If the improper integral of f(x) from 0 to infinity exists, then theres xn such that f(xn) goes to 0.

~~i don't know that this is actually true. how about 1 for 1/(2n)<x<1/(2n+1) and -1 otherwise. i think this would converge...~~this isn't even defined on all of R, dummie

edit: im dumb, look at (-1)\^floor(x^(2)). the oscillation makes it converge by the alternating series test.
so after discussing with a friend, we have come to the answer that yes for lebesgue integrals, and no for riemann integrals.
f(x) = 1/x^2 is a counterexample

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