If I know that $g ∈ R[a,b]$, and $f$ is continuous, how would I show that disc$(f∘g)$ ⊆ disc $(g)$?
I know to use the Lebesgue theorem which states that a bounded function is integrable $⇔$ the discontinuous set of the function has measure zero. But I'm failing to connect the dots.
I was looking at other questions and someone suggested using the fact that if $g$ is continuous at $x$ and f is continuous at $g()$, then $f∘g$ is continuous at $$. Could I use this to show that disc$(f∘g)$ ⊆ disc $(g)$? If not, what is another approach I could take?
Thanks!
False. Let $g(x)=0$ for $x \in (0,1]$ and $g(0)=1$. Let $f(x) =1-x$. Then $g\circ f$ is discontinuous at $1$ even though $g$ is continuous at $1$.
Answer for the edited version: This has nothing to do with Riemann integrability. If $g$ is continuous at $x$ and $f$ is a continuous function then $f(g(x))$ is continuous at $x$; the contra-positive gives $disc(f\circ g) \subseteq disc (g)$.