Exercise 2.4 (Limits of events are preserved in $\sigma$-algebras)

Question

Let $\mathcal{A}$ be a $\sigma$-algebra and $(A_n)_{n\geq 1}$ a sequence of events in $\mathcal{A}$. Show that
$$\liminf_{n	o\infty} A_n; \in \mathcal{A} \quad \limsup_{n	o\infty} A_n \in \mathcal{A};
\quad \liminf_{n	o\infty} A_n \subseteq \limsup_{n	o\infty} A_n.$$

Toğrul Topal · Accepted Answer

\begin{enumerate}
\item Recall that $\limsup$ of the sets is defined as
	$$\limsup_{n	o\infty} A_n := \bigcap_{n=1}^{\infty} \bigcup_{m\geq n} A_m$$
	For any $n\geq 1$, the sets $\bigcup_{m\geq n} A_m$ are contained in $\mathcal{A}$ by the $\sigma$-algebra axioms since the sets $(A_m)_{m\geq n}$ are (closedness under countable unions). Therefore, the intersection $\bigcap_{n=1}^{\infty} \bigcup_{m\geq n} A_m$ of measurable sets $(\bigcup_{m\geq n} A_m)_{m \geq n}$ are contained in $\mathcal{A}$ as well by the $\sigma$-algebra property (closedness under countable intersections).
	
	\item For $\liminf$, defined as
	$$\liminf_{n	o\infty} A_n := \bigcup_{n=1}^{\infty} \bigcap_{m\geq n} A_m$$
	the assertion follows symmetrically.
	
	\item Pick an arbitrary $a\in\liminf_{n	o\infty}A_n$. By definition, there should be at least one $n\in\mathbb{N}$ such that all the sets $A_m$ after $A_n$ contain $a$:
	$$\exists n \in \mathbb{N} \quad \forall m \geq n: \quad a \in A_m.$$
	A weaker implication of this statement is that no matter how big $n'\in\mathbb{N}$ is, if it is greater than our chosen $n$, then all of the $A_m$ after $A_{n'}$ will contain $a$:
	$$\forall n' \geq n \quad \forall m \geq n': \quad a \in A_m.$$
	In particular, we can weaken this to
	$$\forall n' \geq n \quad \exists m \geq n': \quad a \in A_m.$$
	But this is equivalent to the assertion that
	$$a \in \bigcap_{n=n'}^{\infty}\bigcup_{m\geq n'} A_m \subseteq \bigcap_{n=1}^{\infty}\bigcup_{m\geq n'} = \limsup_{n	o\infty} A_n$$
	as desired.
	
\end{enumerate}

Exercise 2.4 (Limits of events are preserved in $\sigma$-algebras)

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Exercise 2.4 (Limits of events are preserved in $\sigma$-algebras)

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