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Lets consider the integral $\int_{0}^{1}logxd\mu$ where measure $\mu$ is equivalent to the Lebesgue measure. What about convergence of this integral? Thanks.

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If $\mu$ and $\nu$ are equivalent measures, then the Radon-Nikodym derivative vanishes nowhere, yes? – Christopher A. Wong Feb 14 '13 at 1:17

Maybe yes, maybe no. Let Lebesgue measure be $\lambda$. If $\mu=\lambda$, the integral is convergent. On the other hand, if $$ \mu(A)=\int_A \frac{1}{|x|} d\lambda(x) \qquad \text{for all measurable sets } A, $$ then it's not.

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In the second example are the measures $\mu$ and $\lambda$ equivalent? – Akmal Feb 15 '13 at 0:34
Yes, they're equivalent. – David Moews Feb 15 '13 at 0:36

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