A continuous function of a integrable function is integrable

Question

If $f: A \to \mathbb{R}$, defined in block $A$, is integrable and $g:[a,b] \to \mathbb{R}$ is continuous in a interval containing $f(A)$, then $g \circ f: A \to \mathbb{R}$ is integrable.

My idea is to use the Lebesgue Theorem, i.e, to prove that the set $D_{g \circ f}$ of all discontinuities points of $g \circ f$ is a null set. But for this, I need to show that $g \circ f$ is bounded in $A$, right?

For the first point, $D_{f}$ is a null set, since $f$ is integrable and, obviously, $D_{g}$ is a null set. How I show that $D_{g \circ f}$ is anull set?

Answer 1

You should show that the set of discontinuities of $g \circ f$ is a null-set.

First show that $D({g\circ f}) \subseteq D(f)$ where $D$ denotes the set of discontinuities. Hence, $D(f)^c$ (the complement) denotes the set of points where $f$ is continuous.

Hover your mouse over the yellow area below to see why $D({g\circ f}) \subseteq D(f)$ is true.

Basically, if $f$ is continuous at some $x$, then so is $g \circ f$ because it's the composition of continuous functions. Hence, $D(f)^c\subseteq D({g\circ f})^c$ which is equivalent to $D({g\circ f}) \subseteq D(f)$.

Now note that a subset of a null-set is a null-set itself. Therefore, $D({g\circ f})$ being a subset of $D(f)$ is a null-set and $g\circ f$ is integrable by the Lebesgue theorem.

Answer 2

The function $g$ is bounded and therefore $g\circ f$ is bounded too.

And the set of points at which $g\circ f$ is discontinuous is a subset of the set of points at which $f$ is discontinuous (since $g$ is continuous).