Let $A_n$ be a sequence of Events. Find the Lim Sup

Question

Let $(\Omega, \mathcal{F})$ be a Measurable Space

Let $\{A_n\}_{n=1}^{\infty}$ be a sequence of events. Consider,$$ F_n = \liminf_k \; (A_n \cap A_k^c ) $$

Prove that $\limsup_n F_n = \emptyset $

This seems almost intuitively obvious but I cant make an arguement to show it.

My reasoning,

$$F_n = \{x \in \Omega : \exists \; k_n ; \; x \in (A_n \cap A_k^c ) \; \forall k \geq k_n\}$$

If I can somehow show now that there is some $n \geq k_n$ then I would be done but I'm finding this difficult to show.

Answer 1

If $x\in\limsup_{k\to\infty}F_k$, then for each positive integer $k$, there exists $j\geqslant k$ such that $x\in F_j$. Hence there is a subsequence $F_{j_n}$ such that $x\in\bigcup_{n=1}^\infty F_{j_n}$. But this means \begin{align} x\in \liminf_{k\to\infty} \left(A_{j_n}\cap A_k^c\right) &= \bigcup_{k=1}^\infty\bigcap_{l=k}^\infty \left(A_{j_n}\cap A_l^c\right)\\ &= A_{j_n} \cap \left(\bigcup_{k=1}^\infty\bigcap_{l=k}^\infty A_l^c \right). \end{align} Therefore there exists $k$ such that $x\in A_l^c$ for $l\geqslant k$. Choosing $k<j_m$ for some $m$ thus implies that $x\in A_{j_m}\cap A_{j_m}^c$, a contradiction.

Answer 2

By definition of $\liminf$ $$ F_n=\bigcup_{l\geqslant 1}\bigcap_{k\geqslant l}( A_{n} \cap A_k^c)=A_{n} \cap \left(\bigcup_{l\geqslant 1}\bigcap_{k\geqslant l} A_k^c \right) $$ So \begin{align} \limsup_{n}F_n&=\bigcap_{m\geqslant 1}\bigcup_{n\geqslant m}\left(A_{n} \cap \bigcup_{l\geqslant 1}\bigcap_{k\geqslant l} A_k^c \right) \\ &=\bigcap_{m\geqslant 1}\bigcup_{n\geqslant m}\left(A_{n} \cap B \right) \\ &=\left(\bigcap_{m\geqslant 1}\bigcup_{n\geqslant m}A_{n}\right) \cap B \tag1 \\ &=\left(\bigcap_{m\geqslant 1}\bigcup_{n\geqslant m}A_{n}\right) \cap \left( \bigcup_{l\geqslant 1}\bigcap_{k\geqslant l} A_k^c \right) \\ &=\left(\bigcap_{m\geqslant 1}\bigcup_{n\geqslant m}A_{n}\right) \cap \left( \bigcup_{m\geqslant 1}\bigcap_{n\geqslant m} A_n^c \right) \\ &=\left(\bigcap_{m\geqslant 1}\bigcup_{n\geqslant m}A_{n}\right) \cap \left( \bigcap_{m\geqslant 1}\bigcup_{n\geqslant m} A_n \right)^c\tag2 \\ &=\varnothing \end{align} $(1)$. $B$ is free of index $m,n$.

$(2)$. By De Morgan's Law.