I don't understand the question.

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SUMMARY

The discussion centers on a specific expression from Vladimir Bogachev's "Measure Theory, Vol 1," where the user questions the derivation of the expression for X-E. The user initially believes their interpretation, which includes an additional term of ((X-B)∩(X-A)), is correct based on De Morgan's laws and distribution. However, they later realize that the disjoint nature of sets S, A, and B clarifies the original expression provided by Bogachev. This highlights the importance of understanding set relationships in measure theory.

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  • Familiarity with set theory concepts, particularly intersections and unions.
  • Understanding of De Morgan's laws as they apply to set operations.
  • Basic knowledge of measure theory, specifically the content of Bogachev's "Measure Theory, Vol 1."
  • Ability to interpret mathematical expressions and proofs in a rigorous manner.
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  • Study the properties of disjoint sets and their implications in measure theory.
  • Review De Morgan's laws in the context of set operations.
  • Explore additional examples from "Measure Theory, Vol 1" to reinforce understanding of set expressions.
  • Practice deriving expressions involving unions and intersections of sets to enhance mathematical rigor.
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Students of mathematics, particularly those studying measure theory, as well as educators and anyone seeking to deepen their understanding of set operations and their applications in mathematical proofs.

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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...:confused:
 
Last edited:
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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
 

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