# Showing that $\mathcal G$ is a $\sigma$-algebra

I'm new to measure theory and got stucked in the next problem. Given a measure space $$(\Omega, \mathcal G, \mu)$$ and the set $$\mathcal M = \{ X \subset \Omega : \exists A \in \mathcal G \space \text{s.t.} X \subset A, \text{and} \space \mu (A) = 0\}$$.

Show that $$\mathcal G^{\mu} := \{ A = B \cup E, \text{where} B \in \mathcal G \space \text{and} \space E \in \mathcal M\}$$ is also a $$\sigma-$$algebra.

So far I've been able to show that $$\mathcal G$$ is non empty, but haven't been able to show that it is closed under complement and countable unions. For the former, considering an arbitrary $$A =B \cup E\in \mathcal G^ \mu$$, I tried to show that $$A^c = B^c \cap E^c \in \mathcal M$$ or $$\in \mathcal G$$ but didn't get anything. For the latter, no idea yet.

Any help is appreciated.

Say $$A=B\cup E$$. Then $$E\subseteq X$$, where $$X$$ has measure $$0$$. Note that $$F=X\setminus E$$ is also a subset of $$X$$, and so $$F\in\mathcal{M}$$.
Now, $$B^c$$ is in $$\mathcal{G}$$. And so is $$B^c\setminus X$$, because both $$B^c$$ and $$X$$ are in $$\mathcal{G}$$. Of course, this is nothing more than $$B^c\cap X^c$$. Since $$E\subseteq X$$, then $$X^c\subseteq E^c$$, so $$B^c\cap X^c\subseteq B^c\cap E^c$$.
Now, what is missing from $$B^c\cap X^c$$ to get $$B^c\cap E^c$$? The elements in $$B^c\cap E^c$$ that are not in $$B^c\cap X^c$$ are precisely the elements that are in $$F = X\cap E^c$$. (Prove it). So we can take $$(B^c\cap X^c)\cup F$$ to get $$B^c\cap E^c$$. But $$(B^c\cap X^c)\cup F$$ is in $$\mathcal{G}^{\mathcal{M}}$$.
Countable unions are even easier: if $$X_1,\ldots,X_n,\ldots$$ all have measure zero, what is the measure of $$X_1\cup X_2\cup\cdots\cup X_n\cup\cdots$$? And if $$E_i\subseteq X_i$$, what measurable set of measure zero can we choose to contain $$E_1\cup\cdots\cup E_n\cup\cdots$$ ?
Complement: Let $$Z = B \cup E$$ where $$E \subset A \in \mathcal{G}$$. Then $$Z^c = (B^c \; \cap A^c ) \cup (B^c \cap A \setminus E) = X \cup Y$$. Show that $$X \in \mathcal{G}$$ and $$Y \in \mathcal{M}$$. (trivial!)
Countable union: Let $$A_i = B_i \cup E_i \Rightarrow \cup_{i \geq 1} A_i = (\cup_{i \geq 1} B_i) \cup (\cup_{i \geq 1} E) = X \cup Y$$. Show that $$X \in \mathcal{G}$$ and $$Y \in \mathcal{M}$$. (trivial!)