# m(A {intersection} U {1 to infinity} Ei)=Sum {1 to infinity} m(A {intersection} Ei)

• Sep 26th 2009, 02:13 PM
Dark Sun
m(A {intersection} U {1 to infinity} Ei)=Sum {1 to infinity} m(A {intersection} Ei)
I have problem here which I am having difficulty proving:

Assume that $\langle E_1\rangle$ is a sequence of disjoint (Lebesgue) measurable sets, and A is any set. w.t.s.:

$m^*(A\cap \bigcup _{i=1}^{\infty } E_i)=\sum _{i=1}^{\infty }m^*(A\cap E_i)$

I have a proof of the finite case as i goes from 1 to n by using induction, however, it requires me to take the intersection of the last term in the sequence, which does not apply to the infinite case.

Also, I believe that $m^*(A\cap \bigcup _{i=1}^{\infty } E_i)\leq \sum _{i=1}^{\infty }m^*(A\cap E_i)$ is trivial since sub-additivity applies to the outer measure, and $A\cap \bigcup _{i=1}^{\infty } E_i=\bigcup _{i=1}^{\infty }(A\cap E_i)$.

So, it may only be neccessary to prove it going the other way.

Thanks a million!
• Sep 26th 2009, 09:04 PM
Dark Sun
Hello, I apologize if my Analysis skills are lacking somewhat. But that's why I'm learning it, after all...

Anyway, something came to me in a flash of insight, perhaps you could tell me if I am mistaken:

It suffices to show that $\exists N$ s.t. $n>N$ implies that $\forall \epsilon >0, \; |m^*(A\cap \bigcup _{i-1}^n\limits E_i)-\sum _{i=1}^n m^*(A\cap E_i)|<\epsilon$.

However, since $m^*(A\cap \bigcup _{i-1}^n\limits E_i)=\sum _{i=1}^n m^*(A\cap E_i)$ by Lemma 9, we have that:

$|m^*(A\cap \bigcup _{i-1}^n\limits E_i)-\sum _{i=1}^n m^*(A\cap E_i)|=0<\epsilon$