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I need to know whether, under some assumptions about the functions behavior , and maybe some values of integration limits $a,b$, the following relation holds:

$\int_a^bdx \ f(x) = \int_a^b dx \ g(x) \rightarrow f(x) = g(x)$

I think there was a theorem about it, not sure if when $(a=0, \ b=\infty)$ or maybe it was when it holds for every $a$ and $b$ values. I was looking on the internet but I could not find it

I would like to know whether is some possibility to write this

$\int_0^\infty dx \ f(x) = \int_0^\infty dx \ g(x) \rightarrow \int_0^\infty dx \ f(x)h(x) = \int_0^\infty dx \ g(x)h(x) $

Thanks in advance!

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up vote 0 down vote accepted

It is not true that:

$$\int_0^\infty dx \ f(x) = \int_0^\infty dx \ g(x) \Rightarrow \int_0^\infty dx \ f(x)h(x) = \int_0^\infty dx \ g(x)h(x)$$

If $f$ only does something on $[0..1]$ and $g$ does the same only a bit later on $[1..2]$ and nothing elsewhere (i.e. is $0$), it isn’t hard to find $h$ which destroys the latter identity.

Your first question is equivalent to whether this is true: $$\int_a^b dx f(x) = 0 \Rightarrow f = 0$$

Depending on your definition on integral, this is true (almost everywhere) for $f ≥ 0$ (almost everywhere). If $f ≥ 0$ isn’t given, it’s easy to find counterexamples.

If you asssume $f$ to be continuous, it easily follows from the mean value theorem for integration.

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Oh, I see your point, thanks a lot! – user2029879 Feb 19 '13 at 15:54

This is true if the functions are continuous and if $f\geq g$ on $[a,b]$ (or of course if $f\leq g$).

This is still true if one or two bounds are infinite, provided the integrals converge.

The theorem is: if $\int_a^b h=0$ for $h$ continuous nonnegative, then $h=0$.

The answer to your second question is no.

Take $f=1_{[0,1]}$ and $g=1_{[1,2]}$. Then consider $h=f$. Continuous counterexamples can be constructed in the same spirit.

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I think he means whether the latter identity is true for general $h$. – k.stm Feb 19 '13 at 15:50
Thanks a lot! Talking about the second question, it holds for every function $h(x)$?, I don't see it obvious ... just because it works for $h(x)=1$ ... – user2029879 Feb 19 '13 at 15:50
@K.Stm Yes, I misunderstood the second question. I'll edit. – 1015 Feb 19 '13 at 15:52
@user2029879 You're welcome. Sorry I first misunderstood your second question. I have edited with a counterexample. – 1015 Feb 19 '13 at 15:56
+1 prime counterexample. – Babak S. Feb 19 '13 at 16:05

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