Is the class of negligible sets a monotone class?

Question

Let $(X,\Sigma,\mu)$ be a measure space we say that the subset $F$ of a set $X$ is negligible set if there exist $G \in\Sigma$ , $F$ is a subset of $G$ And $\mu(G)=0$

My question is:

Is the class of negligible sets a monotone class, and how can I prove that?

Answer 1

It is true.

Take $E_1 \subset E_2 \subset \dots$ a chain of negligible sets. For all $n$ let $E_n \subset G_n$ with $\mu (G_n) = 0$.

Then $\bigcup_n E_n \subset \bigcup_n G_n$ and $\mu \left( \bigcup_n G_n \right) = 0$ (why?). So $\bigcup_n E_n$ is negligible.

On the other hand, if $E_1 \supset E_2 \supset \dots$ is a chain of negligible sets ,there exists $G \supset E_1$ of measure zero, hence every subset of $E_1$ is negligible (because it is contained in $G$). In particular $\bigcap_n E_n$ is negligible.