Suppose we have the triangular array $\{\{X_{in},i=1,\ldots,n\},n=1,2,\ldots\}$:
$$\begin{array}{ccccc}
X_{11}&&&&\\
X_{12}&X_{22}&&&\\
X_{13}&X_{23}&X_{33}&&\\
X_{14}&X_{24}&X_{34}&X_{44}&\\
\vdots&\vdots&\vdots&\vdots&\ddots\\
\end{array}$$
Let the elements of each row be i.i.d., furthermore, let $E[X_{in}]=\mu<\infty$ in this array.
Now let $S_n=\frac{1}{n}\sum_{i=1}^n X_{in}$ be the average of each row.
I am confused as to why almost sure convergence results don't apply to $S_n$, i.e., why can't one claim that $S_n\xrightarrow{\text{a.s.}}\mu$? Why doesn't the strong law of large numbers kick in here?
(I was going to put this in a comment where this came up in one of my other questions, however, I thought that this warrants a separate question.)
Best Answer
Let $X_i$ be an i.i.d. sequence with $\mathbb{P}(X_i=1)=\mathbb{P}(X_i=-1)=1/2$, and set $X_{in}=n X_i$. These random variables satisfy your conditions with $\mu=0$, but ${1\over n}\sum_{i=1}^n X_{in}=\sum_{i=1}^n X_i$ does not converge almost surely as $n\to \infty$.