Composition preserves Riemann integrability

Let $f:[a,b]\to\mathbb{R}$ be Riemann integrable , and let $g:\mathbb{R}\to\mathbb{R}$ be continuous .

Proposition: The composition $g\circ f$ is Riemann integrable on $[a,b]$.

More generally, it suffices that $g$ be continuous on a closed interval containing the compact set $f([a,b])$ (since $f$ is bounded ).

This proposition is used constantly to deduce integrability of $|f|$, $f^2$, $\sin(f)$, etc., from integrability of $f$.

Proof sketch: If $f$ is continuous at $x$, then $g\circ f$ is continuous at $x$ because $g$ is continuous and composition preserves continuity. Hence the discontinuity set of $g\circ f$ is contained in the discontinuity set of $f$. By the Lebesgue criterion , the discontinuity set of $f$ has measure zero, so the same holds for $g\circ f$, which is therefore Riemann integrable.