 Solution

First, $\mathcal{B}\subseteq \mathcal{C}$ by taking $B=A$. If $A\in \mathcal{C}$, thenso $\mathcal{C}$ is closed under complements. Now let $A_i\in \mathcal{C}$ and put $A={\bigcup }_i{A}_i$. Because $\mu (X)<\infty$, continuity from below of a measure gives an $N$ such thatChoose $B_i\in \mathcal{B}$ with $\mu (A_i△B_i)<ϵ/(2N)$ and let $B={\bigcup }_{i=1}^N{B}_i\in \mathcal{B}$. The union bound for symmetric differences givesThus $\mathcal{C}$ is a sigma-algebra containing $\mathcal{B}$, and

Finiteness is essential. Take $X=\mathbb{N}$ with counting measure and let $\mathcal{B}$ be the finite-cofinite Boolean algebra of sets. Then $\sigma (\mathcal{B})=\mathcal{P}(\mathbb{N})$. For $ϵ<1$, the condition $\mu (A△B)<ϵ$ forces $A=B$, so $\mathcal{C}=\mathcal{B}$. Any infinite coinfinite subset of $\mathbb{N}$ belongs to $\sigma (\mathcal{B})$ but not to $\mathcal{C}$.