3.4. Convergence in Measure

중요

Theorem(Almost Everywhere Convergence)

The followings are equivalent;

1. \(f_n\rightarrow f\) a.e.

2. \(\mu(\{ \omega : |f_n(\omega)-f(\omega)|>\epsilon \mbox{ i.o.}(n) \})=0\) for all \(\epsilon>0\).

3. \(\mu(\{ \omega : |f_n(\omega)-f(\omega)|>1/k \mbox{ i.o.}(n) \})=0\) for all integer \(k\ge1\).

• 증명: 2와 3은 당연히 같다. 그러므로 \[\begin{eqnarray*} \{ \omega: f_n(\omega)\nrightarrow f(\omega)\}&=& \cup_{\epsilon>0}\cap_{n\ge1}\cup_{m\ge n}\{\omega:|f_m(\omega)-f(\omega)|>\epsilon \}\\ &=&\cup_{k\ge 1}\cap_{n\ge1}\cup_{m\ge n}\{\omega:|f_m(\omega)-f(\omega)|>1/k\}\\ &=&\cup_{k\ge 1} \limsup_n\{\omega:|f_m(\omega)-f(\omega)|>1/k\}\\ &=&\cup_{k\ge 1} \{\omega:|f_m(\omega)-f(\omega)|>1/k \mbox{ }\mbox{ i.o}(n)\}. \end{eqnarray*}\] Thus by monotonicity, \[\begin{eqnarray*} \{\omega:|f_m(\omega)-f(\omega)|>1/k \mbox{ }\mbox{ i.o}(n)\}&\in& \cup_{k\ge 1} \{\omega:|f_m(\omega)-f(\omega)|>1/k \mbox{ }\mbox{ i.o}(n)\}\\ &=&\{ \omega: f_n(\omega)\nrightarrow f(\omega)\} \mbox{ for all } k\ge 1\\ \implies \mu(\{\omega:|f_m(\omega)-f(\omega)|>1/k \mbox{ }\mbox{ i.o}(n)\})&\le&\mu(\{ \omega: f_n(\omega)\nrightarrow f(\omega)\}), \end{eqnarray*}\] and by subadditivity, \[\begin{eqnarray*} \mu(\{ \omega: f_n(\omega)\nrightarrow f(\omega)\})&=& \mu\left( \cup_{k\ge 1}\{\omega:|f_m(\omega)-f(\omega)|>1/k \mbox{ }\mbox{ i.o}(n)\}\right)\\ &\le& \sum_{k= 1}^\infty\mu(\{\omega:|f_m(\omega)-f(\omega)|>1/k \mbox{ }\mbox{ i.o}(n)\}). \end{eqnarray*}\] Thus, \[\begin{eqnarray*} \mu(\{ \omega: f_n(\omega)\nrightarrow f(\omega)\})=0 \iff \mu(\{\omega:|f_m(\omega)-f(\omega)|>1/k \mbox{ }\mbox{ i.o}(n)\})=0 \mbox{ for all }k\ge 1. \end{eqnarray*}\]

Corollary

If for each \(\epsilon>0\), there exists an \(n\ge 1\) such that \(\sum_{k=n}^\infty \mu[|f_k-f|>\epsilon]<\infty\), then \(f_n\rightarrow f\) a.e.

• By Borel-Cantelli lemma, \[\begin{eqnarray*} \sum_{k=n}^\infty \mu[|f_k-f|>\epsilon]<\infty&\implies& \mu(\limsup_n\{\omega: f_n(\omega)-f(\omega)|>\epsilon\})=0\\ &\implies& \mu(|f_n-f|>\epsilon \mbox{ } \mbox{ } \mbox{ i.o}(n))=0\\ &\implies& f_n\rightarrow f \mbox{ a.e.} \end{eqnarray*}\]

Definition(Convergence in Measure)

A sequence of measurable function \(\{f_n:n\ge 1\}\) converges in measure to a measurable function \(f\) (written \(f_n\stackrel{\mu}\rightarrow f\)) if \[ \mu(\{\omega:|f_n(\omega)-f(\omega)|>\epsilon\})\rightarrow 0 \mbox{ }\mbox{ }\mbox{ }\mbox{ as }n\rightarrow \infty, \mbox{ }\mbox{ for all }\epsilon>0. \] In special case of random variables, we say that \(X_n\) converges in probability to \(X\) (written \(X_n\stackrel{\mu}\rightarrow X\)) if \[ P(|X_n-X|>\epsilon)\rightarrow 0 \mbox{ }\mbox{ }\mbox{ }\mbox{ as }n\rightarrow \infty, \mbox{ }\mbox{ for all }\epsilon>0. \]

• This is weaker than the Almost Everywhere convergence (왜냐하면 a.e. convergence는 \(\mu(|X_n-X|>\epsilon \mbox{ }\mbox{ }\mbox{i.o}(n))=0\)인데 반해 convergence in measure는 이 값이 0으로 근사함을 나타내기 때문이다).

Proposition

If \(f_n\stackrel{\mu}\rightarrow f\) and \(f_n\stackrel{\mu}\rightarrow g\), then \(f=g\) a.e.

• Note that \[ \{ \omega:|f(\omega)-g(\omega)|>\epsilon\}\subset \{ \omega:|f_n(\omega)-f(\omega)|>\epsilon/2\}\cup \{ \omega:|f_n(\omega)-g(\omega)|>\epsilon/2\} \] 왜냐하면 만약 \(\{|f_n(\omega)-f(\omega)|>\epsilon/2\}\cup \{ \omega:|f_n(\omega)-g(\omega)|>\epsilon/2\}\)이라면 \(\{ \omega:|f(\omega)-g(\omega)|>\epsilon\}\)일 수도 있고 부호가 반대일 경우 \(\{ \omega:|f(\omega)-g(\omega)|<\epsilon\}\)일 수도 있기 때문이다.

