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 ($\forall B \subset A, \mu(A) = 0 \implies \mu(B) =0$).
Actually, passing by the disjoint unions makes the things more complicated. Treat the following two cases.
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.
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, $$ 1=\mu\left(X\right)\geqslant \mu\left(\bigcup_{i\in\mathbb N}A_i\right)\geqslant \mu\left(A_{i_0}\right)=1. $$