证明$\int g d{\mu _f} = \int {g \cdot } fd\mu $，其中$ {\mu _f}\left( A \right) = \int_A f d\mu $

要证明:若定义测度$ {\mu _f}\left( A \right) = \int_A f d\mu $ ,（有${\mu _f}\left( A \right) \ll \mu \left( A \right)$), 则$\int g d{\mu _f} = \int {g \cdot } fd\mu $.

1)$ g = 1_{E} $ :

\begin{equation}
\int g d{\mu _f} = \int {{1_E}} d{\mu _f} = {\mu _f}\left( E \right) = \int_E f d\mu = \int {{1_E}} fd\mu = \int {g \cdot } fd\mu
\end{equation}

2) $ g = \sum\limits_{i = 1}^n {{c_i}{1_{{E_i}}}} $ :

\begin{equation}
\int g d{\mu _f} = \int {\sum\limits_{i = 1}^n {{c_i}{1_{{E_i}}}} } d{\mu _f} = \sum\limits_{i = 1}^n {{c_i}} \int {{1_{{E_i}}}} d{\mu _f} = \sum\limits_{i = 1}^n {{c_i}} \int {{1_{{E_i}}}} fd\mu = \int {\left( {\sum\limits_{i = 1}^n {{c_i}} {1_{{E_i}}}} \right)} fd\mu = \int {g \cdot } fd\mu
\end{equation}

3)$ g\ge 0 $:

$$ g_{n} \uparrow g , \int g d{\mu _f}\mathop = \limits^{DEFINITION} {\mkern 1mu} \mathop {\lim }\limits_{n \to \infty } {\mkern 1mu} \int {{g_n}} d{\mu _f} = \mathop {\lim }\limits_{n \to \infty } {\mkern 1mu} \int {{g_n} \cdot f} d\mu $$

With $ g_{n} \uparrow g $ and monotone converge theorem, we have:

$$ g_{n} \cdot f \uparrow g \cdot f$$

$$\mathop {\lim }\limits_{n \to \infty } \int {{g_n}} d{\mu _f} = \mathop {\lim }\limits_{n \to \infty } \int {{g_n} \cdot f} d\mu = \int {g \cdot f} d\mu $$

$$\int g d{\mu _f} = \int {g \cdot f} d\mu $$
4)g: $g={{g}^{+}}-{{g}^{-}}$

\begin{equation}
\int g d{\mu _f} = \int {{g^ + }} d{\mu _f} - \int {{g^ - }} d{\mu _f} = \int {{g^ + } \cdot f} d\mu - \int {{g^ - } \cdot f} d\mu = \int {\left( {{g^ + } - {g^ - }} \right) \cdot f} d\mu = \int {g \cdot f} d\mu
\end{equation}