$D_{\phi \circ f} \subset D_f$ where $D_g$ denote the set of discontinuities of $g$

Question

Let $f:[a,b] \rightarrow [c,d]$ be Riemann integrable and $\phi:[c,d] \rightarrow \Bbb{R}$ is continuous. Then $\phi \circ f$ is Riemann integrable

Ok! This is a standard one in analysis. But in the proof of this statement, the author says, "the discontinuity set of $\phi \circ f$ is contained in the discontinuity set of $f$".

My question is : How?

Let $\phi \circ f$ be discontinuous at $\alpha \in [a,b]$. Then $\lim_{x \rightarrow \alpha}(\phi \circ f)(x)$ dose't exist or not equal to $(\phi \circ f)(\alpha)$ . I dont know how to move further ?

Can I have a hint?

Added:

What is the sufficient condition on $\phi$ so that $\phi \circ f$ and $f$ have equal discontinuity sets for all Riemann integrable functions $f$ ?

Answer 1

You said

Let $\phi \circ f$ be discontinuous at $\alpha \in [a,b]$. Then
$\lim_{x \rightarrow \alpha}(\phi \circ f)(x)$ doesn't exist or not equal to $(\phi \circ f)(\alpha)$.

Since $\phi$ is continuous we have that

$$\lim_{x \rightarrow \alpha}(\phi \circ f)(x)=\phi(\lim_{x \rightarrow \alpha}(f)(x))$$ if both exist. So, if $\lim_{x \rightarrow \alpha}(\phi \circ f)(x)$ doesn't exist the reason is that $\lim_{x \rightarrow \alpha}(f)(x)$ doesn't exist. That is, $f$ is discontinuous at $\alpha.$ And if $$(\phi \circ f)(\alpha)\ne\lim_{x \rightarrow \alpha}(\phi \circ f)(x)=\phi(\lim_{x \rightarrow \alpha}(f)(x))$$ then it must be $$\lim_{x \rightarrow \alpha}(f)(x)\ne f(\alpha).$$ That is, $f$ is discontinuous at $\alpha.$