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In Durrett's "Probability: Theory and Examples" (4th ed., page 244) to state that two measures are either singular or absolutely continuous the author reasons as follow:

(I ommited the entire proof and kept only what's relevant for this question)

First, it proves that

$\mu(A) = \int_A X d\nu$ $+$ $\mu(A\cap\{X=\infty\})$; here $X \geq 0$

It is known that:

$\nu(\{X=\infty\}) = 0 $

$\nu(\{X=0\}) \in \{0, 1\}$

From these facts the author concludes that either $\mu \ll \nu$ or $\mu \bot \nu$ .

I can see that if $\nu(\{X=0\}) = 1$ then $\int_A X d\nu = 0$ $\forall A \in \mathcal{F}$ so in this case $\mu(A) = \mu(A\cap\{X=\infty\})$ and $\mu \bot \nu$ since $\nu(A\cap\{X=\infty\}) = 0$ $\forall A \in \mathcal{F}$

But I cannot see the other implication: if $\nu(\{X=0\}) = 0$ then for some $A$ with $\nu(A) = 0$ I can only say that $\mu(A) = \mu(A\cap\{X=\infty\})$ and from this alone I cannot derive that $\mu(A)=0$

I would greatly appreciate any help, and provide any details if needed. I tried to keep the post stick to what's needed for the proof.

Thank you very much in advance!

EDIT:

Thanks to Nate Eldredge's comment now I see the information is insufficient. I state the hypothesis for the Theorem below:

Both $\mu$ and $\nu$ are measures on $(\mathbb{R}^\mathbb{N}, \mathcal{R}^\mathbb{N})$ that make the coordinates $\xi_n(\omega) = \omega_n$ independent. From the body of the proof I can see that both are probability measures, but this is not stated by the author.

This filtration is defined: $\mathcal{F}_n = \sigma(\xi_m : m \leq n)$ and $\mu_n$ and $\nu_n$ are the restrictions of $\mu$ and $\nu$, respectively, to $\mathcal{F}_n$.

$X_n = \frac{d\mu_n}{d\nu_n}$ is the sequence of the Radom-Nykodim derivatives and $X = \lim \sup X_n$. It is proved that $X_n \to X$, $\nu$-a.s.

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    $\begingroup$ I think we have to dig deeper into the proof. The conclusion doesn't follow from the statements written here. Consider for instance $\Omega = [0,1]$, $\nu = m$ Lebesgue measure, and $\mu = \frac{1}{2} \nu + \frac{1}{2}\delta_0$. Let $X(\omega) = 1/2$ for $0 < \omega \le 1$, and $X(0) = \infty$. Then everything you wrote here is satisfied and $\nu(X=0) = 0$ but we do not have $\mu \ll \nu$. $\endgroup$ Oct 17, 2019 at 14:55
  • $\begingroup$ Thanks for your comment! I updated the question accordingly. $\endgroup$ Oct 17, 2019 at 15:12

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I confess I don't see either how to get the dichotomy from the zero-one law alone. As I noted in a comment above, the absolute continuity doesn't follow from just the decomposition of $\mu$ and the statements $\nu(X=0) = \nu(X=\infty) = 0$.

However, the proof of Durrett's Theorem 5.3.5 does have enough to get the conclusion. In the case $\prod \int \sqrt{q_m}\,dG_m > 0$, it is shown that we actually have $X_n \to X$ in $L^1(\nu)$. Since $\int X_n\,d\nu = \int X_n\,d\nu_n = \int 1\,d\mu_n = 1$ for every $n$, we then have $\int X\,d\nu = 1$ as well. Then taking $A = \{X < \infty\}$ in the decomposition, we have $$\mu(X < \infty) = \int_{\{X < \infty\}} X\,d\nu + \mu(\{X < \infty\} \cap \{X = \infty\}) = 1 + 0$$ since $\nu(X < \infty) = 1$ because $X \in L^1(\nu)$, and so $\int_{\{X < \infty\}} X\,d\nu = \int_{\Omega} X\,d\nu = 1$. So $\mu(X = \infty) = 0$ and we have $\mu(A) = \int_A X\,d\nu$, which is to say that $\mu \ll \nu$ and $\frac{d\mu}{d\nu} = X$.

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  • $\begingroup$ Fair enough! I thikn the original Kakutani's work referred to the case where the condition is met. So this is as much help as I'd expect. Thanks! $\endgroup$ Oct 17, 2019 at 18:23

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