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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.

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2 Answers 2

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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 $E_1\cup\cdots\cup E_n\cup\cdots$ ?

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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!)

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