u-substitution with $u=f(t)$

\begin{align*} \int_0^x (\exp(f(t)))' dt &= \int_{f(0)}^{f(x)} (\exp(u))' du \newline &= \exp(f(x)) - \exp(f(0)) \end{align*}

So that in particular $\exp(f(x)) = \exp(f(0)) + \int_0^x (\exp(f(x))' dx$. Because the derivative of $\exp(f(x))$ exists almost everywhere in $[0,1]$ and because the integral formula holds for this function, $\exp(f(x))$ is absolute continuous on $[0,1]$.

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It may be easier to use the fact that $\exp$ is Lipschitz on the interval $f([0,1])$, so that for example no justification need be given for why integration by substitution is valid in this setting, but rather a direct appeal to the $\varepsilon$-$\delta$ definition of absolute continuity can be made. – Jonas Meyer Dec 8 '11 at 5:06
Thanks, I attempted to prove the problem and it looks like you and I were on the same page – james Dec 8 '11 at 5:25