The question is:
Given $E[|X|]\le M$, for $M$ in real.
Prove $E[X]<\infty$ and $E[X]\ge -M$.
To be honest, I have no idea how to start, what I tried is by using definition, but got stuck:
$E[|X|] = E[X^+]-E[X^-]$.
The minus comes because of absolute value, such that: $$\left|X\right|=\begin{cases}X&X>0\\ -X&X\le 0\end{cases}$$
so:
$$E\left[\left|X\right|\right]=\int _0^{\infty }\left(1-F_X\left(t\right)\right)dt-\:\int _0^{\infty }\:F_X\left(-t\right)dt$$
And then, by using $t=-k$ and $dt=-dk$ we receive:
$$E\left[\left|X\right|\right]=\int _0^{\infty \:}\left(1-F_X\left(t\right)\right)-\int _{-\infty }^0\:F_X\left(t\right)dt$$
And from here sadly, I am stuck...
Thought maybe I can say that if the sum of integrals are bounded by $M$, as given, so of course the partial are also bounded and thus $E[X]<\infty$, I dont know if it is true, but still, I have trouble with the other one - $E[X]\ge -M$
Any tip on how to continue please?
Since $X<|X|$, one has $E X \leq E |X|<\infty$. Moreover, $$E|X| = E X^++EX^- \leq M$$ , implying $E X^- \leq M$. Hence $$ E X = EX^+-EX^- \geq EX^+-M \geq- M$$ We conclude the stronger statement that $E X\geq -M$ with equality if and only if $X\leq 0$ almost surely. Here, $X^+ = \max(X,0)$ and $X^- = \max(-X,0)$