Tuesday, 5 September 2017

Integration a function of a single variable over a 2-dim measure



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:XR 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×YR is B(X){,Y}-measurable function and both measures coincide on this σ-algebra?


Answer



Let πX:X×YX 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(π1X(A)),AB(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×YfπXdμ=X×Yf(x)μ(dx,dy).


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