I have seen many different cases regarding composition of different types of functions, and whether or not the composition is Riemann integrable. I am concerned about the composition of f(g(x)), where f is continuous on [a,b]. I feel intuitively that f(g(x)) cannot be proven to be Riemann integrable, but I am having trouble coming up with a counterexample.
I know that since f is continuous on [a,b], that it is bounded on [a,b], and that it is Riemann integrable on [a,b].
So assuming we have no other information about g, can we make any assumptions?
Your intuition is mostly correct. Consider the following counterexample: Let $f(x)=x$, which is clearly continuous on $[a,b]$ and define $$g(x)=\begin{cases}\frac{1}{x-b}&:a\leq x<b\\ 0&:x=b\end{cases}.$$ Then $$f\circ g(x)=g(x),$$ which clearly has an unbounded integral where $0<a<b<\infty$.