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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?

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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.

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Is there a way to write this using the epsilon-delta definition of continuity, or another formal continuity theorem? – kiwifruit Feb 9 '14 at 1:01

Hint: Argue by counter argument:

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

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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.

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