Let (X,B(X)) and (Y,B(Y)) be two measurable spaces and let μ be a finite measure on the product σ-algebra B(X)⊗B(Y). Let f:X→R be a bounded B(X)-measurable function and define ν(A)=μ(A×Y) to be a measure on B(X). How to show that
∫X×Yf(x)μ(dx×dy)=∫Xf(x)ν(dx).
I guess, this is completely trivial, however I couldn't come with a completely formal solution. Perhaps, that shall follow from some change of variables equality and the fact that ν is a pushforward measure of μ under the projection πX? Or from the fact that f:X×Y→R is B(X)⊗{∅,Y}-measurable function and both measures coincide on this σ-algebra?
Answer
Let πX:X×Y→X be the projection, i.e. πX((x,y))=x for (x,y)∈X×Y. Then ν is the image measure of μ under πX, i.e.
ν(A)=μ(A×Y)=μ(π−1X(A)),A∈B(X).
Integration with respect to ν can be described via integration with respect to μ (but this result relies on a standard argument as described in the comments, see e.g. this answer): If f is bounded and B(X)-meaurable, then
∫Xf(x)ν(dx)=∫Xfdν=∫X×Yf∘πXdμ=∫X×Yf(x)μ(dx,dy).
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