Let $f: [0,1]\to\mathbb{R}$ be a continuous function. How to prove that $f$ is (Riemann) integrable?

  • $\begingroup$ Since [0,1] is a compact set then any continuous function $f$ in this set attain a minimum and maximum... $\endgroup$ – user209663 May 13 at 15:38
  • $\begingroup$ @user209663 I do not see the relation with my question, since in my case $f$ need not be continuous on $[0,1]$. $\endgroup$ – serial May 13 at 15:39
  • $\begingroup$ You are right. I misread the question. $\endgroup$ – user209663 May 13 at 15:44

You can't conclude that $f$ is continuous on $[0,1]$, because it doesn't have to be: $$ f(x) = \cases{1 & if $x \in \{0, 1\}$\\0& otherwise} $$ is convex, and not continuous.

On the other hand, you should be able to show by using the definition of Riemann integrability that (potential) discontinuity at two points doesn't stop a function from being integrable.

Edit: Boundedness

The function value on $(0,1/2)$ cannot at any point go below $\min(f(1/2),2f(1/2) - f(1))$, and the function value on $(1/2, 1)$ cannot at any point go below $\min(2f(1/2) - f(0))$, so the function is bounded below.

If the function ever has a value greater than $\max(f(1),f(2))$, then that violates convexity, so the function must be bounded above.

  • 1
    $\begingroup$ How do I know that $f$ is bounded? I believe this is needed. $\endgroup$ – serial May 13 at 15:37
  • $\begingroup$ @serial I added how to show boundedness. $\endgroup$ – Arthur May 13 at 15:50
  • $\begingroup$ For the first part you mean $[0,1/2]$ and $[1/2,1]$ right? $\endgroup$ – serial May 13 at 17:35
  • $\begingroup$ @serial Not really, but I did mess up some details. Also, the bounded above can be done much easier. Give me a few minutes, and I'll fix. $\endgroup$ – Arthur May 13 at 17:43
  • $\begingroup$ What I mean by "not really" is that I usually consider convexity to be about the function value at three distinct points. In this case, one of the points is $\frac12$, one of the points is either $0$ it $1$, and the third point is between them. $\endgroup$ – Arthur May 13 at 18:03

Your Answer

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

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