The following question is from Folland Real Analysis, chapter 1 problem 3.

Let $\mathcal{M}$ be an infinite $\sigma$-algebra. Prove that

a. $\mathcal{M}$ contains an infinite sequence of disjoint sets.

b. $\text{card}(\mathcal{M}) \ge \mathfrak{c}$.

This is the problem I'm totally stuck at. First, I think there is a missing condition in (a). For (a) to be meaningful, (a) should be corrected : "M contains an infinite collection of disjoint and nonempty sets." But I can't find a way to construct such a collection of sets.

Could anyone show me how to solve it?

  • 1
    $\begingroup$ If $M$ had only finitely many disjoint sets, it would be finite. $\endgroup$
    – coffeemath
    Commented Jan 2, 2015 at 6:06
  • 1
    $\begingroup$ @coffeemath That is not the counterpositive of the statement in part (a). A priory $\mathcal{M}$ could have arbitrarily large collections of disjoint sets without having an infinite disjoint sequence. $\endgroup$
    – Pp..
    Commented Jan 2, 2015 at 6:40
  • $\begingroup$ @Pp.. If the collection of sets is pairwise disjoint, as I thought it meant, such a sequence is immediate. Your answer below takes the more general situation into account, +1 for it. $\endgroup$
    – coffeemath
    Commented Jan 2, 2015 at 7:29

1 Answer 1


Let $X$ be the whole space. First we show that

there is $E\in\mathcal{M}$ such that the restriction of $\mathcal{M}$ to $E^c$ is still infinite.

If no such $E$ existed, then pick any $\emptyset\neq E\in\mathcal{M}$. The restriction of $\mathcal{M}$ to $E^c$ is finite. But the restriction to $E$ must also be finite because otherwise we could take $E^c$ for the role of $E$. Notice that $\mathcal{M}$ would be generated by the two finite, and disjoint, restrictions and that would imply it is itself finite.

Now apply induction to define the infinite sequence. Pick the first $E_0$ with that property, $E_1$ with the same property from the restriction of the $\sigma$-algebra to $E^c$, $E_2$ from the restriction of the $\sigma$-algebra to $E^c\setminus E_1$, and so on ...

For part (b) consider the (uncountable many different) arbitrary unions of elements of the sequence. They must all be elements of $\mathcal{M}$.

  • 5
    $\begingroup$ what do you mean by restriction of sigma algebra?? $\endgroup$ Commented Aug 31, 2016 at 2:52
  • 4
    $\begingroup$ The restriction of $\mathcal{M}$ to $E^c$ is: $\{F \cap E^c: F \in \mathcal{M}\}$. $\endgroup$ Commented Sep 8, 2017 at 0:30
  • $\begingroup$ @Pp.. could you expand on the induction part? How do we know that there exists an $E_2$ in the restriction of the sigma algebra to $E^c$ without $E_1$? Do we need to know that the restriction is a sigma algebra itself? Thank you $\endgroup$
    – MathIsHard
    Commented Aug 22, 2018 at 16:05
  • $\begingroup$ I might just be missing something, but I don't see why we know we can find the second element of the infinite sequence. I feel like all we showed is that we can find one element in an infinite sigma algebra so that the restriction is infinite. I don't know why we can say we can find another in the restriction of the first element and the sigma algebra. Unless maybe the restriction of the first element and the sigma algebra is a sigma algebra itself, but I am not sure. $\endgroup$
    – MathIsHard
    Commented Aug 22, 2018 at 16:27
  • $\begingroup$ Sorry one more thing... I am wondering if we just showed that any infinite set of sets contains an infinite restriction. So that the sigma algebra component isn't needed for the next iteration. $\endgroup$
    – MathIsHard
    Commented Aug 22, 2018 at 16:35

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