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Let $\lambda$ be the Lebesgue measure and $\mu$ be a probability $\sigma$-finite measure on $[0, 1].$ Suppose $\lambda \ll \mu$ and $\mu \ll \lambda.$ What can we say about the convergence of the integral $\int_{0}^{1}\log x\,d\mu$?

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What exactly do you mean by a probability $\sigma$-finite measure? Aren't probability measures already finite by definition? – T. Eskin Feb 15 '13 at 0:53
I guess, we can say that it is divergent. At least we can write it as $\int_0^1\log x\cdot f(x)dx$ for a density function $f:[0,1]\to \Bbb R_+$. – Berci Feb 15 '13 at 1:00
up vote 0 down vote accepted

I assume you want $\mu$ to be a probability measure on $[0,1]$ (all probability measures are $\sigma$-finite, because they are finite.) In this case, the integral could converge or diverge. If $\mu=\lambda$, then $\int_{0}^{1}\log x\,d\mu$ converges and equals $-1$. On the other hand, if $\mu$ is given by the density function $$\frac{d\mu}{d\lambda}=\frac{1}{x(1-\log x)^2}\qquad \text{on } [0,1],$$ then $\mu$ is a probability measure which is equivalent to $\lambda$, but $\int_{0}^{1}\log x\,d\mu$ diverges to $-\infty$.

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Thanks David Moews – Akmal Feb 15 '13 at 1:53

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