# Prove a collection of sets measure 0 or 1 is a $\sigma$-algebra

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 ($$\forall B \subset A, \mu(A) = 0 \implies \mu(B) =0$$).

• If any one of the sets is of measure 1, then so is the union. Commented Apr 27, 2018 at 2:59
• 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? Commented Apr 27, 2018 at 3:53
• Yes it is correct. Commented Apr 27, 2018 at 4:21

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, $$1=\mu\left(X\right)\geqslant \mu\left(\bigcup_{i\in\mathbb N}A_i\right)\geqslant \mu\left(A_{i_0}\right)=1.$$