Definition

The integral of \(f\) over a set \(A\in \mathcal{F}\) is defined by \[ \int_A f \mbox{ }d\mu=\int fI_A \mbox{ }d\mu, \] whenever the latter integral exists. The definition applies even if \(f\) is defined only on \(A\) by taking, e.g., \(f(\omega)=0\) for \(\omega\notin A\).




Proposition

If \(\int f\mbox{ }d\mu\) exists, then so does \(\int f_A\mbox{ }d\mu\) for any \(A\in \mathcal{F}\).




Let \(\nu(A)=\int_A f\mbox{ }d\mu\), \(A\in \mathcal{F}\). This is called indefinite integral of \(f\).

Theorem

If \(\int f\mbox{ }d\mu\) exists and \(A_1,A_2,\ldots\in \mathcal{F}\) are disjoint, then

\[ \nu\left(\biguplus_{n=1}^\infty A_n\right)=\int_{\uplus_{n=1}^\infty A_n}f\mbox{ }d\mu= \sum_{n=1}^\infty\int_{A_n}f\mbox{ }d\mu=\sum_{n=1}^\infty \nu(A_n). \]

\[ \nu(A)= \int fI_A \mbox{ }d\mu= \int \sum_{i=1}^\infty fI_{A_i}d\mu\stackrel{\text{Corollary}}{=}\sum_{i=1}^\infty \int fI_{A_i}d\mu=\sum_{i=1}^\infty \nu(A_i). \]

\[ \nu(A)=\int_A f \mbox{ }d\mu=\int_A f^+ \mbox{ }d\mu- \int_A f^- \mbox{ }d\mu= \sum_{n=1}^\infty\int_{A_n}f^+\mbox{ }d\mu-\sum_{n=1}^\infty\int_{A_n}f^-\mbox{ }d\mu= \sum_{n=1}^\infty\left(\int_{A_n}f^+\mbox{ }d\mu-\int_{A_n}f^-\mbox{ }d\mu\right)=\sum_{n=1}^\infty \int_{A_n}f\mbox{ }d\mu=\sum_{n=1}^\infty\nu(A_n). \]




Suppose that the ingetrals of \(f\) and \(g\) exist and their indefinite integrals are equal, i.e., \[\begin{equation} \int_Af\mbox{ }d\mu=\int_Ag\mbox{ }d\mu\mbox{ }\mbox{ }\mbox{ }\mbox{ for all }A\in \mathcal{F}. \end{equation}\] Under fairly broad conditions this implies that \(f=g\) a.e., as the next theorem shows.

Theorem

Suppose that \(\int f\mbox{ }d\mu\) and \(\int g\mbox{ }d\mu\) exist.

  1. If \(\mu\) is \(\sigma\)-finite on \(\mathcal{F}\), and

    \(\int_Af\mbox{ }d\mu=\int_Ag\mbox{ }d\mu\) for all \(A\in \mathcal{F}\), then \(f=g\) a.e.

  2. If \(f\) and \(g\) are integrable, and

    \(\int_Af\mbox{ }d\mu=\int_Ag\mbox{ }d\mu\) for all \(A\in \mathcal{F}\), then \(f=g\) a.e.

  3. if \(f\) and \(g\) are integrable,

    and \(\int_Af\mbox{ }d\mu=\int_Ag\mbox{ }d\mu\) for all \(A\in \mathcal{P}\), where \(\mathcal{P}\) is a \(\pi\)-system, \(\mathcal{F}=\sigma(\mathcal{P})\), and \(\Omega\) is a finite or countable union of \(\mathcal{P}\)-sets, then \(f=g\) a.e.




Definition

Suppose that \(\delta:\Omega\rightarrow \bar{\mathbb{R}}\) is nonnegative and \(\mathcal{F}\)-measurable, and let \[ \nu(A)=\int_A\delta\mbox{ } d\mu, \mbox{ }\mbox{ }\mbox{ }A\in \mathcal{F}. \] Then, \(\nu\) is a measure on \(\mathcal{F}\), and we say that \(\nu\) has density \(\delta\) with respect to \(\mu\).

\[ \nu\left(\uplus_{n=1}^\infty A_n\right)=\nu(A)= \int \delta I_A \mbox{ }d\mu= \int \sum_{i=1}^\infty \delta I_{A_i}d\mu\stackrel{\text{Corollary}}{=}\sum_{i=1}^\infty \int \delta I_{A_i}d\mu=\sum_{i=1}^\infty \nu(A_i). \]




중요하다

Theorem(Change of Variables)

Suppose that \((\Omega, \mathcal{F}, \mu)\) is a measure space, \((\Omega', \mathcal{F}')\) is a measurable space.

Let \(T:\Omega\rightarrow \Omega'\) is measurable-\(\mathcal{F}/\mathcal{F}'\).

