Is the collection measurable sets of zero measure is a sigma algebra? Given a measure space $(X,\Sigma,\mu)$, let us define the collection $\mathscr{C} = \{S \in \Sigma \:{:}\: \mu(S)=0\}$. I would like to show that $\mathscr{C}$ is a $\sigma$-algebra. I have managed to show that $\mathscr{C}$ is closed under the countable union and that $\emptyset \in \mathscr{C}$. Now the only thing left for me to show is the following.
For every $C \in \mathscr{C}$, the complement of $C$ is also in $\mathscr{C}$. 
But when it comes to taking the complement of $C$, I do not think that I should use $X \setminus C$ as I am not trying to show that $\mathscr{C}$ is a $\sigma$-algebra on $X$. So I suppose that I should use $C^*\setminus C$ where $C^{*}$ is the largest subset of $X$ with $\mu(C^{*})=0$. But how do I know that such $C^{*}$ exists? Is there anyway I can show it using that $C \cap D \in \mathscr{C}$ for every $D \in \Sigma$ which I managed to prove.
Any help will be greatly appreciated.  
 A: As other answers already showed this is not a $\sigma$ algebra. It is however the standard example of a $\sigma$-ideal, where given a measurable space $(X,\mathcal A)$ a $\sigma$-ideal is a subset $\mathcal B$ of $\mathcal A$ such that


*

*$\varnothing\in\mathcal B$

*$X\in\mathcal B$ and $Y\subseteq X$ implies $Y\in\mathcal B$

*$\{X_n\}_{n\in\Bbb N}\subseteq \mathcal B$ implies $\bigcup_{n\in\Bbb N}X_n\in\mathcal B$
A: Usually, there is not a largest set $C$ with $\mu(C) = 0$.  For example, in the real line with Lebesgue measure, $\mu(C) = 0$ for all finite sets, but the union of all finite sets is the whole space, which does not have measure zero.
A: No, if $(X,\Sigma,\mu)$ is a measure space, $\emptyset \in \mathscr{C}$ but if $\mu(X) \neq 0$ (...so in most spaces), $\emptyset^c = X \not \in \mathscr{C}$.
More generally, if $A \in \mathscr{C}$, then $\mu(A^c) = \mu(X) \neq 0$.
A: Because the probability $μ(S)$ is an invertible function, for the probability function properties you have that $μ(∅)=0$.
