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Let $(X,\mu)$ and $E_n\subset X$ s.t $\mu(E_n)=\mu(X)$

can we write $$\mu(X)=\mu(\cap_{n=1}^{\infty}E_n\uplus (\cap_{n=1}^{\infty}E_n)^C)$$?

Shouldn't it be:

$$\mu(X)=\mu(\cup_{n=1}^{\infty}E_n\uplus (\cap_{n=1}^{\infty}E_n)^C)$$?

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$X$ is the disjoint union of $\cap E_n$ and $(\cap E_n)^{c}$ so the first one is correct (without any assumptions on the sets $E_n$).

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  • $\begingroup$ Elementary, I should have looked at it as $X=A\cup A^C$ where $A=\cap E_n$ just as you said $\endgroup$
    – newhere
    Oct 4 '19 at 11:53

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