Let $(X,\mathcal{M},\mu)$ be a complete measure space. Show $\mathcal{J}:=\{A\in\mathcal{M}|\mu(A^{c})=0\}\cup\{\emptyset\}\}$ is a topology Let $(X,\mathcal{M},\mu)$ be a complete measure space. Let $\mathcal{J}:=\{A\in\mathcal{M}|\mu(A^{c})=0\}\cup\{\emptyset\}$
1) Prove that $\mathcal{J}$ is a topology on $X$
Any thoughts on how to start? (We have not had topology course yet..but our instructor says it should not be too difficult, I googled the axioms)
I would appreciate help with this problem.
1) Since $X\in\mathcal{M}$ and $X\subset X$ , $\mu(X)=0$ from the completeness of the measure space, but then $0=\mu(X)=\mu(\emptyset^{c})\Rightarrow \emptyset\in J$
is this ok so far? Seems a bit pointless to do that as by the def of $\mathcal{J}$ the empty set is already there
2) how to show that the entire space $X\in\mathcal{J}$?
3) $\mathcal{J}$ is closed under any union of sets
4) closed under finite interesections 
 A: $\emptyset \in \mathcal J$ by hypothesis.
Since $X^\complement = \emptyset$, we have $\mu(X^\complement) = 0$ and $X \in \mathcal J$.
Let $A, B \in \mathcal J$. If any of $A$ or $B$ is empty, then $A \cap B = \emptyset \in \mathcal J$. Otherwise, suppose $A, B \ne \emptyset$. This means $\mu(A^\complement) = 0$, $\mu(B^\complement) = 0$. Since $(A \cap B)^\complement = A^\complement \cup B^\complement$, it follows that
$$\mu\left((A \cap B)^\complement\right) = \mu(A^\complement \cup B^\complement) \le \mu(A^\complement) + \mu(B^\complement) = 0.$$
Thus, $A \cap B \in \mathcal J$. By induction, the intersection of any finite collection of sets in $\mathcal J$ is also in $\mathcal J$.
Let $\{A_\alpha\} \subset \mathcal J$. If $A_\alpha = \emptyset$ for all $\alpha$, then $\bigcup_\alpha A_\alpha = \emptyset \in \mathcal J$. Otherwise, suppose $A_\beta \ne \mathcal \emptyset$ for some $\beta$. We have
$$
\left(\bigcup_\alpha A_\alpha\right)^\complement = \bigcap_\alpha A_\alpha^\complement \subset A_\beta^\complement.
$$
Since $A_\beta \ne \emptyset$, we must have $\mu(A_\beta) = 0$. Thus $\mu\left(\left(\bigcup_\alpha A_\alpha\right)^\complement\right) = 0$ since $\mu$ is complete. It follows that the union of any collection of sets in $\mathcal J$ is also in $\mathcal J$.
