2
$\begingroup$

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?

  • 0
    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.2015-03-15

1 Answers 1

2

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 Then $f\circ g(x)=g(x),$ which clearly has an unbounded integral where $0.

  • 0
    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.2012-12-05