Let \(\mu_T=\mu\circ T^{-1}\) represent the induced measure on \((\Omega', \mathcal{F}')\), and let \(f:\Omega'\rightarrow \bar{\mathbb{R}}\) is measurable-\(\mathcal{F}'/\bar{\mathcal{R}}\).

Then, \[ \int_\Omega f(T(\omega))d\mu(\omega)=\int_{\Omega'}f(\omega')d\mu_T(\omega'), \] in the sense that if either integral exists then so deos the other and they are equal. More generally, for any \(A'\in \mathcal{F}'\),

\[ \int_{T^{-1}(A')} f(T(\omega))d\mu(\omega)=\int_{A'}f(\omega')d\mu_T(\omega'), \] in the same sense as above.


증명

  1. Suppose \(f\) is nonnegative simple function. Then, \[ 0\le f(\omega')=\sum_{i=1}^m a_i I_{A_i'}(\omega'),\mbox{ }\mbox{ } \omega'\in \Omega'. \]

    Then, note that

\[ f(T(\omega))=\sum_{i=1}^m a_i I_{A_i'}(T(\omega))=\sum_{i=1}^m a_i I(T(\omega)\in A_i') =\sum_{i=1}^m a_i I(\omega\in T^{-1}(A_i'))=\sum_{i=1}^m a_i I_{T^{-1}(A_i')}(\omega),\mbox{ }\mbox{ } \omega\in \Omega.\\ \implies\int_\Omega f(T(\omega))d\mu(\omega)=\sum_{i=1}^m a_i \mu({T^{-1}(A_i')})=\sum_{i=1}^m a_i \mu_T(A_i')= \int_{\Omega'}f(\omega')d\mu_T(\omega'). \]

  1. Suppose now that \(f\) is general nonnegative function. let \(\{f_n\}\) be a sequence of simple functions, \(0\le f_n \uparrow f\). Then, note that \(0\le f_n \circ T \uparrow f \circ T\). By applying the result for simple functions together with the monotone convergence theorem (twice), we have \[ \int_{\Omega}f(T(\omega))d\mu(\omega)\stackrel{\text{M.C.T}}{=} \lim_{n\rightarrow \infty}\int_\Omega f_n(T(\omega))d\mu(\omega)=\lim_{n\rightarrow \infty}\int_{\Omega'} f_n(\omega')d\mu_T(\omega')\stackrel{\text{M.C.T}}{=}\int_{\Omega'} f(\omega')d\mu_T(\omega'). \]
  2. For general \(f\), suppose \((f\circ T)^+=f^+\circ T\), and \((f\circ T)^-=f^-\circ T\). THen, \[ \int_{\Omega}f(T(\omega))d\mu(\omega)= \int_{\Omega}f^+(T(\omega))d\mu(\omega)-\int_{\Omega}f^-(T(\omega))d\mu(\omega)=\int_{\Omega'}f^+(\omega')d\mu_T(\omega')- \int_{\Omega'}f^-(\omega')d\mu_T(\omega')= \int_{\Omega'}f(\omega')d\mu_T(\omega'). \]
  3. Finally, note that \(\int_{A'} f(\omega')d\mu_T(\omega')= \int_{\Omega'}f(\omega')I_{A'}(\omega')d\mu_T(\omega')\). Let \(\tilde{f}=fI_{A'}\). Then, \[\begin{eqnarray*} \int_{T^{-1}(A')} f(T(\omega))d\mu(\omega)&=& \int_\Omega f(T(\omega))I_{T^{-1}(A')}(\omega)d\mu(\omega)\\ &=& \int_\Omega f(T(\omega))I_{A'}(T(\omega))d\mu(\omega) \\ &=&\int_{\Omega}\tilde{f}(T(\omega))d\mu(\omega)\\ &=&\int_{\Omega'}\tilde{f}(\omega')d\mu(\omega')\\ &=&\int_{\Omega'}f(\omega')I_{A'}(\omega')d\mu(\omega')\\ &=&\int_{A'}f(\omega')d\mu_T(\omega'). \end{eqnarray*}\]



Example(Change of Variables)

Suppose that \(X\) is a finite-valued random variable on a probability space \((\Omega, \mathcal{F},P)\), i.e., \(X:\Omega\rightarrow \mathbb{R}\) is \(\mathcal{F}\)-measurable. Then for any Borel measurable \(g\)

Let \(P_X=P\circ X^{-1}\) be the distribution of \(X\) (a probability measure on \((\mathbb{R}, \mathcal{R})\)), and let \(F_X\) be the corresponding distribution function

\[ E[g(X)]= \int g(X)dP= \int_\Omega g(X(\omega))dP(\omega)= \int_\mathbb{R} g(x)dP_{x}(x) \]




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