Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 2 down vote accepted

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 $0<a<b<\infty$.

share|cite|improve this answer
What if we know f(g(x)) is Riemann integrable? – GradStudent Dec 5 '12 at 3:27
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
Yes, that is what I was asking. I like your counterexample - if I limit the domain to [0,1], it is easy to see that g(x) is not Riemann integrable, but that f(g(x)) is. Now just to prove it! Thank you. – GradStudent Dec 5 '12 at 13:47

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.