# Infinite Union of Non-disjoint Sets

1. Nov 15, 2012

### Yagoda

1. The problem statement, all variables and given/known data
To give some context, I'm trying to show that $\mu(\bigcup^{\infty}_{k=1}A_{k})\leq \sum^{\infty}_{k=1}\mu(A_{k})$ where μ is the Lebesgue measure and the A's are a countable set of Borel sets.

Since the A's may not be disjoint, I'm trying to rewrite the left side of the equation to somehow show that the elements that are present in more than one set are not being counted in the same way that they are on the right, which causes the left side to be smaller (or equal if all sets are disjoint).

2. Relevant equations

3. The attempt at a solution
In class we were given the hint to consider elements that are only in one set, present in 2 sets, in 3 sets, etc. and use this to rewrite the inequality.

The elements that are only in one set particular set $A_{k} = A_{k} \setminus \bigcup^{k-1}_{i=1} A_{i}$. When I union all of these, it looks pretty messy.
I'm stuck on figuring out how to write elements that are present in more than one set. Does this approach look like it's on the right track?

(From what I've read it seems to common to build this up using the outer Lebesgue measure, but we haven't covered that. It seem like the way we're going about this is kind of unconventional)

2. Nov 15, 2012

### hedipaldi

use the definition of external measure as infimum.

3. Nov 15, 2012

### Ray Vickson

Can you show $\mu(A \cup B) \leq \mu(A) + \mu(B)?$ If so, you can get by induction that $$V_n \equiv \mu \left( \cup_{i=1}^n A_i \right) \leq \sum_{i=1}^n \mu(A_i) \leq \sum_{i=1}^{\infty} \mu(A_i).$$ The numbers Vn are non-negative, increasing in n and are all <= the infinite sum of μ(Ai). What does that tell you?

RGV

4. Nov 15, 2012

### Yagoda

Yes, I think I can show that μ(A∪B)≤μ(A)+μ(B).

What is Vn, though?

5. Nov 16, 2012

### Ray Vickson

I *defined* Vn to be $\mu \left( \cup_{i=1}^n A_i \right),$; did you miss the $\equiv$ sign?

RGV