Bernoulli infinite product measures singular if $p \not = q$. I am going through Achim Klenke's Probability Theory textbook. In Section 7.4, he discusses what it means for two measures to be singular to one another and gives the following example.
Let $\Omega = \{0, 1\}^\mathbb{N}$ and let $(\mathrm{Ber}_p)^{\otimes \mathbb{N}}$ and $(\mathrm{Ber}_q)^{\otimes \mathbb{N}}$ be the infinite product measures with parameters $p$ and $q$, respectively. For $n \in \mathbb{N}$, let $X_n$ be the $n$-th coordinate map. Then under $(\mathrm{Ber}_r)^{\otimes \mathbb{N}}$, $(X_n)_{n \in \mathbb{N}}$ is independent and Bernoulli distributed with parameter $r$.
Here is the step where I get confused:
Klenke states that one can apply the strong law of large numbers such that for any $r \in \{p, q\}$, there exists a measurable set $A_r \subset \Omega$ with $(\mathrm{Ber}_r)^{\otimes \mathbb{N}}(\Omega \backslash A_r) = 0$ and $\lim_{n \to \infty} n^{-1} \sum_1^n X_i(\omega) = r$ for all $\omega \in A_r$ and therefore in particular $A_p \cap A_q = \emptyset$ if $p \not = q$, and thus $(\mathrm{Ber}_p)^{\otimes \mathbb{N}}$ and $(\mathrm{Ber}_q)^{\otimes \mathbb{N}}$ are singular in that case.
Now I am completely confused by the last paragraph. First off, what guarantees the existence of such a measurable set $A_r$? Then, how does it particular follow that $A_p \cap A_q = \emptyset$ if $p \not = q$? And finally, how does the last imply singularity of the two measures? A nice and clear explanation would be greatly appreciated, thanks!
 A: Consider the sequence space under the measure $P_r = \text{Ber}(r)^{\otimes \mathbb{N}}$. By the strong law of large numbers, $$P_r(\limsup_{n \to \infty}\frac{X_1 + \dots + X_n}{n} = r) = 1.$$
So $A_r = \{\limsup_{n \to \infty}\frac{X_1 + \dots + X_n}{n} = r\}$ is the set Klenke is talking about. The above shows that $P_r$ assigns mass $1$ to $A_r$. Clearly if $p \neq q$ then $A_p \cap A_q = \emptyset$ because $\limsup_{n \to \infty}\frac{X_1 + \dots + X_n}{n}$ cannot simultaneously equal $p$ and $q$.
A: I'm also not sure what the problem is, so I'll argue why there is no problem.
Consider $\{0,1\}^{\mathbb N}$ equipped with the Borel algebra induced by the discrete topology (which is the power set of $\{0,1\}^{\mathbb N}$, cf. this discussion to see why it doesn't matter which product algebra we take, be it induced by projections or rectangles).
For an integer $n>0$ let $a_{n}:\{0,1\}^{\mathbb N}\rightarrow\mathbb R$, $x\mapsto n^{-1}\sum_{m=1}^{n}x_m$, where $\mathbb R$ is equipped with the canonical Borel algebra. Notice that $a_{n}$ is measurable since it is continuous (which is trivial for the discrete topology).
So, for $\bar a\in\mathbb R$ and an integer $k>0$ the set
$$E_{n,k}(\bar a)=\{x\in\{0,1\}^{\mathbb N}:|a_{n}(x)-\bar a|\le 1/k\}\subseteq\{0,1\}^{\mathbb N}$$
is measurable, since this is the preimage of the measurable set $[\bar a-1/k,\bar a+1/k]$ under a measurable function. But then the set $E(\bar a)=\cap_{k>0}\cup_{n_0>0}\cap_{n\ge n_0}E_{n,k}(\bar a)$ is measurable, and we have
$$E(\bar a)=\{x:\forall k\exists n_0\forall n\ge n_0\,|a_{n}(x)-\bar a|\le 1/k\}=\{x:\lim_{n\rightarrow\infty}a_{n}(x)=\bar a\}.$$
Now, the strong law of large numbers (is well-defined and) yields $\mathbb P_r(E(r))=1$, where $\mathbb P_r=\mathrm{Ber}_r^{\otimes\mathbb N}$. And since the limit is unique (by definition of the limit) if it exists, we have $\bigcup_{s\in\mathbb R\setminus\{r\}}E(s)\subseteq \{0,1\}^{\mathbb N}\setminus E(r)$. In particular, we have $\mathbb P_p(E(p))=1\neq 0=\mathbb P_q(E(p))$, so $\mathbb P_p$ and $\mathbb P_q$ are singular (notice that the definition on Wikipedia can be reduced to the existence of an event $A$ such that $\mu(A)=1$, $\nu(A)=0$ since we consider non-negative normalized measures).
I'm fairly confident that this discussion is reasonable. Thus, the underlying $\sigma$-algebra in Klenke's book was overly restrictive (I doubt that), or the wording was intended to be suggestive (e.g. to point towards a definition, result or the likes), but if so, I still didn't get the message.
