Does construction of infinite product measure require axiom of choice? I am learning about infinite (countable) product measure, which the exact statement of the theorem I write below. I was wondering if the theorem requires axiom of choice or not. I would appreciate any assistance! Thank you very much!
Let  $\{X_j\}$  be a sequence of spaces and write
$$
X = \prod_{j=1}^{\infty} X_j.
$$
Let $M_j$ be a $\sigma$-algebra of subsets of $X_j$.
A measurable rectangle with respect to the sequence $\{ M_j \}$
is defined to be a subset $W$ of $X$ which is representable in the
form
$$
W = \prod_{j=1}^{\infty} W_j,
$$
where $W_j \in M_j$ and $W_j = X_j$
except for at most finite set of numbers $j$. 
Then we have the following theorem:
Let $\{ (X_j, M_j, P_j) \}_{j \geq 1} $ be a sequence of measure spaces,
and write
$$
X = \prod_{j=0}^{\infty} X_j.
$$
Let $M$ be the minimal $\sigma$-algebra, of subsets of $X$,
containing every measurable rectangle with respect to the
sequence $\{ M_j \}$.
Then there exists a unique measure $P$ on $M$ with the property
that, for every non-empty measurable rectangle $W$,
$$
P(W) = \prod_{j=1}^{\infty} P_j(W_j),
$$
where the $W_j$ are defined by $W = \prod_{j=0}^{\infty} W_j$,
$W_j \in M_j \ (j \geq 0)  $. 
 A: Well... how do you know that $X$ is non-empty to begin with?
If $X$ is non-empty, then the definition of a measurable rectangle does not require the axiom of choice, since only finitely many arbitrary choices are made. And the definition of the smallest $\sigma$-algebra does not use the axiom of choice either. But it could be larger than you'd expect. And that might cause some issues with defining $P$.
For example if $X$ ends up as a countable union of countable sets, that $\sigma$-algebra might be $\mathcal P(X)$. And in that case, either the measure is atomic, or trivial. And it might be that neither case "fits" the case of $P$.
More concretely, $2^\Bbb N$ is consistently a countable union of countable sets. This means that if $\mu$ is any measure on $2^\Bbb N$, then it is either atomic, or $0$ everywhere. Now consider for each $\{0,1\}$ a probability measure given by a fair coin. The product measure on $2^\Bbb N$ should be neither atomic nor $0$.
A: This theorem implies the axiom of countable choice (and is probably equivalent to it, since I doubt its proof uses any more than countable choice).  To show this, suppose $(Y_j)$ is a countable sequence of nonempty sets, and suppose the product $\prod Y_j$ is empty.  Let $x$ be some point that is not in any $Y_j$ and let $X_j=Y_j\cup \{x\}$.  Let $M_j=\{X_j,Y_j,\{x\},\emptyset\}$ and define a measure $P_j$ on $M_j$ by $P_j(Y_j)=1$, $P_j(\{x\})=0$.  Suppose a measure $P$ exists on the product $X=\prod X_j$ as you specify.  Then $X$ is a nonempty measurable rectangle, so $P(X)=\prod P(X_j)=1$.  For each $j$, let $A_j\subset X$ be the set of points whose $j$th coordinate is $x$.  Then $A_j$ is a nonempty measurable rectangle, and $P(A_j)=0$ since $P_j(\{x\})=0$.  But in fact $\bigcup A_j=X$, since $X\setminus\bigcup A_j$ is exactly $\prod Y_j=\emptyset$.  Thus $1=P(X)\leq \sum P(A_j)=0$, a contradiction.