• 증명:

\[ \mu(\{ \omega:|f(\omega)-g(\omega)|>\epsilon\})\le \mu(\{|f_n(\omega)-f(\omega)|>\epsilon/2\})+\mu(\{ \omega:|f_n(\omega)-g(\omega)|>\epsilon/2\})\rightarrow 0 \mbox{ }\mbox{ }\mbox{ as }n\rightarrow \infty. \]

중요

Theorem

If \(f_n\stackrel{\mu}\rightarrow f\), then there exists a subsequence \(\{f_{n_k}: k\ge 1\}\) for which \(f_{n_k}\rightarrow f\) a.e.

• 증명: For a sequence \(n_1<n_2<\cdots\)

\[\begin{eqnarray*} f_n\stackrel{\mu}\rightarrow f &\implies& f_{n_k}\stackrel{\mu}\rightarrow f \mbox{ }\mbox{ }\mbox{ }\mbox{ as }k\rightarrow \infty\\ &\implies&\mu(|f_{n_k}-f|>\epsilon)\rightarrow 0\mbox{ }\mbox{ }\mbox{ }\mbox{ as }k\rightarrow \infty\\ &\implies&\mu(|f_{n_k}-f|>\epsilon)<2^{-k} \mbox{ for each }k. \end{eqnarray*}\] Thus, \[ \mu(|f_{n_k}-f|>\epsilon)<2^{-k}\mbox{ }\mbox{ }\mbox{ }\forall k>1/\epsilon\\ \implies \sum_{k>1/\epsilon} \mu(|f_{n_k}-f|>\epsilon)<\infty \mbox{ }\mbox{ }\forall \epsilon>0, \] which implies that \(f_{n_k}\rightarrow f\) a.e. by the second corollary in this page.

• 포인트는 \(f_n\stackrel{\mu}\rightarrow f\)인 sequence \(f_n\)이 \(f_{n_k}\rightarrow f\) a.e.를 만족하는 subsequence \(n_k\)를 반드시 갖는다는 것이다.

Theorem(Dominated Convergence in Measure)

If \(f_n\stackrel{\mu}\rightarrow f\) and if there exists an integrable function \(g\) for which \(|f_n|\le g\) a.e. for all \(n\ge 1\), then \(f_n\), and \(f\) are integrable and \(\int f_n\mbox{ }d\mu\rightarrow \int f\mbox{ }d\mu\) as \(n\rightarrow \infty\).

• 증명 :

1. \(|f_n|\le g\)이기 때문에 \(f_n\)들은 integrable하다.

2. \(f_n\stackrel{\mu}\rightarrow f\)기 때문에 for any subsequence \(\{ {n_k}\}\), \(f_{n_k}\stackrel{\mu}\rightarrow f\)이다. 때문에 이전 Theorem에 의해 \(f_{n_{kj}}\rightarrow f\) a.e.인 sub-subsequence \(\{ n_{kj}\}\)가 존재한다.

3. 때문에 D.C.T에 의해 \(\int f_{n_{kj}}\mbox{ }d\mu\rightarrow \int f\mbox{ }d\mu\)이다. 이는 for any subsequence \(\{ {n_k}\}\)는 이를 만족하는 sub-subsequence를 갖는다는 의미이다.

4. Let \(a_{n}:=\int f_{n}\), \(a_{n_{kj}}:=\int f_{n_{kj}}\),and \(a:=\int f\mbox{ }d\mu\). From above results, \[a_n\rightarrow a\] i.e., \(\int f_{n}\mbox{ }d\mu\rightarrow \int f\mbox{ }d\mu\).

• 기존의 Lebesgue D.C.T와 다 똑같고 \(f_n\rightarrow f\) a.e.만 \(f_n\stackrel{\mu}\rightarrow f\) 로 바뀌었을 뿐이다.

• 중요한 것은 더 weaker한 condition인데도(a.e. convergence에서 convergence in measure로 바뀌었는데도) 결론은 똑같다.

Definition

A sequence of measurable functions \(\{f_n:n\ge 1\}\) is Cauchy in measure if for all \(\epsilon>0\), \[ \mu(\{\omega:|f_m(\omega)-f_n(\omega)|>\epsilon \})\rightarrow 0\mbox{ }\mbox{ }\mbox{ as }m,n\rightarrow \infty. \] In special case of random variables, we say that \(X_n\) Cauchy in probability to if for all \(\epsilon>0\), \[ P(|X_m-X_n|>\epsilon)\rightarrow 0 \mbox{ }\mbox{ }\mbox{ }\mbox{ as }m,n\rightarrow \infty. \]

Remark

An equivalent formation is \[ \sup_{m\ge n} \mu(\{\omega:|f_m(\omega)-f_n(\omega)|>\epsilon \})\rightarrow 0\mbox{ }\mbox{ }\mbox{ as }n\rightarrow \infty. \] Similarly, \[ \sup_{m\ge n}P(|X_m-X_n|>\epsilon)\rightarrow 0 \mbox{ }\mbox{ }\mbox{ }\mbox{ as }n\rightarrow \infty. \]

• \(n\)에 \(m\)을 태우고 supremum을 씌워도 똑같다(limit이 존재하기 때문).

Theorem(Completeness for Convergence in Measure)

A sequence of measurable functions \(\{f_n:n\ge 1\}\) is Cauchy in measure

if and only if there exists a measurable function f for which \(f_n\stackrel{\mu}\rightarrow f\) as \(n\rightarrow \infty\).

• 즉 Cauchy in measure와 convergence in measure (say, to \(f\))는 동치이다.

• Cauchy in measure\(\iff\)convergence in measure.

back