Timeline for Prove that the finiteness of $E[X]$ is equivalent to the finiteness of $E[|X|]$

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Oct 6 at 19:17	comment	added	Sextus Empiricus		@whuber it is a bit contrived, but we could create a Cauchy variable $X$ from a uniform variable $Y$ by mapping it piecewise alternatingly to negative and positive pieces of a Cauchy such that the computation of the expectation in the indirect integral converges, eg we can make $$E[X] = \lim_{a \to 1} \int_0^a g(y) f_Y(y) dy = 0$$ what it requires is that the pieces become increasingly small such that the increments of the integral decrease and the limit exists.
Oct 6 at 18:13	comment	added	Sextus Empiricus		@whuber typically we will make computations like that. But if the expectation is defined as the improper Riemann integral $E[X] = \int_{-\infty}{\infty} x f(x) dx$ then it won't work all the time.
Oct 6 at 16:41	comment	added	whuber♦		Ordinarily, we wouldn't be so restrictive. As an example, suppose $Y$ is a non-negative variable with density $f_Y(y)=1/(1+y)^2$ and let $X = y\sin(y).$ Wouldn't you express the expectation as $$E[X]=\int_0^\infty y\sin(y)f_Y(y)\,\mathrm dy = \int_0^\infty \frac{y\sin(y)}{(1+y)^2}\mathrm dy\ \text{?}$$ It is interesting that this has a finite Riemann integral but is not Lebesgue integrable.
Oct 6 at 13:08	history	edited	Sextus Empiricus	CC BY-SA 4.0	edited body
Oct 6 at 13:02	history	answered	Sextus Empiricus	CC BY-SA 4.0