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The following lines are from a proof in my stochastic analysis lecture notes:

Let $(\Omega, \mathcal{F}) = ([0,1], \mathcal{L})$ where $\mathcal{L}$ denotes the $\sigma$-algebra of Lebesgue sets in $[0,1]$.

Define $A \subset [0, \infty) \times \Omega$ by $A = \{(x, x): x \in [0, 0.5]\}$.

Then $A \in \mathcal{B}[0, \infty) \otimes \mathcal{F}$.

I'm struggling to prove this. The product sigma algebra is generated by the inverse image of the projection maps, therefore it's generated by sets $B \subset [0, \infty) \times \Omega)$ of the form $B = \{(t, \omega) \in [0, \infty) \times \Omega): t \in B, \omega \in F \}$ for some $B \in \mathcal{B}[0, \infty), F \in \mathcal{F}$.

I don't see now how $A$ can be expressed as a sequence of countable intersections, unions and complements of sets $B$ since $A$ doesn't have a rectangle-like form and consists of uncountable many singletons.

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Hint: consider the set $A_n$ which is a union of $n$ squares arranged along the diagonal. Show that $A$ is the countable intersection of the sets $A_n$.

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Let $A_n = \bigcup_{k=0}^{n-1} \left[\frac{k}{n}, \frac{k+1}{n}\right) \times \left[\frac{k}{n}, \frac{k+1}{n}\right) \cup \{ 0.5,0.5\}$.

Claim: $$\bigcap_{n=1}^{\infty} A_n = A$$

Proof:

"$\supseteq$": Let $x \in [0, 0.5)$ and $n \in \mathbb{N}$ be arbitrary. Divide $[0,0.5]$ into $n$ intervals of equal length. $x$ must lie in one of those intervals, say the $k$-th interval (i.e. $\left[ k/n, (k+1)/n \right)$). Therefore $\{x, x\}$ must be in $\left[ k/n, (k+1)/n \right) \times \left[ k/n, (k+1)/n \right)$ which by definition is a subset of $A_n$. The proof for the case $x = 0.5$ follows by definition.

"$\subseteq$": Let $x \notin A$. This means that $d(x, A) > c$ for some $c > 0$ where $d$ is the distance w.r.t. the maximum norm in $\mathbb{R}^2$. By definition of $A_n$ we see that for each point $a \in A_n$ we must have $d(A,a) \leq \frac{1}{2}n$ for each $n$. This expression gets arbitrarily small for $n \to \infty$ and therefore there must exist some $n \in \mathbb{N}$ such that $x$ isn't in $A_n$ which completes the proof.

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