# Measure question

• May 7th 2012, 08:37 PM
Wowhapjs
Measure question
Assume $A_j,j\geq 1,j\in\Bbb N$ are measurable sets. Let $m \in N$, and let $E_m$ be the set defined as follows : $x \in E_m \Longleftrightarrow x$ is a member of at least $m$ of the sets $A_k$.

I wanna know how to prove that

1. $E_m$ is measurable.
2. $m\lambda(E_m)\le\sum^{\infty}_{k=1}\lambda(A_k)$.

• May 11th 2012, 12:04 PM
mastermind2007
Re: Measure question
Well, the property of x $\in E_m$ as defined above simply means that if you have a look at at least m intersecting sets,
then x will be in this intersection. And if there is another intersection, then x could also be in there.

In other words: $E_m$ is the set of the union of all intersections including m sets.
Since the $A_j, j \in \mathbb{N}$ are measurable and there is a defined measure $\lambda$,
there must be a $\sigma - algebra \ \Sigma$ which means that if $A_j$ is in $\Sigma$,
so is any countable intersection and countable union.

$m \in \mathbb{N}$ , so $E_m$ lies in the sigma-algebra and is therefore measurable.

(Or otherwise stated: Any countable intersection and union of countable sets is countable.)

2.)

$\lambda(\bigcup_{i = 1}^N A_i) \ \leq \ \sum_{i = 1}^N \lambda(A_i)$

For m = 1 the equation is obvious (use subadditivity).
Since the measure can only become smaller if we raise m, the "at least" does not make any difference to the proof.
The thing is that one has to prove the "m" on the left side.

Let $m \geq 1$ .

Consider $A_i , A_j$ measurable.

Then $\lambda(A_i \cap A_j) \leq \lambda(A_i)$

The same is true for $A_j$.
Is that clear to you? One might have to prove this relation.

So it follows that $2\ \lambda(A_i \cap A_j) \leq \lambda(A_i) + \lambda(A_j)$

Here we set m = 2. By induction with respect to m, you should get to the required relation.