Is the limit of the finite product probability measures equal to the infinite product probability measure?\
$\newcommand{\N}{\mathbb{N}}$
$\newcommand{\A}{\mathfrak{A}}$
To be precise:
Let $(P_n)_{n\in \N}$ be probability distributions on $(\Omega_n, \A_n)$ respectively.
For $I\subset \N$, set $P_I:=\times_{i\in I}P_i$, the product probability measure on $\bigotimes_{i \in I}\A_i =: \A_I$
We know this exists because of Ionescu-Tulcea's Theorem for the infinite case.
Let $\A'_I:=\times_{i\in I}\A_i$ be the cartesian product.
Let $\Omega_I := \times_{i\in I}\Omega_i$
Let $\pi_I:\A_\N \rightarrow \A_I$ be the projection on the I'th components.
For $A \in \A$, we write $\pi_I(A):=\{\pi_I(a)|a\in A\}$
For $n \in \N$, let $\N_n = \{1,\dots,n\}$
and $\N_{> n} = \N\setminus \N_n = \{n+1,\dots\}$.
Set $Q_n(A) := P_{\N_n}(\pi_{\N_n}(A))$
Note that $Q_n$ is only well-defined if $\pi_{\N_n}(A)$ is a measurable set.
Set $Q(A) := lim_n Q_n(A)$
Since $\pi_{\N_n+1}(A)\times\Omega_{\N_{> n+1}}\subseteq \pi_{\N_n}(A)\times\Omega_{\N_{> n}}$ we have that
$Q_n(A)=P_{\N_n}(\pi_{\N_n}(A))=P_{\N}(\pi_{\N_n}(A)\times\Omega_{\N_{> n}})\geq
P_{\N}(\pi_{\N_n+1}(A)\times\Omega_{\N_{> n+1}})=Q_{n+1}(A)$.
So $Q_n(A)$ is non-increasing and bounded by 0. This means the limit exists and Q is well-defined.
Obviously for all rectangles $A=B\times \Omega_{\N{>n}}$ for $n\in \N, B\in \A'_{\N_n}$
we have $Q(A)=Q_n(A)=P_\N(A)$, so if Q is a measure we already know $Q=P_\N$ since they agree on a generator.\
So my first Question is: Is there a counterexample where $\pi_{\N_n}(A)$ is not a measurable set?
My second Question is: Is there a counterexample where $Q_n$ is well-defined, but $Q\neq P_\N$?
My friend came up with a counterexample to the second question, so I might as well share it.
Let $P_i=U(\{0,1\})$
Let $A_{n,m} = \{0,1\}^n \times \{0\}^m \times \{0,1\}^{\mathbb{N}_{>n+m}} \in \mathfrak{R}$, the measurable rectangles.
We have $P(A_{n,m})={1\over 2^m}$
Let $A_n = \bigcap_{m\in\mathbb{N}}A_{n,m} =\{0,1\}^n\times \{0\}^{\mathbb{N}_{>n}} $
We have $P(A_n) = \lim_m P(A_{n,m}) = 0$
Let $A = \bigcup_{n\in\mathbb{N}} A_n = \{b\in\{0,1\}^\mathbb{N}|\exists n_0\in\mathbb{N}:\forall m\in\mathbb{N}_{>n_0}:b_m=0\}$.
Since $A_n \subseteq A_{n+1}$ we have $A_n \big\uparrow A$ so $P(A) = \lim_n P(A_n) = 0$
While $\forall n\in\mathbb{N}:\pi_{\mathbb{N}_n}(A)=\{0,1\}^n$._n So $\forall n\in\mathbb{N}:Q_n(A)=1$, so $Q(A)=1$
We have $P(A)\neq Q(A)$