Interchanging the order of a double infinite sum I'm stuck at a proof of Wald's first equation about interchanging the order of a double infinite sum: 

Suppose $X_n \ge 0$ and $1_{\{\cdot \}}$ be indicator function.
  $$ \sum_{n=1}^\infty \sum_{m=1}^n EX_m 1_{\{N=n\}} = \text{why?} = \sum_{m=1}^\infty \sum_{n=m}^\infty EX_m 1_{\{N=n\}} $$

I know that Fubini's theorem works here, since $X_n \ge 0$, but I don't see how to interchange the summation order with assigning proper indexes.
I consulted two books + wiki, but still don't get a satisfactory answer, they just simply said that we can rearrange the sum in this way. Is there a simple rule for such situations that I can follow?
 A: Note that $$\sum_{k\geqslant 1}\sum_{n\leqslant k}=\sum_{k\geqslant 1}\sum_{n\geqslant 1}[n\leqslant k]=\sum_{n\geqslant 1}\sum_{k\geqslant 1}[n\leqslant k]=\sum_{n\geqslant 1}\sum_{k\geqslant 1}[k\geqslant n]=\sum_{n\geqslant 1}\sum_{k\geqslant n}$$
This is an example of the usefulness of the Iverson bracket.
A: Visualizing things makes the problem much simpler and easier to understand in such cases. Consider the value pairs $(n,m)$ is allowed to take on a 2D grid:
$$
\begin{array}{ccccc}
(1,1) & & & &\\
(2,1) & (2,2) & & & \\
(3,1) & (3,2) & (3,3) &  & \\
(4,1) & (4,2) & (4,3) & (4,4) & \\
\vdots &\vdots&\vdots&\vdots&\ddots
\end{array}
$$
where the rows correspond to different values of $n$, and the columns correspond to different values of $m$. Note that for a fixed $n$ (row), $m$ (column) can take values from $1$ up to $n$, hence the value pairs on this triangular grid indeed correspond to the values that can be taken by the $(n,m)$ pairs on the sum on the left-hand side.
Now, we can look at this triangular grid in a "horizontal-first" manner. Note that for a fixed value of $m$ (column), $n$ (row) can take values from $m$ up to infinity. If you look at the double sum on the right-hand side of your equation, you will see that it exactly implements this order. 
Therefore, the sums on the two sides are equal because they simply sum the argument over the same set of value pairs: the ones on this triangular grid (and, of course, because of Fubini's theorem, as you suggest).
