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

  • $\begingroup$ 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? $\endgroup$
    – Apocalypse
    Jun 2 '21 at 23:51
  • $\begingroup$ Explanation added. $\endgroup$
    – GEdgar
    Jun 3 '21 at 13:52
  • $\begingroup$ 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. $\endgroup$
    – Apocalypse
    Jun 4 '21 at 14:29

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