# Durrett's proof of Kakutani dichotomy

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.

• 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 $\mu \ll \nu$. – Nate Eldredge Oct 17 at 14:55
• Thanks for your comment! I updated the question accordingly. – Cristián Antuña Oct 17 at 15:12

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 $$\frac{d\mu}{d\nu} = X$$.