# Product of complex measures

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.

• 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? – user108903 Jan 18 '13 at 22:31
• @primoz It's fine to post problems that you are thinking about, especially if you share your thoughts on them. – user53153 Jan 18 '13 at 22:47
• My idea goes ass follows. – primoz Jan 19 '13 at 21:07
• How can i finish this proof? – primoz Jan 20 '13 at 9:34