Consider a set $E=\bigcup_{j=1}^{\infty}E_{j}$.
How can I re-express $E$ in such a way that all of the contributions to $E_{n}$ are covered only once. Presumably, such a re-expression of $E$ will be some kind of countable union of disjoint sets.
The end-goal is to answer the following problem: to show that if $m'$ is any map from the Lebesgue measurable sets of $\mathbb{R}^{d}$ to $[0,\infty]$ which obeys countable additivity and $m'(\emptyset)=0$, then $m'$ also obeys monotonicity and sub-additivity.
Showing that $m'$ obeys monotonicity is easy. Sub-additivity seems easy at first glance, until you actually have to prove it. Perhaps there are better ways to do it, but my strategy is to write $E=\bigcup_{j=1}^{\infty}E'_{j}$ where the $E'_{j}$ are all disjoint "components" of $E$. Then applying additivity I could expand out $m\left(\bigcup_{j=1}^{\infty}E'_{j}\right)$ using additivity (I assume the $E'$ will be sets formed by multiple set operations of the original $E_{j}$, so applying additivity could be potentially tricky). Once this expression is obtained, I presume it will be easy to compare it to $\sum_{j=1}^{\infty}m'(E_{j})$ and show that it is smaller, hence proving sub-additivity.
UPDATE
From the answer below...
\begin{align*} m'(E) &=m\left(\bigcup\limits_{j=1}^{\infty}E_{j}\right)\\ &=m\left(\bigcup\limits_{j=1}^{\infty}\left(E_{j}-\bigcup\limits_{k=1}^{j-1}E_{k}\right)\right)&\text{(remove duplicate inersections)}\\ &=\sum\limits_{j=1}^{\infty}m\left(E_{j}-\bigcup\limits_{k=1}^{j-1}E_{k}\right)&\text{(apply assumed additivity)}\\ &\leq\sum\limits_{j=1}^{\infty}m(E_{j}) &\text{(apply aleady proved monotonicity)} \end{align*} so that $m'$ is also sub-additive as required.