# Something in Measure Theory

#### Alan

##### Member
This is a simple question.

On pages 5-6 of Measure Theory,Vol 1, Vladimir Bogachev he writes that:

for $$E=(A\cap S)\cup (B\cap (X-S))$$

Now, he writes that:

$$X-E = ((X-A)\cap S) \cup ((X-B)\cap (X-S))$$

But I don't get this expression, I get another term of $$((X-B)\cap (X-A))$$

i.e, $$X-E =( ((X-A)\cap S) \cup ((X-B)\cap (X-S)))\cup ((X-B)\cap (X-A))$$.

I believe I did it correctly according to De-Morgan rules and distribution.

I am puzzled...

Last edited:

#### Alan

##### Member
Re: Something in Measure Theory.

Nevermind, I got it.

It follows from the fact that S is disjoint to A and B.

Sometimes I wonder how I still can do math...:-D