R&C Analysis-Th 1.29

Suppose \(f:X \rightarrow [0,\infty]\) is measurable, and \[\begin{equation} \varphi(E) = \int_E f d\mu \text{ } [E \in \mathscr{R}] \end{equation}\] Then \(\varphi\) is a measure on \(\mathscr{R}\) \[\begin{equation} \int_X g d\varphi = \int_X gf d\mu \end{equation}\] for every measurable function \(g\) on \(X\) with range in \([0,\infty]\).

In this set up, as we tour through different sets of sigma algebra \(\mathscr{R}\), the integral of measurable function generates a measure.

To prove Measure, need to show

Let \(E_1,E_2,\cdots\) be disjoint members of \(\mathscr{R}\) whose union is \(E\). Then, \[\begin{equation} \chi_E f = \sum_{i=1}^\infty \chi_{E_i}f \end{equation}\] and that \[\begin{equation} \varphi(E) = \int_X \chi_E f \mu \text{ } \varphi(E_j) = \int_X\chi_{E_j}f d\mu \end{equation}\] Then, \[\begin{equation} \sum_{i=1}^\infty \varphi(E_i) = \sum_{i=1}^\infty \int_X \chi_{E_i} f d\mu \end{equation}\] Using previous theorem on summation of integrals, \[\begin{align} \sum_{i=1}^\infty \int_X \chi_{E_i} f d\mu = \int_X\sum_{i=1}^\infty\chi_{E_i} f d\mu \\ = \int_X \chi_E f d\mu \\ = \varphi(E) \end{align}\] Thus \(\sum_{i=1}^\infty \varphi(E_i) = \varphi(E)\) establishing countable additive property. Since \(\phi \in \mathscr{R}\), and \(\varphi(\phi)=0\) the finiteness property of atleast one of the sets in sigma algebra is satisfied. This shows \(\varphi\) is a measure. Each \(\chi\) in above equations is a simple function. Setting \(g=\chi\) If we set \(h=\chi_E\), for any simple measurable function. then \[\begin{align*} hf = \sum_{i=1}^\infty h_if \\ \text{ and } \\ \varphi(E) = \int_E h d\mu \\ \varphi(E_i)=\int_{E_i} h_i f d\mu \\ \text{ then $\varphi(E)$ is a measure. } \\ \int_X hd\mu = \int_X fg d\mu \end{align*}\] Assume, \[\begin{equation*} 0 \leq g_1(x) \leq g_2(x) \leq \cdots \leq g(x) \end{equation*}\] with \(lim_{n \rightarrow \infty} g_i(x) = g(x)\). for each \(i\) the following is true \[\begin{equation} \int_X g_i d\mu = \int_X fg_i d\mu \end{equation}\]

from monotone conv theorem, LHS is \[\begin{equation} lim_{n \rightarrow \infty} \int_X g_i d\varphi = \int_X g d\varphi \end{equation}\] as \(f\) is positive definite, \[\begin{equation} 0 \leq g_1(x)f(x) \leq g_2(x)f(x) \cdots \leq g(x)f(x) \end{equation}\] with \(\int_X g_nf d\mu \rightarrow \int_X gf d\mu\).

Hence, the equation \[\begin{equation} \int_X g d\varphi = \int_X gf d\mu \end{equation}\] holds.