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Something in Measure Theory

Alan

Member
Jul 21, 2012
60
This is a simple question.

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

for [tex]E=(A\cap S)\cup (B\cap (X-S))[/tex]

Now, he writes that:

[tex]X-E = ((X-A)\cap S) \cup ((X-B)\cap (X-S))[/tex]

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

i.e, [tex]X-E =( ((X-A)\cap S) \cup ((X-B)\cap (X-S)))\cup ((X-B)\cap (X-A))[/tex].

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

I am puzzled...:confused:
 
Last edited:

Alan

Member
Jul 21, 2012
60
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