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Let $(X, \mathcal{F}, \mu)$ and $\mu(X) = 1$. Prove that $$ \mathcal{A} = \left\{A \in \mathcal{F} : \mu(A) = 0 \ \text{or} \ \mu(A) = 1\right\} $$ is a $\sigma$-algebra.

I'm having trouble proving it is closed under countable union. My attempt: let $A_1, A_2, \cdots \in \mathcal{A}$, then we can construct $B_n = A_n \backslash \cup_{i=1}^{n-1}A_i$, then $\cup_{i=1}^{\infty}A_i = \cup_{i=1}^{\infty}B_i$ and $B_i$ are pairwise disjoint.

Case 1: If $\mu(A_i) = 0 \ \forall i$, then $B_i \subset A_i \implies \mu(B_i) = 0$. Then $\mu(\cup A_i ) = \mu(\cup B_i) =\Sigma \mu(B_i) = 0$. Then $\cup A_i \in \mathcal{A}$.

Case 2: If $\mu(A_i) = 1 \ \forall i$, then $B_n = A_n \backslash \cup_{i=1}^{n-1}A_i \subset A_{1}^{c}$. Since $ \mu(A_{1}^{c}) = \mu(X) - \mu(A_1) = 0, \mu(B_n) = 0$. Then $\mu(\cup A_i ) = \mu(\cup B_i) =\mu(B_1) + \Sigma_{k=2}^{\infty} \mu(B_k) = 1$. Then $\cup A_i \in \mathcal{A}$.

Case 3: If $\mu(A_i) = 0$ for some $i$. I don't know where to start here. I feel my proof is complicated and it requires $\mu$ to be a complete measure ($\forall B \subset A, \mu(A) = 0 \implies \mu(B) =0$).

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    $\begingroup$ If any one of the sets is of measure 1, then so is the union. $\endgroup$
    – saulspatz
    Commented Apr 27, 2018 at 2:59
  • $\begingroup$ Let $\mu(A_k) = 1$. Since $A_k \subset \cup A_i \subset X$, then$ \mu(A_k) \leq \mu(\cup A_i) \leq \mu(X)$. So $\mu(\cup A_i) = 1$. Is my proof correct? $\endgroup$
    – dunguyen
    Commented Apr 27, 2018 at 3:53
  • $\begingroup$ Yes it is correct. $\endgroup$
    – saulspatz
    Commented Apr 27, 2018 at 4:21

1 Answer 1

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Actually, passing by the disjoint unions makes the things more complicated. Treat the following two cases.

  1. For each $i$, $\mu\left(A_i\right)=0$. Then the measure of a countable union of set of measure zero is zero and $\bigcup_{i\in\mathbb N}A_i$ has measure zero.

  2. There is an index $i_0$ such that $\mu\left(A_{i_0}\right)\neq 0$. Since $A_{i_0}$ is an element of $\mathcal A$, this means that $\mu\left(A_{i_0}\right)=1$. Consequently, $$ 1=\mu\left(X\right)\geqslant \mu\left(\bigcup_{i\in\mathbb N}A_i\right)\geqslant \mu\left(A_{i_0}\right)=1. $$

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