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

Question

I know that there are solutions on the internet, but I tried it myself first and wanted to ask whether my approach is valid.

Prove that the finiteness of $E[X]$ is equivalent to the finiteness of $E[|X|]$. Note that the finiteness of $E[X]$ means that $|E[X]| < \infty$, which implies proving $|E[X]| < \infty \iff E[|X|] < \infty$. Use the fact that $X^+ = X$ for $X \geq 0$ and $X^- = -X$ for $X < 0$.

I tried the following:

$$ X = X^+ - X^- \quad \text{and} \quad |X| = X^+ + X^- $$

$$ |E[X]| = |E[X^+] - E[X^-]| \leq |E[X^+]| + |E[X^-]| $$

$$ E[|X|] = E[X^+ + X^-] = E[X^+] + E[X^-] = |E[X^+] + E[X^-]| \leq |E[X^+]| + |E[X^-]| $$

Using the triangle inequality twice. Then I would argue $|E[X]| < \infty \iff E[|X|] < \infty$ follow, due to the fact that both have the same upper bound. But I am unsure whether the following step was valid: $E[X^+] + E[X^-] = |E[X^+] + E[X^-]|$. My thought was that both $X^+$ and $X^-$ must be positive by definition in the task.

Answer 1

To address your argument:

But I am unsure whether the following step was valid: $E[X^+] + E[X^-] = |E[X^+] + E[X^-]|$. My thought was that both $X^+$ and $X^-$ must be positive by definition in the task.

This is fine: both terms are nonnegative, so their sum is, so the absolute value doesn't change anything.

But just before that:

Then I would argue $|E[X]| < \infty \iff E[|X|] < \infty$ follow, due to the fact that both have the same upper bound.

This doesn't work. If the upper bound is infinite, then one of the quantities can be finite and the other infinite without violating anything so far established.

Answer 2

Turning to a generalized setting, let $(\Omega, \boldsymbol{\mathfrak A}, \mu) $ be a measure space and let $f$ be an extended real-valued $\boldsymbol{\mathfrak A}$–measurable function on $D\in \boldsymbol{\mathfrak A}.$ Then

Result:$f$ is $\mu$–integrable on $D$ if and only if $|f|$ is.

Note if $f^+:=\max(f,0),~f^-:=-\min(f,0),$ $$\int_D |f|~\mathrm d\mu=\int_D (f^++f^-)~\mathrm d\mu=\int_D f^+~\mathrm d\mu+\int_D f^-~\mathrm d\mu<\infty,$$ which shows $|f|$ is $\mu$–integrable on $D.$

Conversely, assume $|f|$ is $\mu$–integrable on $D.$ Since $0\leq f^+, ~f^-\leq |f|, $ their respective integrals $\int_D f^\pm ~\mathrm d\mu\leq \int_D |f|~\mathrm d\mu<\infty, $ that is, $f$ is $\mu$–integrable on $D$ by its definition.

$\blacksquare$

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Reference:

Real Analysis: Theory of Measure and Integration, J. Yeh, World Scientific, $2014, $ pp. $177-178.$

Answer 3

In terms of Riemann integrals:

$$I(a) = \int_{-a}^0 x f(x) dx \\ J(b) = \int_{0}^b x f(x) dx $$

$$ \begin{array}{rccr} E[X] &=& \lim_{(a,b) \to (\infty,\infty)}& I(a)+J(b) \\ E[|X|] &=& \lim_{(a,b) \to (\infty,\infty)} &-I(a)+J(b) \end{array}$$

These double limits are finite if and only if the single limits $ \lim_{a \to \infty} I(a)$ and $ \lim_{b \to \infty} J(b)$ are finite. (And as a consequence $E[X]$ is finite iff $E[|X|]$ is finite)

Note that for the limit $E[X]$ to be finite, it implies that the single terms $I(a)$ and $J(b)$ converge when $a$ and $b$ go to infinity. This is because their sum is bounded for any sufficiently large values.

That is: If $E[X]$ is finite, then for every $\epsilon>0$ there is value $n$ such that for every sufficiently large values, $a>n$ and $b>n$ we have $E[X]-\epsilon \leq I(a) + J(b) \leq E[X]+ \epsilon$.

But such bounds are not possible if $I(a)$ and $J(b)$ diverge (because then we can chooses some $a > n$ and $ b > n$ that make $I(a) + J(b)$ outside the bounds). Therefore $I(a)$ and $J(b)$ must converge if $E[X]$ is finite.