Is the diagonal still measurable in this product measure?

Is the diagonal $$\Delta=\{(x,x)|x \in[0,1]\}$$ still measurable in $$\mathcal{B}([0,1])\times\mathcal{F}$$ where $$\mathcal{F}$$ is the $$\sigma$$-algebra formed by all sets of Lebesgue measure $$0$$ or $$1$$ in$$[0,1]$$?

This problem comes to me when I want to prove that the indicator of $$A$$ is measurable in this question: https://mathoverflow.net/questions/176622/progressively-measurable-vs-adapted

I claim $$\Delta$$ is not $$\mathcal F\mathcal \otimes \mathcal P([0,1])$$-measurable. [Oops. I unintentionally reversed the factors. I used the power set, and not just the Borel algebra.]

If $$A \subseteq [0,1]$$ write the complement $$A^c = [0,1]\setminus A$$.
If $$E \subseteq [0,1]^2$$ write the complement $$E^c = [0,1]^2\setminus E$$.

If $$E \subseteq [0,1]^2$$ and $$y \in [0,1]$$, write the cross-section $$E[y] := \{x \in [0,1] : (x,y) \in E\}$$ If $$N \subseteq [0,1]$$ is a Lebesgue null set, write \begin{align} \mathcal F_N :=& \{A \subseteq [0,1] : A \subseteq N \text{ or } A^c \subseteq N\} \\ =& \{A \subseteq [0,1] : A \subseteq N \text{ or } A \supseteq N^c\} \end{align} so that $$\mathcal F = \bigcup_{N\text{ null}} \mathcal F_N$$ Now for a null set $$N$$ write \begin{align} \mathcal C_N :=& \{E \subseteq [0,1]^2 : \forall y \in [0,1]\; E[y] \in \mathcal F_N\} \\ \mathcal C :=& \bigcup_{N \text{ null}} \mathcal C_N \end{align}

Claim 1: $$\mathcal C$$ is a $$\sigma$$-algebra on $$[0,1]^2$$.
Proof. (1) We claim $$\varnothing \in \mathcal C$$. Indeed, for all $$y \in [0,1]$$ we have $$\varnothing[y] = \varnothing \in \mathcal F_{\varnothing}$$.
(2) We claim: if $$E \in \mathcal C$$, then $$E^c \in \mathcal C$$. Indeed, from $$E \in \mathcal C$$ we conclude there is null $$N$$ with $$E \in \mathcal C_N$$, so for all $$y$$ we have $$E[y] \subseteq N$$ or $$E[y] \supseteq N^c$$. But $$E^c[y] = E[y]^c$$, so then $$E^c[y] \supseteq N^c$$ or $$E^c[y] \subseteq N$$ and thus $$E^c \in \mathcal C_N \subseteq \mathcal C$$.
(3) We claim: if $$E_n \in \mathcal C$$ for all $$n \in \mathbb N$$, and $$E = \bigcup_n E_n$$, then $$E \in \mathcal C$$. Indeed, there are null sets $$N_n$$ so that $$E_n \in \mathcal C_{N_n}$$. Let $$N := \bigcup N_n$$ so that $$N$$ is null. Fix $$y \in [0,1]$$. Then for each $$n$$, either $$E_n[y] \subseteq N_n$$ or $$E_n[y] \supseteq N_n^c$$. There are two possibilities: (3a) $$\forall n\; E_n[y]\subseteq N_n$$ or (3b) $$\exists n_0\;E_{n_0}[y]\supseteq N_{n_0}^c$$. In case (3a), $$\left(\bigcup_n E_n\right)[y] =\bigcup_n E_n[y] \subseteq \bigcup_n N_n = N$$ In case (3b), $$\left(\bigcup_n E_n\right)[y] =\bigcup_n E_n[y] \supseteq E_{n_0}[y] \supseteq N_{n_0}^c \supseteq N^c$$ So in both cases $$\bigcup_n E_n[y] \in \mathcal F_N$$. This is true for all $$y$$, which means $$\bigcup E_n \in \mathcal C_N\subseteq \mathcal C$$. $$\quad\square$$

Claim 2: If $$A \in \mathcal F$$ and $$B \in \mathcal P([0,1])$$, then $$A \times B \in \mathcal C$$.
Proof. Since $$A \in \mathcal F$$, there are two possibilities. (1) $$A$$ is null, and $$A \times B \in \mathcal C_A \subseteq \mathcal C$$.
(2) $$A^c$$ is null, and $$A \times B \in \mathcal C_{A^c} \subseteq \mathcal C$$. $$\quad\square$$

Consequence: $$\mathcal F \otimes \mathcal P([0,1]) = \sigma\big(\{A \times B : A \in \mathcal F, B \in \mathcal P([0,1])\}\big) \subseteq \mathcal C$$.

Claim 3: $$\Delta \not\in \mathcal C$$.
Indeed, let $$N$$ be a null set. There is $$y \not\in N$$. But $$\Delta[y] = \{y\}$$ so that $$\Delta[y] \not\subseteq N$$. On the other hand, $$N^c$$ has measure $$1$$, so $$\Delta[y] \not\supseteq N^c$$. Thus $$\Delta \not\in \mathcal C_N$$. This holds for all null sets $$N$$, so $$\Delta \not\in \mathcal C$$.

Finally, $$\Delta \not\in \mathcal F \otimes \mathcal P([0,1])$$.

• Thanks for the idea! I just don’t get the final step, why is the diagonal not in the sigma algebra you define in your answer? Jun 2 '21 at 23:51
• Explanation added. Jun 3 '21 at 13:52
• Thanks! I think I know where I got confused. In defining $\mathcal{F}_N$, each $N$ is fixed, and this is what rules the diagonal out. Jun 4 '21 at 14:29