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