TITLE: simple question from set theory/measure theory
QUESTION [5 upvotes]: 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...

REPLY [3 votes]: $$\begin{align*}
x\notin E&\iff x\notin (A\cap S)\text{ and }x\notin (B\cap (X-S))\\[0.05in]
&\iff (x\notin A\text{ or }x\notin S) \text{ and }(x\notin B\text{ or }x\notin (X-S))\\[0.05in]
&\iff(x\in (X-A)\text{ or }x\in(X-S)) \text{ and }(x\in (X-B)\text{ or }x\in S)
\end{align*}$$
If $x\in S$, then we must have $x\in (X-A)$ because, in this case,
$$(x\in (X-A)\text{ or }x\in(X-S)) \iff x\in (X-A).$$
Similarly, if $x\in X-S$, we must have $x\in (X-B)$ because
$$(x\in (X-B)\text{ or }x\in S)\iff x\in(X-B).$$