# If $S$ is an infinite $\sigma$ algebra on $X$ then $S$ is not countable

I am going over a tutorial in my real analysis course. There is an proof in which I don't understand some parts of it.

The proof relates to the following proposition:

($$S$$ - infinite $$\sigma$$-algebra on $$X$$) $$\implies$$ $$S$$ is uncountable.

Proof:

Assume: $$S=\{A_{i}\}_{i=1}^{+\infty}$$. $$\forall x\in X: B_{x}:=\cap_{x\in A_{i}}A_{i}$$. [Note: $$B_{x}\in S$$ $$\impliedby$$ ($$B_{x}$$ - countable intersection].

Lemma: $$B_{x}\cap B_{y}\neq\emptyset \implies B_{x}=B_{y}$$.

Proof(of lemma):

$$z\in B_{x}\cap B_{y} \implies B_{z}\subseteq B_{x}\cap B_{y}$$.

1.$$x\not\in B_{z} \implies x\in B_{x}\setminus B_{z} \wedge B_{x}\setminus B_{z} \subset S \wedge B_{x}\setminus B_{z} \subset B_{x}$$ (contradiction:$$\space$$ definition of $$B_{x}$$) $$\implies$$ $$B_{z}=B_{x}$$

2.$$y\not\in B_{z} \implies y\in B_{y} \setminus B_{z} \space \wedge \space B_{y}\setminus B_{z} \subset S \space\wedge\space B_{y} \setminus B_{z}\subset B_{y}$$(contradiction: definition of $$B_{y}$$) $$\implies$$ $$B_{z}=B_{y}$$ $$\implies B_{x}=B_{y} \space \square$$

Consider: $$\{B_{x}\}_{x\in X}$$. If: there are finite sets of the form $$B_{x}$$ then: $$S$$ is a union of a finite number of disjoint sets $$\implies$$ $$S$$ is finite $$\implies$$ there is an infinite number of sets of the form $$B_{x}$$. $$\implies$$ $$|\bigcup\limits_{i\in A \subseteq\mathbb{N}}B_{x_{i}}| \geq \aleph_{0}$$.(contradiction) $$\square$$

There are couple of things I don't understand in this proof:

1. Why the fact that we found a set ($$B_{x}\setminus B_{z}$$) in $$S$$ containing $$x$$ and is strictly contained in $$B_{x}$$ a contradiction ?

2. Why if there are only a finite number of different sets of the form $$B_{x}$$ then $$S$$ is a union of a finite number of disjoint sets and is finite ?

• You may find it helpful to think about the case where the $\sigma$-algebra $S$ separates points; i.e. for every $x,y \in X$ there exists $A \in S$ with $x \in A$, $y \notin A$. In this case, $B_x = \{x\}$, and the problem reduces to showing that the discrete $\sigma$-algebra on an infinite set is uncountable. Mar 4, 2013 at 2:36
• This is the easiest to understand proof I've ever found on this site, thanks!
– Vim
May 8, 2016 at 15:46
• Where did you find this proof? I'm looking for resources to accompany Bass's book Dec 31, 2017 at 1:28
• What is the '$\wedge$' symbol here? Sep 13, 2019 at 16:30
• @Belgi I don't understand the statement in 1. For eg what is $S \wedge x$ ? Sep 13, 2019 at 16:36

1. Because $B_x$ is supposed to be the intersection of all measurable sets containing $x$, but you've found a measurable set containing $x$ strictly inside $B_x$.

2. Because for any measurable set $T$, we have $T=\bigcup_{x\in T}B_x$. Thus, if there are $n$ distinct sets of the form $B_x$, then there are at most $2^n$ elements of $S$.

