There are the theorems which we need for the proof of this theorem:
There is the theorem $1.29$ :
Suppose $f: X \to [0,\infty]$ is measurable, and
$\phi(E) = \int_E f \,d\mu$ ( $E$ $\in$ $\mathfrak M $ ( it's $\sigma$-algebra)
Then $\phi$ is a measure on $\mathfrak M$, and
$\int_X g d\phi$ $=$ $\int_X gf \,d\mu$
for every measurable $g$ on $X$ with range in $[0, \infty]$.
There is the proof:
Let $E_1, E_2, E_3,…$ be disjoint members of $\mathfrak M$ whose union is $E$.
observe that $\chi_E$$f$ $=$ $\sum_{j=1}^\infty$ $\chi_{E_j}$$f$
and that
$\phi(E)$ $=$ $\int_X$ $\chi_E$$f$ $d\mu$
$\phi(E_j)$ $=$ $\int_X$ $\chi_{E_j}$$f$ $d\mu$
It now follows from theorem $1.27$ that
$\phi(E)$ $=$ $\sum_{j=1}^\infty$ $\phi(E_j)$
Since $\phi(\emptyset)$ $=$ $0$, the last equality proofs that $\phi$ is a measure.
Next, $\phi(E) = \int_E f \,d\mu$ ( $E$ $\in$ $\mathfrak M $ ( it's $\sigma$-algebra) shows that $\int_X g d\phi$ $=$ $\int_X gf \,d\mu$ holds whenever $g$ $=$ $\chi_E$ for some $E$ $\in$ $\mathfrak M$. Hence $\int_X g d\phi$ $=$ $\int_X gf \,d\mu$ holds for every simple measurable function g, and the general case follows from the monotone convergence theorem.
I don't understand how does the general case follows from the monotone convergence theorem.
Any help would be appreciated.
Best Answer
First, assume that $g \ge 0$. Choose a sequence of simple measurable functions $\{g_n\}$ such that $g_n \nearrow g$. Then since $f \ge 0$, also $g_n f \nearrow gf$ so that $$\int_X g d\phi = \lim_n \int_X g_n d\phi = \lim_n \int_X g_nf d \mu = \int_X gf d\mu.$$ Here the first and last equality follow from two applications of the monotone convergence theorem, while the middle equality follows because you know the result for simple functions.
If $g$ is real-valued, write it as $g= g^+ - g^-$ with $g^+, g^- \ge 0$ to deduce the result in general.