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I have been struggling to prove that if $f$ is integrable on $B$ and $g$ is continuous and strictly increasing on $A$ such that $Img \subset B$, then $f\circ g$ is integrable on $A$.

I know that this is true, for example, in the case when $f$ is integrable on $B=[0,1]$ and $g=x^n$ with $A=[0,1]$, then $f\circ g=f(x^n)$ is integrable on $A$. This is easily shown by Lebesgue criterion for Riemann integrability, because if $f$ is integrable on $[0,1]$, then it is discontinuous on the subset of $[0,1]$ of measure zero. Then since $g:[0,1] \rightarrow [0,1]: x \mapsto x^n$ is continuous bijection, the set of discontinuities of $f \circ g$ has measure zero. So, $f \circ g$ is integrable on $[0,1]$ by Lebesgue criterion again.

Is my argument above for the special case example correct? If yes, can one generalize it somehow?

Renat Sergazinov
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    The composition of integrable functions is integrable and monotonic functions are integrable. The latter is a consequence of Froda's theorem, which states that a monotonic function has at most countably many discontinuities (and that such a function is integrable is easier to demonstrate than the Lebsgue criterion in full generality). – Thorgott May 06 '19 at 17:22
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    The composition of integrable functions is not always integrable. – cmk May 06 '19 at 19:44

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I have been struggling to prove that if $f$ is integrable on $B$ and $g$ is continuous and strictly increasing on $A$ such that $Img \subset B$, then $f\circ g$ is integrable on $A$.

I don't think this is generally true.

Consider this counterexample:

Let $f$ be the characteristic function on the ternary cantor set $C$. Since $C$ is measure 0 and $f$ is continuous at points outside of $C$, $f$ is riemann integrable. Let $g$ be an ambient homeomorphism of $\mathbb{R}$ that carries the fat cantor set $F$ to $C$. We can do this since every cantor space on $\mathbb{R}$ are ambiently homeomorphic. Since $g$ is a continuous bijection on $\mathbb{R}$ (it's a homeomorphism of $\mathbb{R}$ to itself), it must be strictly monotone. However, $f \circ g$ restricted to $[0,1]$ is not riemann integrable since it's discontinuity set is $F$, which is not measure 0 by construction.

jet457
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