Proof of Image Measures and Measures with a Density: Measures, Integrals and Change of Variables

Throughout, write $\nu=\mu\circ T^{-1}$ (Image Measure (i)); for $B\in\mathcal{G}$ we have $T^{-1}(B)\in\mathcal{F}$ (Measurable Map, and the Set of Measurable Maps (i)), and preimages are computed pointwise by Preimages under a Map: Membership, the Preimage Map, Sequences of Preimages, and the Set of Sets with Preimage in a Given Set (ii). By Integrals over Measurable Subsets (i), $h\mathbf{1}_{A}\in\mathcal{M}^{+}(X,\mathcal{F})$ and

$$(h\mu)(A)=\int_{A}h\,d\mu=\int_{X}h\mathbf{1}_{A}\,d\mu\qquad(A\in\mathcal{F}).$$

The image measure. Since $T^{-1}(\emptyset)=\emptyset$, $\nu(\emptyset)=\mu(\emptyset)=0$. Let $(B_{m})$ be a pairwise disjoint sequence in $\mathcal{G}$. The preimages $T^{-1}(B_{m})$ form a sequence in $\mathcal{F}$ (Preimages under a Map: Membership, the Preimage Map, Sequences of Preimages, and the Set of Sets with Preimage in a Given Set (iv)); it is pairwise disjoint, since a point of $T^{-1}(B_{m})\cap T^{-1}(B_{k})$ would be mapped by $T$ into $B_{m}\cap B_{k}=\emptyset$; and $T^{-1}\bigl(\bigcup_{m}B_{m}\bigr)=\bigcup_{m}T^{-1}(B_{m})$ (Preimages of Unions, Intersections and Complements (i)). Countable additivity of $\mu$ (Measure, Measure Space, and Probability Measure (i)) gives

$$\nu\Bigl(\bigcup_{m\in\mathbb{N}}B_{m}\Bigr)=\mu\Bigl(\bigcup_{m\in\mathbb{N}}T^{-1}(B_{m})\Bigr)=\sum_{m\in\mathbb{N}}\mu\bigl(T^{-1}(B_{m})\bigr)=\sum_{m\in\mathbb{N}}\nu(B_{m}),$$

so $\nu$ is a measure on $(Y,\mathcal{G})$. As $T^{-1}(Y)=X$ (Preimages of Unions, Intersections and Complements (iv)), $\nu(Y)=\mu(X)$; hence $\nu(Y)<\infty$ if $\mu(X)<\infty$, and $\nu(Y)=1$ if $\mu(X)=1$ (Measure, Measure Space, and Probability Measure (iii), Measure, Measure Space, and Probability Measure (v)). This proves (i).

Change of variables for simple functions. Let $s\in\mathcal{S}^{+}(Y,\mathcal{G})$, let $C=s[Y]$, a finite set, and for $c\in C$ let $B_{c}=s^{-1}(\{c\})\in\mathcal{G}$ (Simple Functions, Nonnegative Simple Functions, and the Simple Integral (iv)). The sets $T^{-1}(B_{c})$, $c\in C$, lie in $\mathcal{F}$ and are pairwise disjoint, and a point $x\in X$ lies in exactly one of them, the one with $c=s(T(x))$. Hence all terms but one of the following sum vanish (Finite Sums: Vanishing Terms, Finite Additivity of Measures, and Finite Sums of Series in [0,∞] (iii)), and

$$(s\circ T)(x)=\sum_{c\in C}c\,\mathbf{1}_{T^{-1}(B_{c})}(x)\qquad(x\in X).$$

By Properties of Simple Functions and of the Simple Integral (iii), $s\circ T\in\mathcal{S}^{+}(X,\mathcal{F})$ and, with the definition of the simple integral (Simple Functions, Nonnegative Simple Functions, and the Simple Integral (iv)),

$$I_{\mu}(s\circ T)=\sum_{c\in C}c\cdot\mu\bigl(T^{-1}(B_{c})\bigr)=\sum_{c\in C}c\cdot\nu(B_{c})=I_{\nu}(s).$$

Integrals of nonnegative simple functions are their simple integrals (Basic Properties of the Integral of Nonnegative Measurable Maps (i)), so

$$\int_{Y}s\,d\nu=\int_{X}s\circ T\,d\mu .\tag{1}$$

Change of variables for nonnegative functions. Let $g\in\mathcal{M}^{+}(Y,\mathcal{G})$. Then $g\circ T\in\mathcal{M}^{+}(X,\mathcal{F})$ (Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (ii)). For $m\in\mathbb{N}$ the dyadic approximations satisfy $s_{m}(g)\circ T=s_{m}(g\circ T)$ (Precomposition Commutes with Pointwise Operations, and Sections of Functions Are Precompositions (iii)). By Approximation of Measurable Functions by Dyadic Simple Functions (i), $(s_{m}(g))_{m}$ is a nondecreasing sequence in $\mathcal{S}^{+}(Y,\mathcal{G})$ with pointwise supremum $g$, and $(s_{m}(g\circ T))_{m}$ one in $\mathcal{S}^{+}(X,\mathcal{F})$ with pointwise supremum $g\circ T$. The monotone convergence theorem (Monotone Convergence Theorem (ii)) on both sides and (1) give

