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If $f(x) = \sum_{j=1}^n c_j 1_{A_j}(x)$ ie a simple function. Why is $$\prod_{j=1}^n \exp(t[\exp(i(u,c_j))-1]\mu(A_j)) =\exp[t\int_A[\exp(i(u,f(x)))-1]\mu(dx)].$$ The notation $(u,c)$ is dot product, $1_{A}(x)$ is the indicator function and $i=\sqrt{-1}$.

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    apologies, I have edited the question.2010-10-12

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Because $e^a \cdot e^b = e^{a+b}$.

The only thing $t(e^{i\langle u,c_j\rangle}-1)$ does (for the purpose of evaluating the integral) is just to make the manipulations a bit more ugly.