Sets with measure 0, $\mu (G)=0$ for some set G.

Question

I am relatively new to Measure theory and I am really struggling with the concept of measure. I'm interested to know how a set which isn't an empty set can have measure of 0.. that is $\mu (G)=0$ and what would be the $\sigma$-algebra generated by the family of sets such that $\{G \in \Sigma : \mu (G) =0\}$. For $(X,\Sigma, \mu)$ a measure space. Can someone please explain to me in noob terms please so I can finally understand this concept. Thanks in advanced.

Answer 1

If $\mu$ is countably additive then $H=\mu^{-1}\{0\}=\{G: \mu(G)=0\}$ is closed under countable unions and countable intersections. (That is if $J\subset H$ and $J$ is countable then $\cup J\in H$ and $\cap J\in H$.)

A $\sigma$-algebra $S$ on $X$ such that $S\supset H$ must satisfy $S\supset H^*=\{X\setminus G: G\in H\}.$

If $\mu$ is countably additive then $H\cup H^*$ $is$ the $\sigma$-algebra generated by $H.$