I am having trouble proving this part (for a bigger homework problem).
I have to prove that the set
$$C=\{A \cup N : A \in \mathcal{A}, N \in \mathcal{N}\}$$
is a $\sigma$-algebra where $(X,\mathcal{A},\mu)$ is a measure space and $\mathcal{N}$ is the collection of all null sets with respect to $\mathcal{A}$ and $\mu$.
I am stuck in the part of proving closure under complements. I already completed closure under unions.
Let $A \in \mathcal{A}$ and $N \in \mathcal{N}$. Then $(A \cup N)^C = A^C \cap N^C$. But I don't know how to proceed. I thought the complement of a null set would have measure equal to that of the set $X$ (Assuming it is part of the sigma algebra $\mathcal{A}$).
Is it better to prove also closure under intersections and then use that and closure under unions to prove closure under complements?
Best Answer
Not sure what is meant by null set, I assume the following $$\mathcal N := \{N\subseteq X \mid \exists F\in\mathcal A, N\subseteq F, \mu (F) = 0\}. $$ Take $U\in\mathcal C$, then $U = A\cup N$ and pick $F \in \mathcal A$ with $N \subseteq F$ and $\mu(F)=0$ and note that $$U^c = (A\cup N)^c = (A\cup F)^c \cup (F\setminus U) $$ Convince yourself, this satisfies the criteria of $\mathcal C$.