Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given: $f$ is Riemann integrable on $[a,b]$ and $f(x)\geq 0$ for all $x$.

Prove that if \begin{equation} \int_a^b f(x) dx=0 \end{equation} and $f$ is continuous, then $f(x)=0$ for all $x$. My idea: Find the lower sum, which must be $0$ since the infimum is $0$. Since f is integrable, the supremum is also $0$. Hence it must be that $f(x)=0$ for all $x$. However, I do not use the definition of continuity. Am I missing something?

share|improve this question
add comment

3 Answers

up vote 4 down vote accepted

The upper sum need not be $0$ just because the integral is $0$- it could happen that the upper sum converges to $0$. For example, if $f$ is $0$ everywhere except a point, then the integral is $0$ but $f$ is not zero.

So we need to use continuity.

Use the fact that if $f$ is continuous and non-zero at a point, then it's non zero in an interval. $f$ must achieve its minimum value in the interval, which is $>0$, so the integral is non zero- contradiction.

share|improve this answer
    
Is there a way to write this using the epsilon-delta definition of continuity, or another formal continuity theorem? –  kiwifruit Feb 9 at 1:01
add comment

Hint: Argue by counter argument:

If $f(x) > 0$ for some $x$ then what can you say by continuity?

share|improve this answer
add comment

Suppose that for a point $x$ in $[a,b]$, $f(x)>0$. So, by continuity, there is $epsilon>0$ s.t. for t in $[x-\epsilon,x+\epsilon]$ $f(t)>0$ and the integral must be grater than zero wich is a contradiction.

share|improve this answer
add comment

Your Answer

 
discard

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.