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specifically for an improper integral, but I'm also wondering about for definite integrals.

I'd guess that it's true, but I feel like there must exist different functions that integrate to the same value of the same range (flipping something, for example. That is maybe $f(s) = g(-s)$ for all $s$?)

Also, I am interested more in lebesgue integration than Riemann, but also wonder about both.

Thanks.

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  • $\begingroup$ Does $f(a)+f(b)=0$ imply $f=0$? $\endgroup$ – A.S. Feb 1 '16 at 22:46
  • $\begingroup$ $\int_a^b f(x) dx = \int_a^b g(x) dx$ for every $a,b$ implies $f(x) = g(x)$ almost everywhere (that $\int_C |f(x)|dx = \int_C |g(x)|dx = 0$ with $ C$ being the set where they are different) $\endgroup$ – reuns Feb 1 '16 at 23:10
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This is false

There are myriad counter examples, but an easy one, consider a function $f$ that is zero outside some interval $[a,b]$ and nonnegative on that interval. Define a new function $g(x)=f(x-10)$. These functions clearly have the same integral, because the shape above the x-axis is the same, but the interval on which the shape appears is different.

Your idea produces a similar example: take some function not symmetric across the $y$ axis and look at $f(-x)$. If the integral of $f(x)$ across $\mathbb{R}$ is finite, then the integral of $f(-x)$ is finite and in fact the same value. If you're integrating on a subset of $\mathbb{R}$ this might not work. For example, $f(x)=x$ on $[-1,3]$ has a different integral than $g(x)=f(-x)=-x$ on $[-1,3]$. The first is $4$ but the second is $-5$.

Another counter example that has a different underlying reason is to consider some function $f$ and to take one point of $f$ and make a jump discontinuity at that point. It turns out that this doesn't change the integral at all, but obviously changes the function.

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