# My proof of very basic measure theory theorem

1. Aug 26, 2011

### gunitinug

Hi. I have a proof of a very basic measure theory theorem related to the definition of a measure, and would like to ask posters if the proof is wrong.

Theorem: If $E$ is measurable, then $\overline{E}$ is measurable and conversely.

My Proof:
Let's try the converse version first.

$m(E)=m(E \cap \overline{E})+m(E \cap E)$
$=m(E \cap \overline{E})+m(E)$
So $m(E \cap \overline{E})=0$. By this we've shown that $\overline{E}$ is measurable. Converse is true by similar method.

$m(\overline{E})=m(\overline{E} \cap \overline{E})+m(\overline{E} \cap E)$
$=m(\overline{E})+m(E \cap \overline{E})$
$=m(\overline{E})+0=m(\overline{E}).$

Last edited: Aug 26, 2011
2. Aug 26, 2011

### Citan Uzuki

That is in fact, completely wrong. In order to show that $\overline{E}$ is measurable, you have to show that for any set A, $m(A) = m(A \cap \overline{E}) + m(A \cap \overline{\overline{E}})$. You seem to be trying to do this only in the case where A=E, which is not sufficient.

3. Aug 26, 2011

### micromass

And the converse is not true. The thing is that all closed sets are measurable. So $\overline{E}$ is always measurable. But E doesn't need to be.

4. Aug 26, 2011

### Citan Uzuki

I took the meaning of the bar to be compliment, rather than closure. gunitinug, can you confirm that that's what the notation means?

5. Aug 26, 2011

### micromass

Aah, yes. That would make sense...