Proof of Expectations Computed from the Law

Throughout, $P_{X}=P\circ X^{-1}$ is the image measure of $P$ under $X\in\mathrm{Meas}(\Omega,\mathcal{F};S,\mathcal{B}(S))$ (The Law of a Random Element (i), Probability Spaces, Events and Random Elements (iii)), and expectations are integrals against $P$ (Expectation (i), Expectation (ii), Expectation (iii)).

Nonnegative functions. By Image Measures and Measures with a Density: Measures, Integrals and Change of Variables (ii) with $T=X$, $g\circ X\in\mathcal{M}^{+}(\Omega,\mathcal{F})$ and

$$\int_{S}g\,dP_{X}=\int_{\Omega}g\circ X\,dP=\mathbb{E}\,g(X).$$

This is (i).

Real functions. By Image Measures and Measures with a Density: Measures, Integrals and Change of Variables (iii) with $T=X$, $g\circ X:\Omega\to\mathbb{R}$ is measurable, that is, a random variable (Probability Spaces, Events and Random Elements (iv)); $g\in\mathcal{L}^{1}(P_{X})$ if and only if $g\circ X\in\mathcal{L}^{1}(P)$; and then $\int_{S}g\,dP_{X}=\int_{\Omega}g\circ X\,dP=\mathbb{E}\,g(X)$. This is (ii).

Vector-valued functions. The function $g\circ X$ is measurable by Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (ii), so it is a random vector in $\mathbb{R}^{m}$ (Probability Spaces, Events and Random Elements (v)). Its components are $(g\circ X)_{k}=g_{k}\circ X$ ($k\in[m]$), and each $g_{k}:S\to\mathbb{R}$ is measurable by Vector- and Matrix-Valued Measurable Maps: Components, Entries and Continuous Images (i). By Integrable Vector- and Matrix-Valued Functions and Their Integrals (i) and (ii) applied to each $g_{k}$,

$$g\circ X\in\mathcal{L}^{1}(P;\mathbb{R}^{m})\iff g_{k}\circ X\in\mathcal{L}^{1}(P)\ \ (k\in[m])\iff g_{k}\in\mathcal{L}^{1}(P_{X})\ \ (k\in[m])\iff g\in\mathcal{L}^{1}(P_{X};\mathbb{R}^{m}).$$

In that case, by Integrable Vector- and Matrix-Valued Functions and Their Integrals (ii) and (ii) again, for every $k\in[m]$,

$$\bigl(\mathbb{E}\,g(X)\bigr)_{k}=\int_{\Omega}g_{k}\circ X\,dP=\int_{S}g_{k}\,dP_{X}=\Bigl(\int_{S}g\,dP_{X}\Bigr)_{k},$$

so $\mathbb{E}\,g(X)=\int_{S}g\,dP_{X}$. This is (iii).