0
$\begingroup$

I know the composition of Riemann integrable functions is not necessarily Riemann integrable. But I am not finding any argument how to conclude this for self composition, or how to find a counterexample. $f$ is Riemann integrable on $[a,b]$ means that the discontinuity set of $f$ is of measure zero. But the discontinuity of the composition is a larger set, so we cannot conclude anything.

Similarly for the self composition of measurable functions: is it measurable or not? I am unable to find a counterexample.

$\endgroup$
6
  • $\begingroup$ Your second paragraph is a very different question and should be asked in a separate post (though it may also have an answer on this site already). You should also clarify what you mean by "measurable": Borel measurable? Lebesgue measurable? On what domain? $\endgroup$ May 14 at 5:11
  • 1
    $\begingroup$ If you know two Riemann integrable functions $g,h$ whose composition $h \circ g$ is not Riemann integrable, try creating a piecewise function of horizontally and vertically shifted versions of it, so that $f \circ f$ looks like $h \circ h$ up to shifts. $\endgroup$ May 14 at 5:13
  • $\begingroup$ Domain is same [a,b] and Lebesgue measureable $\endgroup$
    – user1043248
    May 14 at 5:34
  • $\begingroup$ I mean $h \circ g$ in previous comment. $\endgroup$ May 14 at 5:35
  • $\begingroup$ For Riemann Integrable answer is no I guess, I understand just not able to find the precise counterexample. $\endgroup$
    – user1043248
    May 14 at 5:38

1 Answer 1

6
$\begingroup$

Suppose you have Riemann integrable functions $g,h : [0,1] \to [0,1]$ such that $h \circ g$ is not Riemann integrable. Define $f : [0,3] \to \mathbb{R}$ by $$f(x) = \begin{cases} g(x) + 2, & 0 \le x \le 1 \\ 123.456 & 1 < x < 2 \\ h(x-2), & 2 \le x \le 3. \end{cases}$$ Then $f \circ f = h \circ g$ on $[0,1]$. I leave it to you to verify that $f$ is again Riemann integrable, that $f \circ f$ is not, and to make adjustments if you don't like the domain and range I chose.

$\endgroup$
1
  • $\begingroup$ For me, this is a great answer (+2). $\endgroup$ May 14 at 8:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy