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 $N \in \mathscr{I}$.

| cite | improve this answer | |
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.