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Let $f:[0,1]\to \mathbb{R}$ and $g:[0,1]\to[0,1]$ be two Riemann integrable functions. Assume that $|g(x)-g(y)|\geq a|x-y|$ for any $x,y\in [0,1]$ and some fixed $a\in (0,1)$. Show that $f\circ g$ is Riemann integrable.

I've learnt that if $g:[0,1]\to[0,1]$ integrable and $f:[0,1]\to\mathbb{R}$ continuous, then $f\circ g$ is integrable. Because the set of discontinuity of $f\circ g$ is contained in the set of discontinuity of $g$. Thus it follows from Lebesgue criterion. I feel like this problem is somehow similar, but what does $|g(x)-g(y)|\geq a|x-y|$ tell us?

Fluffy Skye
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1 Answers1

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There is a set $E$ of measure $0$ such that $f$ is continuous at all points of $E^{c}$ and set $D$ of measure $0$ such that $g$ is continuous at all points of $D^{c}$. Let $F=g^{-1}(E)$. If we show that $F$ has measure $0$ then it is clear that if $x \notin F\cup D$ implies $f\circ g$ is continuous at $x$. [ $x_n \to x$ implies $g(x_n) \to g(x)$ and $f(g(x_n)) \to f(g(x))$]. The hypothesis $|g(x)-g(y)|\geq a|x-y|$ is required to assert that $F$ has measure $0$: it follows from this hypothesis that $g$ is one-to-one and its inverse is a Lipschitz function on the range of $g$. It is well known and, in fact easy to show, that a Lipschitz function maps a set of measure $0$ to set of measure $0$.

Kavi Rama Murthy
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