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$$\int_{0}^{1} |f(t)|dt=0$$

Does this equation imply that $f(t)=0$ for every $t \in [0,1]$? I need a proof of whether the answer is yes or no. I couldn't prove this; I started thinking, if it were zero then the integral is definitely zero, but what if there is another function (other than zero) which also has its integral equal to zero?

ki3i
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HHH
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    $f$ could be non-zero at a single point (or more). – David Mitra Mar 28 '15 at 15:20
  • where is the proof? – HHH Mar 28 '15 at 15:22
  • What have you tried? Where did you encounter the question? On this site, we look for this kind of context, not just for a statement of a problem. You can edit your post to improve it. – Carl Mummert Mar 28 '15 at 15:25
  • i edited it now see what i have tried or thought – HHH Mar 28 '15 at 15:28
  • @CarlMummert i encountered this question while proving that this integral mentioned above is a norm on E which is a vector space of real valued continous functions defined on [0,1] and i encountered this while proving that: N(x)=0 equivalent x=0 – HHH Mar 28 '15 at 15:31

1 Answers1

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This is not true. Consider the function $$f(x) = \begin{cases} 1 & x = \frac{1}{2} \\ 0 & x \in [0, \frac{1}{2}) \cup (\frac{1}{2}, 1] \end{cases}$$

Then $\int \limits_{0}^{1} f(x) \,dx = 0$ (can you show why this is?), but $f(x)$ is not $0$ everywhere.

But, we do know that $f(x) = 0$ almost everywhere. Do you know what the concept of "almost everywhere" is? If so, I can give you a proof that $f$ will be equal to $0$ almost everywhere.

layman
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