Show that $g\circ f$ is Riemann-Integrable too.

75 Views Asked by At

Let $g:\Bbb R\to \Bbb R$ be a continuous function and $f:\Bbb R\to \Bbb R$ be a Riemann-Integrable function.

Show that $g\circ f$ is Riemann-Integrable too.

ATTEMPT:

Investigating the set of points of discontinuity of $g\circ f$ we get that if $f$ is discontinuous at $a$ then $g\circ f$ is discontinuous at $g(a)$ since

$f$ is discontinuous at $a\implies \lim _{x\to a} f(x)\neq f(a)\implies g(\lim _{x\to a} f(x))\neq g(f(a))\implies \lim _{x\to a} g\circ f(x)\neq g(f(a))$ since $g$ is continuous.

So the set of all points of discontinuity of $g\circ f\subset $ set of all points of discontinuity of $f$.

Since set of all points of discontinuity of $f$ has measure $0$ so does the set of all points of discontinuity of $g\circ f$. So $g\circ f$ is Riemann-Integrable.

Is the proof correct?Please help