Friday, 18 October 2019

analysis - Prove that region under graph of function is measurable



In the measure theory book that I am studying, we consider the 'area' under (i.e. the product measure of) the graph of a function as an example of an application of Fubini's Theorem for integrals (with respect to measures).




The setting: (X,A,μ) is a σ-finite measure space, λ is Lebesgue measure on (R,B(R)) (Borel σ-algebra), f:X[0,+] is A-measurable, and we are considering the region under the graph of f,



E={(x,y)X×R|0y<f(x)}.



I need to prove EA×B(R). I thought to write E=g1((0,+])(X×[0,+]) where g(x,y)=f(x)y but I can't see why g must be A×B(R)-measurable. Any help would be appreciated.


Answer



g=kh where h(x,y)=(f(x),y) and k(a,b)=ab. [ Here h:X×RR2 and k:R2R]. k:R2R is Borel measurable because it is continuous. To show that h is measurable it is enough to show that h1(A×B)A×B(R) for A,BB(R). This is clear because h1(A×B)=f1(A)×B.



I have assumed that f takes only finite values. To handle the general case let g(x)=f(x) if f(x)< and 0 if f(x)=. Let $F=\{(x,y):0\leq y . Then E=(f1{}×[0,))[(f1{R}×R)F].



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