Convergence Using Law of large numbers $Z_1, Z_2, Z_3,...$ are independent and identically distributed R>V.s s.t. $E(Z_i)^- < \infty$ and $E(Z_i)^+ = \infty$. Prove that $$\frac {Z_1+Z_2+Z_3+\cdots+Z_n} n \to \infty$$ almost surely.
What does $E(Z_i)^+$ $E(Z_i)^-$ mean? I believe it is integrating fromnegative infinity to zero and zero to positive infinity for - and x resepctively. But LLN wont apply here since expectation doesnt exist? like E|Zi|=$\infty$.
p.s. another post i made is similar but diferent it is $E(Z)_i^+$ in the other which stands for max and min for the minus sign.
 A: If $x$ is a real number, $x^+$ denotes the positive part of $x$, that is, $x^+=\max\{x,0\}$ and $x^-=\max\{x,0\}-x$ so that $x=x^+-x^-$. Write $Z_i=Z_i^+-Z_i^-$. An application of the strong law of large number to $\left(Z_i^-\right)_{i\geqslant 1}$ shows that $n^{-1}\sum_{i=1}^nZ_i^-\to \mathbb E\left[Z_1^-\right] $ almost surely. For the part involving $Z_i^+$, let $Y_{i,M}:=Z_i^+\mathbf 1\left\{Z_i^+\leqslant M\right\}$. Then for all $n$ and $M$, 
$$
\frac 1n\sum_{i=1}^nZ_i^+\geqslant\frac 1n\sum_{i=1}^n Y_{i,M}
$$
hence by the strong law of large number applied to the i.i.d. sequence $\left(Y_{i,M}\right)_{i\geqslant 1}$,
$$
\liminf_{n\to +\infty}\frac 1n\sum_{i=1}^nZ_i^+\geqslant\liminf_{n\to +\infty}\frac 1n\sum_{i=1}^n Y_{i,M}=\mathbb E\left[Y_{1,M}\right],
$$
and by assumption, $\mathbb E\left[Y_{1,M}\right]\to +\infty$ as $M$ goes to infinity.
A: The positive part of a random variable $X$ is the random variable $X^+ = \begin{cases} 0 & \text{if } X<0, \\ X & \text{if } X\ge 0. \end{cases}$
The negative part $X^-$ is defined similarly. If $F(x) = \Pr(X\le x)$ then you have $\displaystyle\operatorname E(X^+) = \int_0^\infty x\, dF(x)$ and $\displaystyle\operatorname E(X^-) = \int_{-\infty}^0 x \, dF(x).$ If the distribution is one that has a density function $f$ then these are the same as $\displaystyle \int_a^b xf(x)\,dx$ for $(a,b) = (0,\infty) \text{ or } (-\infty,0)$ respectively.
The law of large numbers cannot be used in the usual way if at all. This calls for a different sort of proof.
