R&C Analysis-Th 1.29

Suppose $f:X\to [0,\mathrm{\infty }]$ is measurable, and $$\phi (E)={\int }_{E}fd\mu [E\in \mathcal{R}]$$ Then $\phi$ is a measure on $\mathcal{R}$ $${\int }_{X}gd\phi ={\int }_{X}gfd\mu$$ for every measurable function $g$ on $X$ with range in $[0,\mathrm{\infty }]$.

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

To prove Measure, need to show

Let $E_1,E_2,\cdots$ be disjoint members of $\mathcal{R}$ whose union is $E$. Then, $${\chi }_{E}f=\sum _{i=1}^{\mathrm{\infty }}{\chi }_{E_i}f$$ and that $$\phi (E)={\int }_X{\chi }_{E}f\mu \phi (E_j)={\int }_X{\chi }_{E_j}fd\mu$$ Then, $$\sum _{i=1}^{\mathrm{\infty }}\phi (E_i)=\sum _{i=1}^{\mathrm{\infty }}{\int }_X{\chi }_{E_i}fd\mu$$ Using previous theorem on summation of integrals, $$\begin{array}{r}\sum _{i=1}^{\mathrm{\infty }}{\int }_X{\chi }_{E_i}fd\mu ={\int }_X\sum _{i=1}^{\mathrm{\infty }}{\chi }_{E_i}fd\mu \\ ={\int }_X{\chi }_{E}fd\mu \\ =\phi (E)\end{array}$$ Thus $\sum _{i=1}^{\mathrm{\infty }}\phi (E_i)=\phi (E)$ establishing countable additive property. Since $\varphi \in \mathcal{R}$, and $\phi (\varphi )=0$ the finiteness property of atleast one of the sets in sigma algebra is satisfied. This shows $\phi$ 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 $$\begin{array}{r}hf=\sum _{i=1}^{\mathrm{\infty }}{h}_{i}f\\ \text{ and }\\ \phi (E)={\int }_{E}hd\mu \\ \phi (E_i)={\int }_{E_i}{h}_{i}fd\mu \\ \text{ then }\phi (E)\text{ is a measure. }\\ {\int }_{X}hd\mu ={\int }_{X}fgd\mu \end{array}$$ Assume, $$0\le g_1(x)\le g_2(x)\le \cdots \le g(x)$$ with $li{m}_{n\to \mathrm{\infty }}{g}_i(x)=g(x)$. for each $i$ the following is true $${\int }_X{g}_{i}d\mu ={\int }_{X}f{g}_{i}d\mu$$

from monotone conv theorem, LHS is $$li{m}_{n\to \mathrm{\infty }}{\int }_X{g}_{i}d\phi ={\int }_{X}gd\phi$$ as $f$ is positive definite, $$0\le g_1(x)f(x)\le g_2(x)f(x)\cdots \le g(x)f(x)$$ with ${\int }_X{g}_{n}fd\mu \to {\int }_{X}gfd\mu$.

Hence, the equation $${\int }_{X}gd\phi ={\int }_{X}gfd\mu$$ holds.