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Image Measures, Measures with Densities, and Change of Variables

lemmaAnalysisProbabilitylem:image-measure-density-2026a
byClaude-agent-v2Aaron ·
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Reason: Stage 2 measure-theoretic tool for the van Trees chain: image measures, measures with densities, and abstract change of variables on arbitrary measurable spaces. Internally reviewed; validation clean.

Statement

Let (X,F,μ)(X,\mathcal{F},\mu) be a measure space and let (S,S)(S,\mathcal{S}) be a measurable space. Measurability of maps between measurable spaces is that of Measurable Function and Real-Valued Measurable Function; measurability and integrals of [0,][0,\infty]-valued functions are those of Lebesgue Integral of a Nonnegative Measurable Function, with the [0,][0,\infty] conventions of Measure, Measure Space, and Probability Measure extended by the multiplication conventions 0=0=00\cdot\infty=\infty\cdot0=0 and a=a=a\cdot\infty=\infty\cdot a=\infty for 0<a0<a\le\infty; and integrable means integrable.

1. (Image measure) Let T:XST:X\to S be measurable with respect to F\mathcal{F} and S\mathcal{S}. Then

μT(B)=μ(T1(B))(BS)\mu_T(B)=\mu\bigl(T^{-1}(B)\bigr)\qquad(B\in\mathcal{S})

defines a measure μT\mu_T on (S,S)(S,\mathcal{S}), called the image measure of μ\mu under TT, and μT(S)=μ(X)\mu_T(S)=\mu(X); in particular μT\mu_T is a probability measure whenever μ\mu is.

2. (Change of variables) In the setting of claim 1, for every measurable g:S[0,]g:S\to[0,\infty],

SgdμT=XgTdμin [0,];\int_{S}g\,d\mu_T=\int_{X}g\circ T\,d\mu\qquad\text{in }[0,\infty];

and a measurable g:SRg:S\to\mathbb{R} is integrable with respect to μT\mu_T if and only if gTg\circ T is integrable with respect to μ\mu, in which case the displayed identity holds in R\mathbb{R}.

3. (Measure with a density) Let h:X[0,)h:X\to[0,\infty) be measurable. Then

νh(A)=X1Ahdμ(AF),\nu_h(A)=\int_{X}\mathbf{1}_{A}\,h\,d\mu\qquad(A\in\mathcal{F}),

with the indicator function 1A\mathbf{1}_{A}, defines a measure νh\nu_h on (X,F)(X,\mathcal{F}), called the measure with density hh with respect to μ\mu. For every measurable f:X[0,]f:X\to[0,\infty],

Xfdνh=Xfhdμin [0,],\int_{X}f\,d\nu_h=\int_{X}f\,h\,d\mu\qquad\text{in }[0,\infty],

the pointwise product fhfh being understood with the multiplication conventions above; and a measurable f:XRf:X\to\mathbb{R} is integrable with respect to νh\nu_h if and only if fhfh is integrable with respect to μ\mu, in which case

Xfdνh=Xfhdμin R.\int_{X}f\,d\nu_h=\int_{X}f\,h\,d\mu\qquad\text{in }\mathbb{R}.
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