# 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!

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 $$q$$.
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.