This seems to be really basic and might be a case of begging the question here, but I can't seem to prove the that the right hand side implies the left hand side. Let $J$ be the set of all $j$ and $K(j)$ be the set of all $k$ that are valid for each $j\in J$. $K, J(k)$ are defined likewise.
So suppose I was working out a summation and I want to sum first on $j$ instead of $k$. Then by working out the set $K$ of valid $k$ from the LHS I arrive at the RHS. I also work out the valid $K(j)$ from the conditions of the LHS. So I can say that $J, K(j)\Rightarrow K,J(k)$.
Intuitively, this suggests the inverse is clearly true, but can it be proven rigorously?
$\sum_{j\in J}\sum_{k\in K(j)}f(j,k) = \sum_{k\in K}\sum_{j\in J(k)}f(j,k)$
All you have to prove is $\left\{(j,k)\,|\, j\in J\text{ and }k\in K(j)\right\} = \left\{(j,k)\,|\, k\in K\text{ and }j\in J(k)\right\}$.