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.


  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.