Suppose ${X_n}$ are iid and $E(X_n^+)<\infty$ and $E(X_n{^-})=\infty$. Show that if $S_n := \sum_{j=1}^nX_n$ then $\frac{S_n}{n} \rightarrow -\infty$ almost surely.
A hint in my book says to use truncation so here is what I tried doing.
Since $E(X_n^+)<\infty$ then $E(X_n^+)=M$ where $M$ is some real number. Then $E(X_n^+1_{[|X_n|\le N]})=M$ for any $N\ge M$. Then since $E(X_n{^-})=\infty$ there is going to exist some $N_0 \ge M$ such that for all $N \ge N_o \;$ we have that $E(X_n{^-}1_{[|X_n|\le N]})=N$
So for all $N \ge N_o$ we have that $\begin{align} \\ E(X_n{}1_{[|X_n|\le N]}) &= E(X_n{^+}1_{[|X_n|\le N]}) -E(X_n{^+}1_{[|X_n|\le N]}) \\ &= M - N \end{align}$
So be the strong law of large numbers $\dfrac{\sum_{j=1}^nX_n{}1_{[|X_n|\le N]}}{n} \rightarrow M-N$ almost surely. So as $N$ goes to infinity we get that $\dfrac{\sum_{j=1}^nX_n}{n} \rightarrow -\infty$ almost surely