Kingman (1973) writes on Page 11:

Suppose that $S$ is a measurable space. ... by making the weak assumption that the diagonal $D = \{(x, y); x = y\}$ is measurable in the product space $S\times S$. This automatically implies that every singleton set $\{x\}$ in $S$ is measurable.

Formally, why is the measurability of individual points in $S$ automatically implied by the measurability of the diagonal in $S\times S$?

  • $\begingroup$ Fubini-Tonelli would give that $\{ x \}$ is measurable for a.e. $x$... $\endgroup$ Dec 19, 2017 at 23:55
  • $\begingroup$ @DanielSchepler I guess Fubini-Tonelli might be an overkill here... $\endgroup$
    – p-value
    Dec 20, 2017 at 4:46
  • $\begingroup$ @DanielSchepler, if $\mu$ were a $\sigma$-finite measure on $S$, then Fubini-Tonelli would imply $\{x\}$ is $\mu$-measurable (i.e. in a possibly larger $\sigma$-algebra than the original). $\endgroup$
    – user81375
    Dec 20, 2017 at 6:01

1 Answer 1


For definiteness, let $(S,\mathcal{A})$ be a measure space and $\mathcal{A} \otimes \mathcal{A}$, the corresponding product $\sigma$-algebra on $S \times S$. Recall that if $E \in \mathcal{A} \otimes \mathcal{A}$ and $y \in S$, then the set $E_{y} \subseteq S$ defined by $$E_{y} = \{x \in S \, \mid \, (x,y) \in E\}$$ satisfies $E_{y} \in \mathcal{A}$. Indeed, if we let $\mathcal{G}$ denote the family $$\mathcal{G} = \{E \subseteq S \times S \, \mid \, E_{y} \in \mathcal{A}\},$$ then $\mathcal{G}$ is a $\sigma$-algebra on $S \times S$ containing the collection of measurable rectangles $\{A \times B \, \mid \, A,B \in \mathcal{A}\}$. Therefore, $\mathcal{A} \otimes \mathcal{A} \subseteq \mathcal{G}$, proving the claim.

Now if $\Delta \subseteq S \times S$ is the diagonal, then $\Delta_{y} = \{y\}$. Therefore, $$\forall y \in S \quad \{y\} = \Delta_{y} \in \mathcal{A}.$$

  • $\begingroup$ Thank you for the clean and rigorous proof! $\endgroup$
    – p-value
    Dec 20, 2017 at 21:43

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