Thread: m(E_1 U E_2)+m(E_1 ^ E_2)=mE_1+mE_2

1. m(E_1 U E_2)+m(E_1 ^ E_2)=mE_1+mE_2

Hello everyone. I've been hitting my head against the chalk-board trying to figure this one out.

Suppose $\displaystyle E_1$ and $\displaystyle E_2$ are Lebesgue measurable. Then we want to show that:

$\displaystyle m(E_1\cup E_2)+m(E_1\cap E_2)=mE_1+mE_2$.

Recall: $\displaystyle mA=\mbox{inf }_{A\subset \cup I_n} \sum \ell (I_n)$

Also Recall: E is measurable means for any set $\displaystyle A, \; mA=m(A\cap E)+m(A\cap \widetilde E)$.

Any help on this problem would be greatly appreciated!

2. It might help to look at it this way:
$\displaystyle m(E_1\cup E_2)=mE_1+mE_2-m(E_1\cap E_2)$

Then use the measurability condition for the measurable set $\displaystyle E=E_1\setminus (E_1\cap E_2)$ on $\displaystyle A=E_1\cup E_2$

3. What are A and In ?

4. Thanks! This proof is fairly self-explanatory, although it is missing a critical piece, and I cannot find the lemma anywhere in any textbook, and I am having difficulty proving it.

Assume that $\displaystyle E_1$ and $\displaystyle E_2$ are measurable. w.t.s. $\displaystyle m(E_1\cup E_2)+m(E_1\cap E_2)=mE_1+mE_2$.

Set $\displaystyle E=E_1\sim (E_1\cap E_2)$ and $\displaystyle A=E_1\cup E_2$. Clearly, $\displaystyle E$ is measurable.

Now, $\displaystyle m(E_1\cup E_2)=m((E_1\cup E_2)\cap E)+m((E_1\cup E_2)\cap \widetilde E)=$

$\displaystyle mE+mE_2=mE_1-m(E_1\cap E_2)+mE_2,$ and;

$\displaystyle m(E_1\cup E_2)+m(E_1\cap E_2)=mE_1+mE_2$

Now, I'm having trouble with $\displaystyle mE=mE_1-m(E_1\cap E_2)$, this seems to be the last road block to tackle.

P.S.: A is any set for which m is defined. And, $\displaystyle \bigcup In$ is a covering of A.

5. But of course since $\displaystyle E_1=E\cup (E_1\cap E_2)$ and $\displaystyle E, (E_1\cap E_2)$ are disjoint,

$\displaystyle mE+m(E_1\cap E_2)=m(E_1)$ by additivity of a measure.

6. You're a genius! Thanks a million!