Composition of Functions - Riemann Integrable or Not?

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?

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Notice that if $g$ is Riemann-integrable, then the composite is too. But if $f$ and $g$ are only Riemann-integrable, you can't assure that the composite is. – Aloizio Macedo Mar 15 '15 at 18:26

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 $0<a<b<\infty$.
Are you asking what can we say about $g(x)$ if we know $f\circ g(x)$ is integrable? If so, we still can't say anything. Take $g$ as above, but let $f(x)=1$. – Clayton Dec 5 '12 at 3:32