# Function (Riemann) integrable

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

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

• How do I know that $f$ is bounded? I believe this is needed. – serial May 13 at 15:37
• @serial I added how to show boundedness. – Arthur May 13 at 15:50
• For the first part you mean $[0,1/2]$ and $[1/2,1]$ right? – serial May 13 at 17:35
• @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. – Arthur May 13 at 17:43
• 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. – Arthur May 13 at 18:03