Let

f be a bounded integrable non-negative function continous on [a, b].

Prove:

integral from a to b f(x) dx = 0 implies f(x) = 0 for every x

I believe I understand why it is true, I'm just not sure how to write the formal proof.

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- February 23rd 2010, 08:17 PMamm345Proof involving integrals equal to 0.
Let

f be a bounded integrable non-negative function continous on [a, b].

Prove:

integral from a to b f(x) dx = 0 implies f(x) = 0 for every x

I believe I understand why it is true, I'm just not sure how to write the formal proof.

- February 23rd 2010, 08:30 PMDrexel28
What have you tried? Here is a weird approach.

Suppose that then there exists some such and since this set is open there exists some open ball and by continuity there exists some such that . In other words, there exists some open ball around the point where which is entirely positive. Take the concentric closed ball . This is a closed and bounded interval and thus, by the Heine-Borel theorem, compact. Thus, since is continuous we know that assumes a minimum on . In particular, . Thus, . This of course is a contradiction. - February 23rd 2010, 08:33 PMamm345
That was sort of my original approach, but when I checked with my instructer he said we could not use that theorem and I can't think of anything else.

- February 23rd 2010, 08:39 PMDrexel28
- February 23rd 2010, 08:40 PMamm345
Oh sorry, the Heine-Borel theorem.

- February 23rd 2010, 08:42 PMDrexel28
- February 23rd 2010, 08:44 PMamm345
I have no idea.. I couldn't think of anything else it is kinda ridiculous!

- February 23rd 2010, 08:48 PMDrexel28
- February 23rd 2010, 08:57 PMamm345
We started integration last week and we've mostly just covered the riemann integral, basic principles, the fundamental theorem and continuity.

- February 23rd 2010, 09:03 PMDrexel28
- February 24th 2010, 12:48 AMMoo
Hello,

Why can't we just say that (without HB theorem) :

if there exists such that , then by definition of the continuity,

In particular, if we let ,

So there exists such that ,

Hence

So

which is a contradiction.

So m can't exist, and hence