# Countable mutually disjoint measurable sets

• Oct 19th 2011, 12:01 PM
Countable mutually disjoint measurable sets
I have an end of section problem from my analysis book that I would like a hint on, it reads: Prove that if $E_1,E_2,E_3,...$are mutually disjoint measurable sets on the line then for any set A, $m^* \left(\displaystyle\bigcup_{k=1}^{\infty} (A \cap E_k) \right) = \displaystyle\sum_{k=1}^{\infty} m^*(A \cap E_k)$
I have seen the proof for a finite number, n, of measurable sets where induction is used on n, could the same method be used and extended to the countably infinite case? Or should I be trying to show that the intersections of the E_k sets with A are measurable (regardless if A is measurable) and use the proposition stating the union of a countable collection of measurable sets is measurable then I can say the outer measure is equal to the lebesgue measure? I'm just not sure which direction I should take this, any suggestions?
• Oct 19th 2011, 08:53 PM
Re: Countable mutually disjoint measurable sets
If for some $i:\,m^*E_i = \infty$ then it is trivial so we consider only sets of finite outer measure. Note that
$m^*\left(\displaystyle\bigcup_{k=1}^{\infty}(A \cap E_k)\right) \ge m^*\left(\displaystyle\bigcup_{k=1}^n(A \cap E_k\right)=\displaystyle\sum_{k=1}^n m^*(A \cap E_k)$
$\Longrightarrow \displaystyle\lim_{n\to\infty} m^*\left(\displaystyle\bigcup_{k=1}^{\infty}(A \cap E_k)\right) \ge \displaystyle\lim_{n\to\infty} \displaystyle\sum_{k=1}^n m^*(A \cap E_k)$
$\Longrightarrow m^*\left(\displaystyle\bigcup_{k=1}^{\infty}(A \cap E_k)\right) \ge \displaystyle\sum_{k=1}^{\infty}m^*(A \cap E_k) \ge m^*\left(\displaystyle\bigcup_{k=1}^{\infty}(A \cap E_k)\right)$

where the last inequality on the right comes from the proposition that states outer measure is countably subadditive.

So does this seem reasonable?
• Oct 20th 2011, 12:55 PM
Drexel28
Re: Countable mutually disjoint measurable sets
Quote:

If for some $i:\,m^*E_i = \infty$ then it is trivial so we consider only sets of finite outer measure. Note that
$m^*\left(\displaystyle\bigcup_{k=1}^{\infty}(A \cap E_k)\right) \ge m^*\left(\displaystyle\bigcup_{k=1}^n(A \cap E_k\right)=\displaystyle\sum_{k=1}^n m^*(A \cap E_k)$
$\Longrightarrow \displaystyle\lim_{n\to\infty} m^*\left(\displaystyle\bigcup_{k=1}^{\infty}(A \cap E_k)\right) \ge \displaystyle\lim_{n\to\infty} \displaystyle\sum_{k=1}^n m^*(A \cap E_k)$
$\Longrightarrow m^*\left(\displaystyle\bigcup_{k=1}^{\infty}(A \cap E_k)\right) \ge \displaystyle\sum_{k=1}^{\infty}m^*(A \cap E_k) \ge m^*\left(\displaystyle\bigcup_{k=1}^{\infty}(A \cap E_k)\right)$

where the last inequality on the right comes from the proposition that states outer measure is countably subadditive.

So does this seem reasonable?

This looks like the right way to do it to me--this was a homework problem of mine a long time ago, and I think this is how I did it.