$j_\infty$ notation in probability (or set) theory This is a problem I found in a recent ``Probability using theorems and exercises'' book written by Shiryaev, Ehrlich, and Yaskov. What I'm confused about is the notation $j_\infty$ they introduce. Here's the problem:
"Let $J_i$, where $i\in I$, be arbitrary sets. Prove that the following is always true:
$$\bigcup_{i\in I}\bigcap_{j\in J_i}A_{ij}=\bigcap_{j_\infty}\bigcup_{i\in I}A_{ij_i},$$
where the intersection is taken over all "paths" $j_\infty=(j_i,i\in I)$, $j_i\in J_i$."
I think there's some kind of typo or mistake in the definition of $j_\infty$, because I just can't figure out what it is. Here's a couple of examples I'm thinking about.
$\textbf{Example 1.}$ Let $J_1=\{a,b\}$, $J_2=\{\alpha,\beta\}$, $A_{1a}=A_{1b}=A_{2\alpha}=\{\omega_1\}$, $A_{2\beta}=\{\omega_2\}$. Then $\bigcup_{i\in I}\bigcap_{j\in J_i}A_{ij}=\{\omega_1\}$.
$\textbf{Example 2.}$ Let $J_1=\{a,b\}$, $J_2=\{\alpha,\beta\}$, $A_{1a}=\{\omega_1,\omega_2\},A_{1b}=A_{2\alpha}=\{\omega_1\}$, $A_{2\beta}=\{\omega_2\}$. Then again $\bigcup_{i\in I}\bigcap_{j\in J_i}A_{ij}=\{\omega_1\}$.
All my guesses about $j_\infty$ failed one of these examples and I didn't get the required equality. Honestly, I'm not even sure about the order of operations. Say $J_i$'s are non-intersecting, then $j_i$ pins down $i$, so how can we even take a union $\bigcup_{i\in I}A_{ij_i}$? And if we fix "path" (the way I understand it) $(i,j_i)$, then again $i$ is fixed and there's nothing to take a union over.
Can someone tell me what $j_\infty$ above is?
 A: A path here is a function $f\colon I \to \bigcup_{i\in I}J_i$ such that each $f(i) \in J_i.$  The symbol $j_\infty$ is intended to denote one of these functions, but I think that's a confusing notation, so I'll just stick to writing $f$ instead to make it clear that we're talking about a function.
If we write $P$ for the set of all paths, then the equality means
$$\bigcup_{i\in I}\bigcap_{j\in J_i}A_{i,j}=\bigcap_{f\in P}\bigcup_{i\in I}A_{i,f(i)}.$$
We'll show using the axiom of choice that an arbitrary $x$ belongs to the set on the left iff $x$ belongs to the set on the right.  (Without the axiom of choice, there may not even be any paths.)
Note first that if $x$ belongs to the set on the left, then there exists $i_0\in I$ such that for all $j\in J_{i_0},$ $x\in A_{i_0,j}.$  It follows that for every path $f,$ since $x$ belongs to every $A_{i_0,j}$ for $j \in J_{i_0},$ we have that $x$ belongs in particular to $A_{i_0,f(i_0)}.$  So for every path $f,$ $x\in \bigcup_{i\in I}A_{i,f(i)}.$  It follows that $x\in \bigcap_{f\in P}\bigcup_{i\in I}A_{i,f(i)}.$
Conversely, if $x$ does not belong to the set on the left, then for every $i\in I,$ there exists $j\in J_i$ such that $x \not\in A_{i,j}.$  Using the axiom of choice, let $f\colon I \to \bigcup_{i\in I}J_i$ be a function with the property that each $f(i)$ belongs to $J_i$ but $x$ does not belong to $A_{i,f(i)}.$ Then $f$ is a path such that $x$ does not belong to $\bigcup_{i\in I} A_{i,f(i)}.$  It follows that $x$ does not belong to the set on the right.
