# How to show that $\mathscr F$ is a Sigma Algebra?

Let $$\left( \Omega, \mathscr A, \mu \right)$$ be a measure space. Define: $$\mathscr I = \{ N \subseteq \Omega \, | \, \exists A \in \mathscr A \; s.t. \; \mu(A) = 0 \text{ and } N \subseteq A \}$$ (a) Show that $$\mathscr I$$ is closed under countable unions and any subset of a set in $$\mathscr I$$ is also in $$\mathscr I$$.
(b) Define $$\mathscr F = \{ A \Delta N \, | \, A \in \mathscr A$$ and $$N \in \mathscr I \}$$. Show that $$\mathscr F$$ is a sigma algebra and $$\mathscr A \subseteq \mathscr F$$.

I have been able to show (a) and that $$\mathscr F$$ contains $$\Omega$$ and $$\emptyset$$, and that $$\mathscr F$$ is closed under complementation. But closure under countable unions is becoming difficult to manage directly. Any help would be appreciated.

The fact that $$\mathscr A \subseteq \mathscr F$$ is obvious, by taking $$N=\emptyset$$.

Let $$(F_j)_{j \in \mathbb{N}} \subseteq \mathcal{F}$$, i.e. $$F_j = A_j \, \Delta \, N_j$$ $$A_j \in \mathcal{A}$$ and $$N_j \in \mathscr{I}$$. By the definition of $$\mathscr{I}$$, there is for each $$j \geq 1$$ a set $$B_j \in \mathcal{A}$$ such that $$\mu(B_j)=0$$ and $$N_j \subseteq B_j$$. Set

$$F := \bigcup_{j \geq 1} F_j$$

and define

$$A := \underbrace{\left( \bigcup_{j \geq 1} A_j \right)}_{=: \tilde{A}} \backslash \underbrace{\left( \bigcup_{j \geq 1} B_j \right)}_{=: \tilde{B}} \quad \text{and} \quad N := F \backslash A.$$

Since $$A$$ and $$N$$ are pairwise disjoint, we have $$F = A \cup N = A \, \Delta \, N.$$ Moreover, since $$A_j \in \mathcal{A}$$ and $$B_j \in \mathcal{A}$$ for all $$j \geq 1$$ we clearly have $$A \in \mathcal{A}$$. It remains to show that $$N \in \mathscr{I}$$. To this end, we note that

$$F \subseteq \bigcup_{j \geq 1} A_j \cup \bigcup_{j \geq 1} B_j = \tilde{A} \cup \tilde{B}$$

and so

\begin{align*} N \subseteq \left( \tilde{A} \cup \tilde{B} \right) \backslash \left( \tilde{A} \backslash \tilde{B} \right) &\subseteq (\tilde{A} \backslash (\tilde{A} \backslash \tilde{B})) \cup \tilde{B} \subseteq \tilde{B}. \end{align*}

Since $$\tilde{B} \in \mathcal{A}$$ and

$$\mu (\tilde{B}) \leq \sum_{j \geq 1} \mu(B_j) = 0$$

we conclude that $$N \in \mathscr{I}$$.

• I got it... Thanks for the help... – Neel Mar 10 '19 at 5:56
• @Neel You are welcome. – saz Mar 10 '19 at 7:15