Proof of Integrals with Respect to the Restriction of a Measure to a Sub-σ-Algebra

The restriction as an image measure. Let $\iota:X\to X$ be the identity. For $G\in\mathcal{G}$ we have $\iota^{-1}(G)=G\in\mathcal{F}$, so $\iota$ is $\mathcal{F}/\mathcal{G}$-measurable, and its image measure $\mu\circ\iota^{-1}$ is the function on $\mathcal{G}$ with

$$(\mu\circ\iota^{-1})(G)=\mu(G)=\mu|_{\mathcal{G}}(G)\qquad(G\in\mathcal{G}).$$

Hence $\mu\circ\iota^{-1}=\mu|_{\mathcal{G}}$. Moreover $f\circ\iota=f$ for every function $f$ on $X$.

Nonnegative functions. Let $f\in\mathcal{M}^{+}(X,\mathcal{G})$. Since $f=f\circ\iota$ is the composite of an $\mathcal{F}/\mathcal{G}$-measurable and a $\mathcal{G}/\mathcal{B}([0,\infty])$-measurable function, $f\in\mathcal{M}^{+}(X,\mathcal{F})$ (Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (ii)). The change-of-variables formula Image Measures and Measures with a Density: Measures, Integrals and Change of Variables (ii), with $T=\iota$ and $(Y,\mathcal{G})=(X,\mathcal{G})$, gives

$$\int_{X}f\,d(\mu|_{\mathcal{G}})=\int_{X}f\,d(\mu\circ\iota^{-1})=\int_{X}f\circ\iota\,d\mu=\int_{X}f\,d\mu ,$$

which is (i).

Real functions. Let $f:X\to\mathbb{R}$ be $\mathcal{G}/\mathcal{B}(\mathbb{R})$-measurable. As before, $f=f\circ\iota$ is $\mathcal{F}/\mathcal{B}(\mathbb{R})$-measurable (Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (ii)). By Image Measures and Measures with a Density: Measures, Integrals and Change of Variables (iii), again with $T=\iota$, $f\in\mathcal{L}^{1}(\mu\circ\iota^{-1})=\mathcal{L}^{1}(X,\mathcal{G},\mu|_{\mathcal{G}})$ if and only if $f=f\circ\iota\in\mathcal{L}^{1}(\mu)$, and then

$$\int_{X}f\,d(\mu|_{\mathcal{G}})=\int_{X}f\circ\iota\,d\mu=\int_{X}f\,d\mu ,$$

which is (ii).

Integrals over sets of $\mathcal{G}$. Let $G\in\mathcal{G}$; then also $G\in\mathcal{F}$. For $f\in\mathcal{M}^{+}(X,\mathcal{G})$, the product $f\mathbf{1}_{G}$ lies in $\mathcal{M}^{+}(X,\mathcal{G})$ and $\int_{G}f\,d(\mu|_{\mathcal{G}})=\int_{X}f\mathbf{1}_{G}\,d(\mu|_{\mathcal{G}})$, by Integrals over Measurable Subsets (i) in the measure space $(X,\mathcal{G},\mu|_{\mathcal{G}})$; by (i) this equals $\int_{X}f\mathbf{1}_{G}\,d\mu$, which is $\int_{G}f\,d\mu$ by the same clause in $(X,\mathcal{F},\mu)$. This proves (iii).

Now let $f$ be as in (iv). The indicator $\mathbf{1}_{G}$ is $\mathcal{G}/\mathcal{B}(\mathbb{R})$-measurable (Basic Properties of the Integral of Nonnegative Measurable Maps (ii) in $(X,\mathcal{G},\mu|_{\mathcal{G}})$, with Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (vii)), hence so is the real product $f\mathbf{1}_{G}$ (Arithmetic of Measurable Real- and [0,∞]-Valued Maps (i)). By Integrals over Measurable Subsets (ii), applied in $(X,\mathcal{G},\mu|_{\mathcal{G}})$ and in $(X,\mathcal{F},\mu)$, $f$ is integrable on $G$ with respect to $\mu|_{\mathcal{G}}$ if and only if $f\mathbf{1}_{G}\in\mathcal{L}^{1}(X,\mathcal{G},\mu|_{\mathcal{G}})$, and integrable on $G$ with respect to $\mu$ if and only if $f\mathbf{1}_{G}\in\mathcal{L}^{1}(\mu)$; by (ii) applied to $f\mathbf{1}_{G}$ these two conditions are equivalent, and then

$$\int_{G}f\,d(\mu|_{\mathcal{G}})=\int_{X}f\mathbf{1}_{G}\,d(\mu|_{\mathcal{G}})=\int_{X}f\mathbf{1}_{G}\,d\mu=\int_{G}f\,d\mu .$$

This proves (iv).