Product Measurability, Parameter Integrals, and a Fubini Counterexample
نویسنده
چکیده
Negative results: 1) A convolution R g(?y)h(y) dy need not be measurable with respect to the {algebra generated by the translates of g. 2) There exist a Borel set A R and two {{nite measures , such that R R 1 A (x + y) dd(x)dd(y) 6 = R R 1 A (x + y) dd(y)dd(x). Positive result: A function of two variables, measurable with respect to a product {algebra A B and partially measurable with respect to A 0 A and B 0 B, is {almost measurable with respect to A 0 B 0 , for , {{nite measures on A, B.
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تاریخ انتشار 1999