$\Phi : \mathbb R \rightarrow \mathbb R$ , $f$ is R.I. $\rightarrow \Phi \circ f$ is R.I.

Question

$\Phi : \mathbb R \rightarrow \mathbb R$ is continuous and $\Phi(0)=0$. Suppose that $f$ is Riemann Integrable. Show that $\Phi \circ f$ is also R.I.

I assumed $f$ has bounded support $[a,b]$. Thus $f:[a,b]\rightarrow [m,M]$

Since $\Phi$ is continuous on the whole real line, it's also continuous on $[m,M]$, meaning it's also R.I.

Is this correct? Why do we need the $\Phi(0)=0$ condition?

Answer 1

I would like to expand a little on Adelafif's correct but IMO overly concise answer. For starters we have Baby Rudin 11.33: "Suppose $f$ is bounded on $[a,b]$. Then $f\in{\mathcal R}$ [the Riemann-integrable functions] on $[a,b]$ if and only if $f$ is continuous almost everywhere on $[a,b]$." Here "almost everywhere" means "except on a set of Lebesgue measure zero."

OK, suppose now that $A$ is that set of Lebesgue measure zero where $f$ is not continuous. For $x\in[a,b]\setminus A$, we have $f$ continuous at $x$, and by hypothesis $\Phi$ is continuous at $f(x)$ (Baby Rudin 4.7), so $\Phi \circ f$ is continuous at $x$ too. Thus $\Phi \circ f$ is continuous on $[a,b]$ except on the same set of measure zero, namely $A$, and is thus Riemann integrable.

Answer 2

continuous of continuous is continuous. A function is R-integrable iff it is bounded and a.e cont. The result follows.