若${{\mu }_{X}}\left( B \right)=P\left( X\left( \omega \right)\in B \right)$是与随机变量$X\left( \omega \right)$相关联的测度，其中$B\in \mathcal{B}$, $B\subseteq \mathbb{R}$，则可测函数$g:\mathbb{R}\to \mathbb{R}$，有$\int{g\left( x \right)d{{\mu }_{X}}=\int{g\left( X\left( \omega \right) \right)d}}\mathbb{P}=\int{g}\circ X\left( \omega \right)d\mathbb{P}$
(1) $ g\left( x \right)={{1}_{E}} $:
$$\int{gd{{\mu }_{X}}}=\int{{{1}_{E}}d{{\mu }_{X}}={{\mu }_{X}}\left( E \right)=P\left( X\left( \omega \right)\in E \right)}$$
而$g\circ X\left( \omega \right)={{1}_{E}}\left( X\left( \omega \right) \right)=\left\{ \begin{align}
& 1 \quad \text{ if: X}\left( \omega \right)\in E \\
& 0 \quad \text{ others}
\end{align} \right.={{1}_{X\left( \omega \right)\in E}}\left( \omega \right)$，则
$$\int{g}\circ X\left( \omega \right)d\mathbb{P}=\int{{{1}_{X\left( \omega \right)\in E}}\left( \omega \right)d}\mathbb{P}=P\left( X\left( \omega \right)\in E \right)$$
于是有$\int{gd{{\mu }_{X}}}=\int{g}\circ X\left( \omega \right)d\mathbb{P}$。
(2) $g=\sum\limits_{i=1}^{n}{{{c}_{i}}{{1}_{{{E}_{i}}}}},{{c}_{i}}\ge 0$，其中${{E}_{i}}$互不相交：
$$\int{gd{{\mu }_{X}}}=\int{\sum\limits_{i=1}^{n}{{{c}_{i}}{{1}_{{{E}_{i}}}}}}d{{\mu }_{X}}=\sum\limits_{i=1}^{n}{{{c}_{i}}\int{{{1}_{{{E}_{i}}}}}d{{\mu }_{X}}}=\sum\limits_{i=1}^{n}{{{c}_{i}}\int{{{1}_{{{E}_{i}}}}}\circ X\left( \omega \right)d\mathbb{P}}=\sum\limits_{i=1}^{n}{{{c}_{i}}\int{{{1}_{X\left( \omega \right)\in {{E}_{i}}}}}d\mathbb{P}}$$
$$g\circ X\left( \omega \right)=\sum\limits_{i=1}^{n}{{{c}_{i}}{{1}_{{{E}_{i}}}}}\left( X\left( \omega \right) \right)=\left\{ \begin{align}
& {{\text{c}}_{i}} \quad \quad \text{ if X}\left( \omega \right)\in {{E}_{i}} \\
& 0 \quad \quad \text{ others} \\
\end{align} \right.=\sum\limits_{i=1}^{n}{{{c}_{i}}{{1}_{X\left( \omega \right)\in {{E}_{i}}}}\left( \omega \right)}$$
$$\int{g}\circ X\left( \omega \right)d\mathbb{P}=\int{\sum\limits_{i=1}^{n}{{{c}_{i}}{{1}_{X\left( \omega \right)\in {{E}_{i}}}}\left( \omega \right)}d\mathbb{P}}=\sum\limits_{i=1}^{n}{{{c}_{i}}}\int{{{1}_{X\left( \omega \right)\in {{E}_{i}}}}\left( \omega \right)d\mathbb{P}}$$
于是有$$\int{gd{{\mu }_{X}}}=\int{g}\circ X\left( \omega \right)d\mathbb{P}$$
（3）$g\ge 0,{{g}_{n}}\uparrow g$:
通过单调收敛定理可得：$$\int{gd{{\mu }_{X}}}=\underset{n\to \infty }{\mathop{\lim }}\,\int{{{g}_{n}}d{{\mu }_{X}}}=\underset{n\to \infty }{\mathop{\lim }}\,\int{{{g}_{n}}\circ X\left( \omega \right)d\mathbb{P}}$$
而${{g}_{n}}\left( x \right)\uparrow g\left( x \right),\forall x\in \mathbb{R}$
所以有${{g}_{n}}\left( X\left( \omega \right) \right)\uparrow g\left( X\left( \omega \right) \right),\forall \omega \in \Omega $，即${{g}_{n}}\circ X\left( \omega \right)\uparrow g\circ X\left( \omega \right)$
通过单调收敛定理可得：$$\int{g}\circ X\left( \omega \right)d\mathbb{P}=\int{\underset{n\to \infty }{\mathop{\lim }}\,\left( {{g}_{n}}\circ X\left( \omega \right) \right)d\mathbb{P}}=\underset{n\to \infty }{\mathop{\lim }}\,\int{{{g}_{n}}\circ X\left( \omega \right)d\mathbb{P}}$$
即可得到：$\int{gd{{\mu }_{X}}}=\int{g}\circ X\left( \omega \right)d\mathbb{P}$
(4)$g$ integrable function, $g={{g}^{+}}-{{g}^{-}}$,
$$\int{gd{{\mu }_{X}}}=\int{{{g}^{+}}d{{\mu }_{X}}}-\int{{{g}^{-}}d{{\mu }_{X}}=}\int{{{g}^{+}}}\circ X\left( \omega \right)d\mathbb{P}-\int{{{g}^{-}}}\circ X\left( \omega \right)d\mathbb{P}=\int{\left( {{g}^{+}}\circ X\left( \omega \right)-{{g}^{-}}\circ X\left( \omega \right) \right)}d\mathbb{P}$$
只要验证${{g}^{+}}\circ X\left( \omega \right)-{{g}^{-}}\circ X\left( \omega \right)=g\circ X\left( \omega \right)$即完成证明。
而$${{g}^{+}}\circ X\left( \omega \right)-{{g}^{-}}\circ X\left( \omega \right)=\left\{ \begin{align}
& {{g}^{+}}\circ X\left( \omega \right)-0=g\circ X\left( \omega \right)\text{ }\quad \quad g\circ X\left( \omega \right)\ge 0 \\
& 0-{{g}^{-}}\circ X\left( \omega \right)\text{=}g\circ X\left( \omega \right)\text{ } \quad \quad g\circ X\left( \omega \right)<0
\end{align} \right.$$
可知$\forall \omega \in \Omega ,{{g}^{+}}\circ X\left( \omega \right)-{{g}^{-}}\circ X\left( \omega \right)=g\circ X\left( \omega \right)$，得证。