Let $X(\omega)$ be a continuous random variable defined on a probability space $(\Omega,\mathcal{F},\mathbb{P})$.

How does the measurability condition of this random variable (i.e. $\forall r\in\mathbb{R}, \{\omega:X(\omega)\leq r\}\in\mathcal{F}$) cohere with the $\sigma$-additivity of the $\sigma$-algebra $\mathcal{F}$?

I understand that a $\sigma$-algebra requires that the union of countable many elements of $\mathcal{F}$ must also be contained in $\mathcal{F}$. But if we consider a continuous random variable, then the events that the $\sigma$-algebra $\mathcal{F}$ must contain are uncountable, since $r$ can be any value in $\mathbb{R}$ and $X(\omega)$ is a continuous r.v.

What am I missing?

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    – Andrei
    Commented Jan 22, 2022 at 11:25

1 Answer 1


A $\sigma$-algebra is closed under countable unions. This does not mean that its elements cannot be uncountable, or that it's impossible that some (or even all) uncountable unions of elements in the $\sigma$-algebra are in the $\sigma$-algebra.

For example, if $\Omega$ is an arbitrary nonempty set (possibly an uncountable one, like $\mathbf R$ or $\mathbf R^{\mathbf R}$), both $\{\emptyset,\Omega\}$ and $\mathcal P(\Omega)$ (the whole power set) are perfectly good $\sigma$-algebras of subsets of $\Omega$, and both of them are closed under arbitrary (even uncountable) unions and have uncountable elements, if $\Omega$ is not countable.

In fact, unless $\Omega$ is countable, there is always an uncountable elements in any $\sigma$-algebra: namely, $\Omega$ itself.

Further, if $\mathcal F$ is a $\sigma$-algebra of subsets of an uncountable set $\Omega$, and $\mathcal F$ contains singletons, then there is always at least one uncountable disjoint union of elements of $\mathcal F$ which is in $\mathcal F$, namely, $\Omega=\bigcup_{\omega\in \Omega} \{\omega\}$.

  • $\begingroup$ Thank you. I am studying probability theory, and is it correct to conclude that we require a $\sigma$-algebra to be closed only under countable unions because otherwise under uncountable unions we could only assign probability = 0 to every element of $\mathcal{F}$? $\endgroup$
    – Andrei
    Commented Jan 22, 2022 at 11:24
  • $\begingroup$ @Andrei: No, that is not quite correct: for more or less the reason you say, the measure is only countably additive (that is, otherwise, if singletons have zero measure, then all sets would have zero measure, so the only measures left would be the weighted counting measures). The $\sigma$-additivity of the $\sigma$-algebra itself is related more closely to the fact that if we allow arbitrary unions, we get all kinds of strange, pathological sets we have no hope to be able to sensibly assign a measure to. $\endgroup$
    – tomasz
    Commented Jan 22, 2022 at 13:48

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