3.3. Limit Sets

매우 중요하다.

들어가기에 앞서, set들의 supremum(least upper bound)은 합집합(제일 큰 집합), set들의 infimum(greatest lower bound)은 교집합(제일 작은 집합)임을 염두에 두자.

Definition

For a sequence \(\{A_n: n\ge 1\}\) of subsets of \(\Omega\), we define the limit superior and the limit inferior by \[ \limsup_{n\rightarrow \infty}A_n = \bigcap_{n=1}^\infty \bigcup_{k=n}^\infty A_k\mbox{ }\mbox{ }\mbox{ and }\mbox{ }\mbox{ } \liminf_{n\rightarrow \infty}A_n=\bigcup_{n=1}^\infty\bigcap_{k=n}^\infty A_k. \] If \(\liminf_nA_n = \limsup_nA_n=A\), then we say that the limit of \(A_n\) exists and we write \(\lim_n A_n=A\).

• 중요 : \(\bigcap_{k=n}^\infty A_k\)은 increasing function, \(\bigcup_{k=n}^\infty A_k\)는 decreasing function이다. 또한 \[ \bigcap_{k=n}^\infty A_k \subset A_n \subset \bigcup_{k=n}^\infty A_k \] 이고, 만약 \(\liminf_nA_n = \limsup_nA_n=A\)라면 \[ \bigcap_{k=n}^\infty A_n \uparrow \lim_n A_n=A, \mbox{ }\mbox{ }\mbox{ }\mbox{ } \bigcup_{k=n}^\infty A_n \downarrow \lim_n A_n=A. \]

Remark

1. Note that \(\omega \in \limsup_n A_n\) if and only if \(\omega \in A_n\) infinitely often(i.o) in \(n\): \[ \limsup_n A_n =\{ \omega \in \Omega: \forall n\ge 1\mbox{ } \exists k\ge n \mbox{ s.t. }\omega \in A_k \}=\{\omega\in \Omega: \omega\in A_n \mbox{ i.o.}(n) \}; \] and \(\omega \in \liminf_n A_n\) if and only if \(\omega \in A_n\) for all but finitely many(almost all) \(n\): \[ \liminf_n A_n =\{ \omega \in \Omega: \exists n\ge 1\mbox{ } \mbox{ s.t. }\omega \in A_k \mbox{ }\forall k\ge n \}=\{\omega\in \Omega: \omega\in A_n \mbox{ a.a.}(n) \}. \]

2. From 1. it is clear that \(\liminf_nA_n\subset \limsup_nA_n\) and that \[ I_{\limsup_nA_n}=\limsup_nI_{A_n}\mbox{ }\mbox{ }\mbox{ }\mbox{ }\mbox{ and }\mbox{ }\mbox{ }\mbox{ }\mbox{ } I_{\liminf_nA_n}=\liminf_nI_{A_n} \]
   • Note: \(I_{\limsup_nA_n}=I_{\cap_n \{\cup_{k\ge n} A_n\}}=\wedge_n I_{\cup_{k\ge n} A_n}=\wedge_n\vee_{k\ge n}I_{A_n}=\limsup I_{A_n}\).

3. By de Morgan’s law, \((\limsup_n A_n)^c=\liminf_n A_n^c\) or equivalently \((\liminf_n A_n)^c=\limsup_n A_n^c\).

   • Note: \((\limsup_n A_n)^c=(\cap_n\cup_{k\ge n}A_n)^c=\cup_n\cap_{k\ge n}A_n^c=\liminf_n A_n^c\).

4. The limit superior and the limit inferior can be expressed as monotone limits:

\[ \bigcap_{k=n}^\infty A_n \uparrow \liminf_n A_n, \mbox{ }\mbox{ }\mbox{ }\mbox{ } \bigcup_{k=n}^\infty A_n \downarrow \limsup_n A_n\mbox{ }\mbox{ }\mbox{ }\mbox{ }\mbox{ }\mbox{ as }n\rightarrow \infty. \]

• _{k=n}^A_n은 어차피 점점 작아지는 decreasing function이기 때문에 \(\cap_{n=1}^{\infty}\)를 취해도 결국은 decreasing하는 것은 똑같다고 생각할 수 있다. 때문에 \(\lim_{n\rightarrow \infty}\bigcup_{k=n}^\infty A_n=\bigcap_{n=1}^{\infty}\bigcup_{k=n}^\infty A_n=\limsup_nA_n\). \(\liminf\)도 똑같이 생각하면 된다.

