Let $(X,\mathcal{E},\mu)$ be an ordinary measure space and let $\mathcal{A}\subset\mathcal{E}$ be an algebra. Denote by $\mathcal{N}$ the set of $\mu$-null sets.
Is it true that, for all $E\in\sigma(\mathcal{A}\cup\mathcal{N})$ with $\mu(E)<+\infty$, there exists $E'\in\sigma(\mathcal{A})$ such that $\mu(E\triangle E')=0$? If I prove $$\mu(E)=\inf\{\mu(F)\mid F\supset E\text{ and }F\in\mathcal{A}\cup\mathcal{N}\}$$ for all $E\in\sigma(\mathcal{A}\cup\mathcal{N})$, then $ \mu(E)=\inf\{\mu(F)\mid F\supset E\text{ and }F\in\mathcal{A}\}$. Hence, for all $n\in\mathbb{N}$ there exists $F_n\in\mathcal{A}$ such that $$\mu(F_n\setminus E)\leq\frac{1}{n}.$$ The set $\bigcap_{n\in\mathbb{N}}F_n\in\sigma(\mathcal{A})$ satisfies the required property. Anyway, I'm not sure that my claim is correct, since $\mathcal{A}\cup\mathcal{N}$ in general is neither an algebra, nor a ring of sets.
Let $\mathcal{C} = \{ E \in \mathcal{E} : (\exists E' \in \sigma(\mathcal{A}) : \mu(E \triangle E') = 0)\}$. Clearly $\mathcal{A} \subset \mathcal{C}$ (take $E' = E$) and $\mathcal{N} \subset \mathcal{C}$ (take $E' = \emptyset$).
I claim $\mathcal{C}$ is a $\sigma$-algebra:
We have $\emptyset \in \mathcal{C}$ (take $E' = \emptyset$).
If $E \in \mathcal{C}$, so that there exists $E' \in \sigma(\mathcal{A})$ with $\mu(E \triangle E') =0$, then we have $E'^c \in \sigma(\mathcal{A})$ and $\mu(E^c \triangle E'^c) = 0$ so $E^c \in \mathcal{C}$.
If $E_1, E_2, \dots \in \mathcal{C}$, so that there are $E_1', E_2', \dots \in \sigma(\mathcal{A})$ with $\mu(E_n \triangle E_n') = 0$ for each $n$, then set $E = \bigcup_n E_n$, $E' = \bigcup_n E_n'$. We have $E' \in \sigma(\mathcal{A})$, and you may check that $E \triangle E' \subset \bigcup_n E_n \triangle E_n'$. Hence $\mu(E \triangle E') = 0$ by countable additivity and so $E \in \mathcal{C}$.
Thus $\sigma(\mathcal{A} \cup \mathcal{N}) \subset \mathcal{C}$ which is the desired statement.
Note we did not need to assume that $\mathcal{A}$ is an algebra.