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I'm aware through a plethora of counterexamples that for two functions $f$, $g$ that are absolutely continuous, their composition is not necessarily so. However, before seeing these examples, I had written out what I thought was a proof confirming that it was true, and I'm not sure where my error is. My proof is as follows:

Assume that $f$ and $g$ are absolutely continuous. Let $\epsilon_f > 0$. Since $f$ is AC, there exists a $\delta_f$ such that for any finite disjoint collection $\{(a_k,b_k)\}$

$$ \sum_{k=1}^n [b_k - a_k] < \delta_f \Longrightarrow \sum_{k=1}^n |f(b_k) - f(a_k)| < \epsilon_f$$

Since $g$ is absolutely continuous as well, let $\epsilon_g = \delta_f$. Then there exists $\delta_g$ such that

$$ \sum_{k=1}^n [b_k - a_k] < \delta_g \Longrightarrow \sum_{k=1}^n |g(b_k) - g(a_k)| < \epsilon_g = \delta_f \Longrightarrow \sum_{k=1}^n |f(g(b_k)) - f(g(a_k)| < \epsilon_f $$

Thus, the composition is AC. I realize that the only difference between this is that in the composition, for the domain of $f$, I'm assuming that the sum $|g(b_k) - g(a_k)|$ in absolute value is less than $\delta_f$. However, I'm not quite sure why that absolute value should matter in the case of continuity. It's left out of the original definition because it's assumed that $b_k > a_k$. Other than that, though, I don't see any errors in logic. Thoughts?

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First, note that in the definition we take $\sum_{k=1}^n |b_k-a_k|<\delta$ and not $>$. The other problem is that $(g(a_k),g(b_k))$ are not necessarily disjoint intervals, as required in the definition of absolute continuity.

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  • $\begingroup$ Good catch on the inequality. A typo proliferated through copy and pasting. Ok, excellent, that is exactly what I was not seeing. Thanks for your answer! $\endgroup$
    – cnolte
    Aug 10 '16 at 3:53

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