Theorem

If \((\Omega, \mathcal{F},P)\) is a probability space, and \(A_n\in \mathcal{F}\), \(n\ge 1\), then

1. \(P(\liminf_nA_n)\le \liminf_nP(A_n)\le \limsup_n P(A_n)\le P(\limsup_n A_n)\);

2. if \(A_n\rightarrow A\), then \(P(A_n)\rightarrow P(A)\).

• Note : \[\begin{eqnarray*} \cap_{k=n}^\infty A_n \subset A_n\mbox{ }\mbox{ }\mbox{ }\forall n\ge 1&\stackrel{\text{monotone}}{\implies}& P(\cap_{k=n}^\infty A_n)\le P(A_n)\\ &\implies&\lim_nP( \cap_{k=n}^\infty A_n)\le \liminf_n P(A_n)\\ \cap_{k=n}^\infty A_n \uparrow \liminf_n A_n &\stackrel{\text{ctn from below}}{\implies}& P(\cap_{k=n}^\infty A_n)\uparrow P( \liminf_n A_n)\\&\implies& \lim_nP(\cap_{k=n}^\infty A_n)= P( \liminf_n A_n)\\ \end{eqnarray*}\]
  \(\therefore P( \liminf_n A_n)\le \liminf_n P(A_n)\). Also, \[\begin{eqnarray*} \cup_{k=n}^\infty A_n \supset A_n\mbox{ }\mbox{ }\mbox{ }\forall n\ge 1&\stackrel{\text{monotone}}{\implies}& P(\cup_{k=n}^\infty A_n)\ge P(A_n)\\ &\implies&\lim_nP( \cup_{k=n}^\infty A_n)\ge \limsup_n P(A_n)\\ \cup_{k=n}^\infty A_n \downarrow \limsup_n A_n &\stackrel{\text{ctn from above}}{\implies}& P(\cup_{k=n}^\infty A_n)\downarrow P( \limsup_n A_n)\\&\implies& \lim_nP(\cup_{k=n}^\infty A_n)= P( \limsup_n A_n)\\ \end{eqnarray*}\]
  \(\therefore \limsup_n P(A_n)\le P( \limsup_n A_n)\).
  
  

• Let \(A_n\rightarrow A\). Then, \(\limsup_nA_n=\liminf_nA_n=A\). Then, from 1. \[ P(\liminf_nA_n)\le \liminf_nP(A_n)\le \limsup_n P(A_n)\le P(\limsup_n A_n)\\ \implies P(A)\le \liminf_nP(A_n)\le \lim_nP(A_n) \le \limsup_n P(A_n)\le P(A). \] Thus, \(\lim_nP(A_n)=P(A)\).

이 chapter의 핵심 Theorem

Theorem (Borel-Cantelli lemma : convergence half)

If \(\sum_{k=n}^\infty \mu(A_k)<\infty\) for some \(n\ge 1\), then \(\mu(\limsup_n A_n)=0\).

• 우선 \(\sum_{k=n}^\infty \mu(A_k)<\infty\) for some \(n\) \(\implies \lim_{n\rightarrow\infty}\sum_{k=n}^\infty \mu(A_k)=0\)(꼬리부분은 0이다).

• Note that \(\limsup_{n\rightarrow \infty}A_n = \bigcap_{n=1}^\infty \bigcup_{k=n}^\infty A_k \subset \bigcup_{k=n}^\infty A_k\) \(\forall n\ge 1\). Thus, by monotonicity and subadditivity, \[ \mu(\limsup_n A_n)\le \mu(\cup_{k=n}^\infty A_k) \le \sum_{k=n}^\infty \mu(A_k)\mbox{ }\mbox{ }\mbox{ for all }n\ge 1\\ \implies \mu(\limsup_n A_n)\le \lim_{n\rightarrow \infty} \sum_{k=n}^\infty \mu(A_k)=0. \]

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