# Composition of Integrable Function and Strictly Increasing Continuous Function

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?

• 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). May 6, 2019 at 17:22
• The composition of integrable functions is not always integrable.
– cmk
May 6, 2019 at 19:44

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 $$A$$.
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.