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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 $N=\emptyset$.

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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 $N \in \mathscr{I}$.

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

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