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Let $\lambda, \mu$ be complex measures on $(X,\alpha)$ and $(Y,\beta).$ Prove there exists a unique complex measure $\lambda\times \mu$ on the sigma algebra $\alpha\otimes \beta,$ such that $(\lambda\times \mu)(A\times B)=\lambda(A)\mu(B),$ for every $A\in \alpha, B\in \beta.$

This question is from an old exam I am studying for.

My idea goes as follows. Because $\mu\ll|\mu|,$ by Radon Nikodym we have $$ \mu(A)=\int_A fd|\mu|,$$ also $\lambda(A)=\int_B gd|\lambda|.$ Now we have $$\mu(A)\lambda(B)=\int_A fd|\mu|\int_B gd|\lambda|=\int_B\int_Af(x)g(y)d|\mu|(x)d|\lambda|(y)=\int_{A\times B}f(x)g(y)d(|\mu|\times|\lambda|).$$ The expression $$\int_{A\times B}f(x)g(y)d(|\mu|\times|\lambda|)$$ defines a measure on $A\times B.$ I know $|\mu|\times|\lambda|$ is a unique positive measure on $A\times B$ such that $(|\mu|\times|\lambda|)(A\times B)=|\mu|(A)\cdot|\lambda|(B).$ I cant finish the proof from here on, but I suppose my idea is on the right way.

If I knew that $(|\mu|\times|\lambda|)(A\times B)=|\mu|(A)\cdot|\lambda|(B),$ then the uniqueness is obvious. But is that true? How can I prove that? If it is not true, how can I finish the proof? Please help.

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  • $\begingroup$ How is $\alpha\otimes\beta$ defined? Presumably it's the sigma algebra generated by something or other. In that case, do you have any theorems about extending set functions (uniquely) to a measure on a generated sigma algebra? $\endgroup$ – user108903 Jan 18 '13 at 22:31
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    $\begingroup$ @primoz It's fine to post problems that you are thinking about, especially if you share your thoughts on them. $\endgroup$ – user53153 Jan 18 '13 at 22:47
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    $\begingroup$ My idea goes ass follows. $\endgroup$ – primoz Jan 19 '13 at 21:07
  • $\begingroup$ How can i finish this proof? $\endgroup$ – primoz Jan 20 '13 at 9:34
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Use the real and imaginary parts of the component measures and the positive and negative parts of these.

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    $\begingroup$ I sense that you are suggesting that the problem can be reduced to one involving finite positive measures, but I could not discern from your terse one-line post how (or whether) you propose to prove the equality highlighted by the final paragraph of the Question. $\endgroup$ – hardmath Nov 27 '15 at 22:28

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