In measure theory, one starts with a measure $\mu_{0}$ on an algebra $\mathcal{A}$ (of subsets of $X$) and then defines an outer measure $\mu^{*}$ on $2^{X}$. Then, one proves the set of $\mu^{*}$-measurable sets, denoted here by $\sigma(\mu^{*})$ is a complete $\sigma$-algebra in which the restriction $\mu^{*}$ to it is a measure. Here, complete means that $\sigma(\mu^{*})$ contains all null sets, where a null set is defined as follows.
Definition: Let $(\Omega, \mathcal{F},\mu)$ be a measure space. A set $N\subseteq \Omega$ is called a null set (with respect to $\mu$) if there exists $\tilde{N}\in \mathcal{F}$ with $N \subseteq \tilde{N}$ and $\mu(\tilde{N}) = 0$.
As mentioned before, $\sigma(\mu^{*})$ is complete by Carathéodory's Theorem. It means that it contains all null sets. But suppose $N \subseteq X$ is such that its outer measure $\mu^{*}(N) = 0$. Does it imply $N \in \sigma(\mu^{*})$?
Add: For completeness, if $A \subseteq X$: $$\mu^{*}(A) := \inf\{\sum_{n\in \mathbb{N}}\mu_{0}(A_{n}): \hspace{0.1cm} \mbox{$A \subseteq \bigcup_{n\in \mathbb{N}}A_{n}$, $(\forall n \in \mathbb{N}) A_{n}\in \mathcal{A}$}\}$$
Suppose $\mu^*(N) = 0$. To show that $N$ is $\mu^*$ measurable, we need to check that for any set $Y \subset \Omega$, $$\mu^*(Y) \geq \mu^*(Y \cap N) + \mu^*(Y \cap N^c).$$ We have $\mu^*(Y \cap N) \leq \mu^*(N) = 0$ and $\mu^*(Y \cap N^c) \leq \mu^*(Y)$ by monotonicity, so $N$ is indeed measurable.