Let $X_{j,t}$ be a double-indexed sequence of random variables whose sample average over the $t$'s converges to the constant $\alpha_j$. Assume further that the $\alpha_j$'s are summable:
$$\frac{1}{N}\sum_{t =1}^{N}X_{j, t} \xrightarrow[]{p}\alpha_j, \text{ and } \sum_{j =1}^{\infty}\alpha_j = \alpha$$
Does this imply the following, as $N\rightarrow\infty$?
$$\frac{1}{N}\sum_{j =1}^{2^N}\sum_{t =1}^{N}X_{j, t} \xrightarrow[]{p}\alpha$$
My hunch says yes, though I'm not sure. Any hints?
The answer is no.
Let $S_{j,N}=\frac{1}{N}\sum_{t =1}^{N}X_{j, t}$ and $A_{j,N}=\{\omega:|S_{j,N}-\alpha_j|>\epsilon_j\}$. Then $\lim_{N\to\infty}P(A_{j,N})=0$. But $$\frac{1}{N}\sum_{j =1}^{2^N}\sum_{t =1}^{N}X_{j, t} =\sum_{j =1}^{2^N}S_{j,N}\xrightarrow[]{p}\alpha$$ means $$ \lim_{N\to\infty}P\left(\bigcup_{j=1}^{2^N}A_{j,N}\right)=\lim_{N\to\infty}\sum_{j=1}^{2^N}P(A_{j,N})=0 $$ for $A_{j,N}$ are disjoint. This is clearly not true if $P(A_{j,N})=\dfrac1{N}$.
However we can prove that if $$ \frac{1}{N}\sum_{t =1}^{N}X_{j, t}\xrightarrow[]{a.s}\alpha_j $$ Then $$ \frac{1}{N}\sum_{j =1}^{2^N}\sum_{t =1}^{N}X_{j, t} \xrightarrow[]{a.s}\alpha $$ Let $A_{j,N}=\{\omega:|S_{j,N}-\alpha_j|>\epsilon_j\}$. Since for any $N$, $P(A_{j,N})=0$ $$ P\left(\bigcup_{j=1}^{2^N}A_{j,N}\right)\leqslant \sum_{j=1}^{2^N}P(A_{j,N})=0 $$ Hence $$ \lim_{N\to\infty}P\left(\bigcup_{j=1}^{2^N}A_{j,N}\right)=0 $$