$$\int_{Y}g\,d\nu=\sup_{m}\int_{Y}s_{m}(g)\,d\nu=\sup_{m}\int_{X}s_{m}(g)\circ T\,d\mu=\sup_{m}\int_{X}s_{m}(g\circ T)\,d\mu=\int_{X}g\circ T\,d\mu ,$$

which is (ii).

Change of variables for real functions. Let $g:Y\to\mathbb{R}$ be measurable. Then $g\circ T$ is measurable (Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (ii)), and $(g\circ T)^{+}=g^{+}\circ T$ and $(g\circ T)^{-}=g^{-}\circ T$ (Precomposition Commutes with Pointwise Operations, and Sections of Functions Are Precompositions (i)). As $g^{+},g^{-}\in\mathcal{M}^{+}(Y,\mathcal{G})$ (Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (iv)), (ii) gives

$$\int_{Y}g^{\pm}\,d\nu=\int_{X}g^{\pm}\circ T\,d\mu=\int_{X}(g\circ T)^{\pm}\,d\mu .$$

So the integrals of $g^{+}$ and $g^{-}$ with respect to $\nu$ are finite exactly when those of $(g\circ T)^{+}$ and $(g\circ T)^{-}$ with respect to $\mu$ are, which is the asserted equivalence (Integrable Functions and Their Lebesgue Integral (i)); and then, by Integrable Functions and Their Lebesgue Integral (iii),

$$\int_{Y}g\,d\nu=\int_{Y}g^{+}\,d\nu-\int_{Y}g^{-}\,d\nu=\int_{X}(g\circ T)^{+}\,d\mu-\int_{X}(g\circ T)^{-}\,d\mu=\int_{X}g\circ T\,d\mu .$$

This proves (iii).

Composition of image measures. $S\circ T\in\mathrm{Meas}(X,\mathcal{F};Z,\mathcal{H})$ by Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (ii), and $\nu\circ S^{-1}$ is defined since $\nu$ is a measure by (i). For $E\in\mathcal{H}$ we have $(S\circ T)^{-1}(E)=T^{-1}(S^{-1}(E))$, so

$$(\nu\circ S^{-1})(E)=\nu\bigl(S^{-1}(E)\bigr)=\mu\bigl(T^{-1}(S^{-1}(E))\bigr)=\mu\bigl((S\circ T)^{-1}(E)\bigr)=\bigl(\mu\circ(S\circ T)^{-1}\bigr)(E).$$

Both sides of (iv) are functions on $\mathcal{H}$ with the same values, hence equal.

The measure with a density. The function $h\mathbf{1}_{\emptyset}$ is identically $0$, so $(h\mu)(\emptyset)=0$ (Basic Properties of the Integral of Nonnegative Measurable Maps (v)). Let $(A_{m})$ be a pairwise disjoint sequence in $\mathcal{F}$ and $A=\bigcup_{m}A_{m}$. For $x\in X$,

$$\sum_{m\in\mathbb{N}}h(x)\mathbf{1}_{A_{m}}(x)=h(x)\mathbf{1}_{A}(x):$$

if $x\notin A$, every term is $0$ (Small Facts about [0,∞]: Zero, Finiteness and Sums of Zeros (iv)); if $x\in A_{k}$, then $x\notin A_{m}$ for $m\neq k$, and the sum equals its $k$-th term (Sums of Sequences in [0,∞]: Real Series, Linearity, Comparison, Rearrangement, Double Series and Monotone Convergence (vii), and Finite Sums: Vanishing Terms, Finite Additivity of Measures, and Finite Sums of Series in [0,∞] (iii) for the partial sum up to $k$). By the series clause of linearity, Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (ii),

$$(h\mu)(A)=\int_{X}h\mathbf{1}_{A}\,d\mu=\int_{X}\sum_{m\in\mathbb{N}}h\mathbf{1}_{A_{m}}\,d\mu=\sum_{m\in\mathbb{N}}\int_{X}h\mathbf{1}_{A_{m}}\,d\mu=\sum_{m\in\mathbb{N}}(h\mu)(A_{m}),$$

so $h\mu$ is a measure (Measure, Measure Space, and Probability Measure (i)). This proves (v).

