If \((\Omega, \mathcal{F})\) and \((\Omega', \mathcal{F}')\) are measurable spaces, then the mapping \(T:\Omega \rightarrow \Omega'\) is said to be measurable-\(\mathcal{F}/\mathcal{F}'\)(or \(\mathcal{F}/\mathcal{F}'\)-measurable) if \[ T^{-1}(A')\in \mathcal{F} \mbox{ for all } A' \in \mathcal{F}'. \]
If \((\Omega, \mathcal{F}, P)\) is a probability space, then function \(X:\Omega\rightarrow \mathbb{R}\) is called a random variable if \(X\) is \(\mathcal{F}/\mathcal{R}\)-measurable.
즉, function \(X:\Omega\rightarrow \mathbb{R}\)에 대해 \(X^{-1}(A')\in \mathcal{F}\) for any \(A'\in \mathcal{R}\)이면 \(X\)를 Random Variable이라 한다.
여기서 Random variable은 반드시 measurable해야 한다는게 중요하다.
A composition of measurable maps is measurable, i.e., if \((\Omega, \mathcal{F})\), \((\Omega', \mathcal{F}')\), \((\Omega'', \mathcal{F}'')\) are measurable spaces and the mapping \(T:\Omega\rightarrow \Omega'\) and \(T':\Omega'\rightarrow \Omega''\) are measurable (in the obvious senses), then the mapping \(T'\circ T:\Omega\rightarrow \Omega''\) is measurable.
If \((\Omega, \mathcal{F})\) and \((\Omega', \mathcal{F}')\) are measurable spaces and if \(\mathcal{F}'=\sigma(\mathcal{A})\), then a mapping \(T:\Omega \rightarrow \Omega'\) is measurable if and only if \[ T^{-1}(A')\in \mathcal{F} \mbox{ for all } A' \in \mathcal{A}. \]
*만약 \(\mathcal{A}\)가 \(\sigma\)-field \(\mathcal{F}\)를 generate하는 class of set이라면, (\(\mathcal{F}'=\sigma(\mathcal{A})\)), mapping \(T\)의 measurability를 확인하기 위해서는 \(\mathcal{A}\)에 대해서만 다루면 된다. (basis의 개념과 비슷하다.)
Let \((\Omega, \mathcal{F})\) be a measurable space and let \(f:\Omega \rightarrow \mathbb{R}\). Then, \(f\) is measurable-\(\mathcal{F}/\mathcal{R}\) if and only if \[ \{\omega: f(\omega)\le x\}= f^{-1}((-\infty,x])\in \mathcal{F},\mbox{ } \forall -\infty<x<\infty. \]
If \(f_j:\Omega\rightarrow \mathbb{R}\), \(j=1,2,\ldots,k\) are \(\mathcal{F}\)-measurable, then so are each of;
\(\sum_{i=1}^k f_i\);
\(\prod_{i=1}^k f_i\);
\(\max_{1\le i\le k} f_i\);
\(\min_{1\le i\le k} f_i\);
Similarly, if \(f:\Omega\rightarrow \mathbb{R}\) is \(\mathcal{F}\)-measurable, then so are \(e^f\), \(\sin(f)\), and \(f^2\). The function \(1/f\) is also \(\mathcal{F}\)-measurable as long as \(f(\omega)\ne 0\) for all \(\omega \in \Omega\).
Suppose that \(f,g:\Omega\rightarrow \mathbb{R}\) are measurable-\(\mathcal{F}/\mathcal{R}\). Then, each of the following functions is measurable-\(\mathcal{F}/\mathcal{R}\):
\(c\) where \(c\) is any constant;
\(cf\) where \(c>0\) is a constant;
\(-f\);
\(1/f\) if it is well defined (i.e., \(f(\omega)\ne 0\) for all \(\omega \in \Omega\));
\(f+g\) if it is well defined (i.e., never \(\infty-\infty\) or \(-\infty+\infty\));
\(f\vee g =\max(f,g)\) and \(f \wedge g= \min(f,g)\);
\(f^+=f\vee 0\) and \(f^-=-(f\wedge 0)\);
\(|f|\);
\(fg\) where we define \(0\times \infty=0\);
\(f/g\) if it is well defined (i.e., \(g(\omega)\ne 0\) for all \(\omega \in \Omega\)), where we take \(\infty/\infty=0\).
if \(f,g:\Omega \rightarrow \mathbb{R}\) are both measurable-\(\mathcal{F}/\mathcal{R}\), then \[ \{\omega: f(\omega)<g(\omega) \}, \{f(\omega)\le g(\omega) \}, \{f(\omega)=g(\omega)\}\in \mathcal{F}. \]
Suppose that \(f,g:\Omega\rightarrow \mathbb{R}\) are measurable-\(\mathcal{F}/\mathcal{R}\). Then, each of the following functions is measurable-\(\mathcal{F}/\mathcal{R}\):
\(\inf_n f_n\) and \(\sup_n f_n\) ;
\(\liminf_n f_n\) and \(\limsup_n f_n\) ;
\(\lim_n f_n\) if it exists.