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I'm given a measurable space $(\Omega, \mathcal F)$ and two probability measures $\mu,\nu$ on it with $\mu\ne\nu$. Consider the product measures $P:=⊗_{k\in \Bbb N}\mu$ and $Q:=⊗_{k\in \Bbb N}\nu$ on $(\prod_{k\in \Bbb N}\Omega,⊗_{k\in \Bbb N}\mathcal F)$. My goal is to show that $P$ and $Q$ are singular.

Since our current topic in the lecture is related to convergence of random variables I assume that possible solution might deal with the Laws of Large Numbers. However, in order to apply them I would have to construct a suitable sequence of random variables and I'm really stuck with it. What would be their state space, for example ?

Any hint would be highly appreciated.

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    Choose $B$ in $\mathcal F$ such that $\mu(B)\ne\nu(B)$ and, for every $n$ in $\mathbb N$, define $X_n$ on the product $E=\prod\Omega$ by $X_n(\omega)=\omega_n$ for every $\omega=(\omega_k)_{k\in\mathbb N}$. Finally consider the event $A$ in $E$ defined by $A=\{N_n\to\mu(B)\}$ where $N_n=\frac1n\sum\limits_{k=1}^n\mathbf 1_{X_k\in B}$. Then $P(A)=1$ (why?) and $Q(A)=0$ (why?) hence you are done (why?). – 2017-01-06
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    Yes, this is what I wrote, and? – 2017-01-06
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    $P(A)=1$ (why?) - I guess that's the point where we apply the Strong Law of Large Numbers. $Q(A)=0$ (why?) - frankly, I don't really know why for now. hence you are done - by the definition of a singular measures while $P(A)=Q(A^C)=1$ – 2017-01-06
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    And also I think that it makes sense to consider event $A \in ⊗_{k\in \Bbb N}\mathcal F$ rather than in $\prod_{k\in \Bbb N}\Omega$ ? – 2017-01-06
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    Yes, $A\subset\prod\Omega$ and $A\in ⊗\mathcal F$ -- as I have written. – 2017-01-06
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    I'm still unable to fully understand why $N_n\to\mu(B)$ is a sure event. As far as I understand, $N_n$ counts how many (normalized by $n$) of those $X_k$ are in $B$ but why does this equal (in the sense of limit) to the probability of the event $B$ ? – 2017-01-07
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    Law. Of. Large. Numbers. – 2017-01-07

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