Integrals against a density: simple functions. Let $s\in\mathcal{S}^{+}(X,\mathcal{F})$, let $C=s[X]$ and $A_{c}=s^{-1}(\{c\})\in\mathcal{F}$ for $c\in C$. A point $x\in X$ lies in $A_{c}$ exactly for $c=s(x)$, so (Finite Sums: Vanishing Terms, Finite Additivity of Measures, and Finite Sums of Series in [0,∞] (iii) in $[0,\infty]$)

$$s(x)h(x)=\sum_{c\in C}c\,h(x)\mathbf{1}_{A_{c}}(x)\qquad(x\in X).$$

By Basic Properties of the Integral of Nonnegative Measurable Maps (iv), $c\,h\mathbf{1}_{A_{c}}\in\mathcal{M}^{+}(X,\mathcal{F})$ with $\int_{X}c\,h\mathbf{1}_{A_{c}}\,d\mu=c\,(h\mu)(A_{c})$. The integral is additive over finite sums in $\mathcal{M}^{+}(X,\mathcal{F})$ (from Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (i) by induction on the number of terms, Induction over Finite Sets, and Predicative Formulas for the Segments [n] and for Finiteness (iii), the empty sum being the zero function, with integral $0$ by Basic Properties of the Integral of Nonnegative Measurable Maps (v)). With the definition of the simple integral for the measure $h\mu$ (Simple Functions, Nonnegative Simple Functions, and the Simple Integral (iv)) and Basic Properties of the Integral of Nonnegative Measurable Maps (i), this gives

$$\int_{X}sh\,d\mu=\sum_{c\in C}c\cdot(h\mu)(A_{c})=I_{h\mu}(s)=\int_{X}s\,d(h\mu).\tag{2}$$

Integrals against a density: nonnegative functions. Let $f\in\mathcal{M}^{+}(X,\mathcal{F})$. Then $fh\in\mathcal{M}^{+}(X,\mathcal{F})$, and likewise $s_{m}(f)h$ for $m\in\mathbb{N}$ (Arithmetic of Measurable Real- and [0,∞]-Valued Maps (iii), Basic Properties of the Integral of Nonnegative Measurable Maps (i)). By Approximation of Measurable Functions by Dyadic Simple Functions (i), $s_{m}(f)\in\mathcal{S}^{+}(X,\mathcal{F})$ and $s_{m}(f)(x)$ increases in $m$ with supremum $f(x)$; multiplying by $h(x)$ preserves this (Arithmetic in [0,∞]: Monoid Laws, Distributivity, Monotonicity and Cancellation (iv), Suprema and Finite Sums in [0,∞] (ii), Arithmetic in [0,∞]: Monoid Laws, Distributivity, Monotonicity and Cancellation (ii)):

$$s_{m}(f)(x)h(x)\le s_{m+1}(f)(x)h(x),\qquad\sup_{m}s_{m}(f)(x)h(x)=f(x)h(x)\qquad(x\in X).$$

The monotone convergence theorem (Monotone Convergence Theorem (ii)) for $h\mu$ and for $\mu$, together with (2), gives

$$\int_{X}f\,d(h\mu)=\sup_{m}\int_{X}s_{m}(f)\,d(h\mu)=\sup_{m}\int_{X}s_{m}(f)h\,d\mu=\int_{X}fh\,d\mu ,$$

which is (vi).

Integrals against a density: real functions. Let $h(x)<\infty$ for every $x\in X$ and let $f:X\to\mathbb{R}$ be measurable. As a real function $h$ is measurable (Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (vii)), so $fh$ is measurable (Arithmetic of Measurable Real- and [0,∞]-Valued Maps (i)). Since $h(x)\ge0$,

$$(fh)^{+}(x)=\max\{f(x)h(x),0\}=f^{+}(x)h(x),\qquad(fh)^{-}(x)=f^{-}(x)h(x)\qquad(x\in X),$$

and these real products of nonnegative numbers are their products in $[0,\infty]$ (Addition and Multiplication on [0,∞] (ii)). As $f^{+},f^{-}\in\mathcal{M}^{+}(X,\mathcal{F})$ (Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (iv)), (vi) gives

$$\int_{X}f^{\pm}\,d(h\mu)=\int_{X}f^{\pm}h\,d\mu=\int_{X}(fh)^{\pm}\,d\mu .$$

By these equalities, $\int_{X}f^{+}\,d(h\mu)<\infty$ if and only if $\int_{X}(fh)^{+}\,d\mu<\infty$, and $\int_{X}f^{-}\,d(h\mu)<\infty$ if and only if $\int_{X}(fh)^{-}\,d\mu<\infty$; this is the asserted equivalence (Integrable Functions and Their Lebesgue Integral (i)), and subtracting (Integrable Functions and Their Lebesgue Integral (iii)) gives

$$\int_{X}f\,d(h\mu)=\int_{X}(fh)^{+}\,d\mu-\int_{X}(fh)^{-}\,d\mu=\int_{X}fh\,d\mu .$$

This proves (vii).