is a $\mu$-null set necessarily in the $\sigma$-algebra

Question

Let $(X, \mathcal{A}, \mu)$ be a measure space. A $\textbf{$\mu\text{-null set}$}$ is a subset $N \subseteq X$ such that there exists an $M\in \mathcal{A}$ such that $N\subseteq M$ and $\mu(M)=0$.

Is it necessarily true that $N\in \mathcal{A}$?

Answer 1

Hint. Let $X$ be a set, and $Y$ be any subset of $X$. Then $\Sigma = \{\emptyset, Y, X \setminus Y, X\}$ is a $\sigma$-algebra over $X$.

Answer 2

No, it is not necessarily true. Let $\mathcal{A}$ be Borel set that is Lebesgue measurable. Let $M$ be Cantor set. Then $M$ is Borel set, but there are subset of Cantor set which is not Borel set. So there is $N\subset M$ not in the $\sigma$-algebra, i.e. $N\notin \mathcal{A}$.