Countable mutually disjoint measurable sets

• October 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?
• October 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?
• October 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
$m^*\left(\displaystyle\bigcup_{k=1}^{\infty}(A \cap E_k)\right) \ge m^*\left(\displaystyle\bigcup_{k=1}^n(A \cap E_k\right)=\displaystyle\sum_{k=1}^n m^*(A \cap E_k)$
$\Longrightarrow \displaystyle\lim_{n\to\infty} m^*\left(\displaystyle\bigcup_{k=1}^{\infty}(A \cap E_k)\right) \ge \displaystyle\lim_{n\to\infty} \displaystyle\sum_{k=1}^n m^*(A \cap E_k)$
$\Longrightarrow m^*\left(\displaystyle\bigcup_{k=1}^{\infty}(A \cap E_k)\right) \ge \displaystyle\sum_{k=1}^{\infty}m^*(A \cap E_k) \ge m^*\left(\displaystyle\bigcup_{k=1}^{\infty}(A \cap E_k)\right)$