# Thread: Outer measure and measurable sets.

1. ## Outer measure and measurable sets.

If $\displaystyle \mu$ is an outer measure on X and $\displaystyle \{ A_n \} ^ \infty _{n=1}$ is a sequence of disjointed $\displaystyle \mu$ measurable sets, prove that $\displaystyle \mu ( E \cap ( \bigcup ^ \infty _{n=1} A_n )) = \sum ^ \infty _{n=1} \mu (E \cap A_n ) \ \ \ \forall E \subset X$

Proof so far.

Since $\displaystyle \mu$ is an outer measure on $\displaystyle X$, we know that:

1. $\displaystyle \mu (A) \leq \mu (B)$ if $\displaystyle A \subset B$

2. $\displaystyle \mu ( \bigcup _{n=1} ^ \infty A_n ) \leq \sum ^ \infty _{n=1} \mu (A_n)$

And since $\displaystyle \{ A_n \} ^ \infty _{n=1}$ is $\displaystyle \mu$-measurable, then for each n, we have $\displaystyle \mu (E) = \mu (E \cap A_n )+ \mu (E \cap A_n ^c ) \ \ \ \forall E \subset X$

Now, $\displaystyle \mu (E \cap ( \bigcup _{n=1} ^ \infty A_n)) \leq \sum _{n=1} ^ \infty \mu (E \cap A_n )$ since $\displaystyle \mu$ is an outer measure.

So what is left to prove is that $\displaystyle \mu (E \cap ( \bigcup _{n=1} ^ \infty A_n)) \geq \sum _{n=1} ^ \infty \mu (E \cap A_n )$

So I have $\displaystyle \sum _{n=1} ^ \infty \mu (E \cap A_n ) = \mu (E \cap A_1)+ \mu (E \cap A_2) + . . .$

And I know I will have to use the measurable property, but I don't seem to be making any progress here...

Any hints? Thank you.

If $\displaystyle \mu$ is an outer measure on X and $\displaystyle \{ A_n \} ^ \infty _{n=1}$ is a sequence of disjointed $\displaystyle \mu$ measurable sets, prove that $\displaystyle \mu ( E \cap ( \bigcup ^ \infty _{n=1} A_n )) = \sum ^ \infty _{n=1} \mu (E \cap A_n ) \ \ \ \forall E \subset X$

Proof so far.

Since $\displaystyle \mu$ is an outer measure on $\displaystyle X$, we know that:

1. $\displaystyle \mu (A) \leq \mu (B)$ if $\displaystyle A \subset B$

2. $\displaystyle \mu ( \bigcup _{n=1} ^ \infty A_n ) \leq \sum ^ \infty _{n=1} \mu (A_n)$

And since $\displaystyle \{ A_n \} ^ \infty _{n=1}$ is $\displaystyle \mu$-measurable, then for each n, we have $\displaystyle \mu (E) = \mu (E \cap A_n )+ \mu (E \cap A_n ^c ) \ \ \ \forall E \subset X$

Now, $\displaystyle \mu (E \cap ( \bigcup _{n=1} ^ \infty A_n)) \leq \sum _{n=1} ^ \infty \mu (E \cap A_n )$ since $\displaystyle \mu$ is an outer measure.

So what is left to prove is that $\displaystyle \mu (E \cap ( \bigcup _{n=1} ^ \infty A_n)) \geq \sum _{n=1} ^ \infty \mu (E \cap A_n )$

So I have $\displaystyle \sum _{n=1} ^ \infty \mu (E \cap A_n ) = \mu (E \cap A_1)+ \mu (E \cap A_2) + . . .$

And I know I will have to use the measurable property, but I don't seem to be making any progress here...

Any hints? Thank you.
Look at
$\displaystyle \mu ( E \cap ( \bigcup ^k _{n=1} A_n )) = \sum ^k _{n=1} \mu (E \cap A_n ) \ \ \ \forall E \subset X$
Do you know that this is true (for all natural numbers k)?
If you don't, then try prove it.
If you do, then you will notice this is a sequence of real numbers indexed by k. This sequence is bounded above (by what?). Use some results about sequences of real numbers to get the other inequality.
I hope I have given you enough hints.