Given a measure space $(X,\mathbb{E},\mu)$, where $\mathbb{E}$ is the $\sigma$-algebra on $X$ and $\mu$ is the Lesbegue measure. My professor has been writing the following (not exactly his words - what I wrote down during the lecture regarding this topic):
Let us define $$\mathbb{N}_\mu=\{N\subseteq X\;|\;\exists E\in\mathbb{E}:N\subseteq E,\;\; \mu(E)=0\}$$ Let $\mathbb{E}_\mu$ be the paving [system of sets on X] such that: $$\mathbb{E}_\mu=\{E\cup N|E\in\mathbb{E},N\in N_\mu\}$$
Show that $\mathbb{E}\subseteq \mathbb{E}_\mu$ - this confused me.
How can $\mathbb{E}$ not contain the sets with length $\mu(E)=0$? Because as I understand it, the $\mathbb{N}_\mu$ is the sets of $\mathbb{E}$ with length $0$ - so if $\mathbb{E}_\mu$ is the union of $\mathbb{E}$ and all the sets in $\mathbb{E}$ with length $0$ - how can $\mathbb{E}_\mu$ be a superset to $\mathbb{E}$?
Secondly he has been writing, let $\mathbb{D}=\mathbb{E}\cup\mathbb{N}_\mu$ to show that $\mathbb{E}_\mu$ is the smallest $\sigma$-algebra containing $\mathbb{E}$ and $\mathbb{N}_\mu$ - however, as I understand it, the above should just be $\mathbb{E}\cup\mathbb{N}_\mu=\mathbb{E}_\mu$, but this cannot be true obviously.
Any help understanding this set containing all sets with length $0$ will be greatly appreciated.