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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.

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up vote 1 down vote accepted

You define $$ F_1=E_1,\ F_2=E_2\setminus E_1,\ \ldots,\ F_{n+1}=E_{n+1}\setminus\bigcup_{j=1}^nE_j,\ \ldots $$

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I figured it'd be something trivial. I had originally something similar: $$E=\bigcup_{j=1}^{\infty}E_{j}=\bigcup_{j=1}^{\infty}\left(E_{j}-\bigcup\limits‌​_{k=1}^{j-1}E_{k}\cap E_{j}\right).$$ I suppose if I had just thought for a little longer, I would've obtained the answer by replacing the intersection with just $E_{k}$. Instead, I didn't think, and I spent (literally) 2 hours messing with set-theoretic manipulations! It was terrible. –  Taylor Martin Oct 22 '12 at 8:28

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