• No problem, glad to help! Mar 4, 2013 at 1:45
• Could someone explain why we must have strict containment? I see how it's possible, but what if nothing is removed ? May 26, 2014 at 5:49
• @NS248 $S$ is essentially generated by the class $\{B_x\}$. Each element in $S$ can only be a countable Union of members in that class, since intersection can only lead to the null set.
– Vim
Jun 26, 2016 at 7:51
• Do you know where I can find the a proof of the $2^n$ i know its supposed to be trivial but I'm not able to show that the space of all strings formed by complement, union, and intersect quotient set-wise equality is of the same size or less than the size of power set $B_x$ Jan 12, 2018 at 22:28
• i don't understand the last part of the proof... how having infinite number of x can conclude S being uncountable? Apr 5, 2018 at 20:27

One can actually change the argument of @Josef slightly to obtain a correct proof. This is probably also what @ncmathsadist is hinting at, although he does not indicate how to obtain the countably infinite collection of disjoint subsets.

Incidentially, the argument below even shows that an infinite $$\sigma$$-algebra is not only uncountable, but it has at least the cardinality of the continuum.

Let $$(A_n)_{n \in \Bbb{N}}$$ be a sequence of pairwise distinct (not necessarily disjoint) sets in $$S$$.

For an arbitrary subset $$M \subset X$$ let $$M^1 := M$$ and $$M^{-1} := M^c$$.

For a sequence $$\omega = (\omega_n)_{n} \in \{\pm 1\}^\Bbb{N}$$, define

$$B^\omega := \bigcap_{n \in \Bbb{N}} A_n^{\omega_n}.$$

Observe that $$B^\omega \in A$$, because it is a countable intersection of elements of $$A$$.

ALso note that $$B^\omega \cap B^\gamma = \emptyset$$ for $$\gamma = (\gamma_n)_n \neq \omega$$, because there is some $$n$$ such that $$\gamma_n \neq \omega_n$$, so that

$$B^\omega \cap B^\gamma \subset A_n^{\omega_n} \cap A_n^{\gamma_n} = \emptyset.$$

Finally, note that for $$n \in \Bbb{N}$$ arbitrary, we have (why?)

$$A_n = \bigcup_{\omega \text{ with } \omega_n = 1} B^\omega .$$

Hence, if the set $$\{B^\omega \mid \omega \in \{\pm 1\}^\Bbb{N} \}$$ were finite, it would easily follow that there could only be finitely many distinct $$A_n$$, a contradiction.

Hence, there is an infinite family $$\omega^{(n)}$$ with $$B^{\omega^{(n)}} \neq B^{\omega^{(m)}}$$ for $$n \neq m$$.

Let $$C_n := B^{\omega^{(n)}}$$. By discarding (at most) one set, we can assume $$C_n \neq \emptyset$$ for all $$n$$. Using the pairwise disjointness of the $$C_n \neq \emptyset$$, it is now easy to see that the map

$$\Gamma : \{0,1\}^\Bbb{N} \to A, (\alpha_n) \mapsto \biguplus_{n \text{ with } \alpha_n = 1} C_n$$

is injective, because

$$\Theta : A \to \{0,1\}^\Bbb{N}, M \mapsto \left(n \mapsto \begin{cases} 1, & \text{if }M\cap C_{n}\neq\emptyset\\ 0, & \text{if }M\cap C_{n}=\emptyset \end{cases} \right)$$

is a left inverse for $$\Gamma$$.

Proof by picture:

Take a collection of finitely many disjoint elements of the sigma algebra. Then the picture below shows gives some intuition as to why one can always add a element to this set of pairwise dijoint sets. Once you have a infinite collection of pairwise disjoint sets one can identify each of these as distinct elements where unions of sets are also distinct. So by taking all countable unions on this collection one would generate a set with uncountably many elements as it is the power set of a countable set.

• How do we know we don't get stuck in choosing some finite family of sets that happens to cover everything, making the choice of the "next" element impossible? Oct 4, 2014 at 15:35
• Easy, we have an infinite collection of sets in our $\sigma$ algebra. There are only a finite number of sets that can be unions of sets from our collection $2^k$ for a collection of k disjoint sets. Thus we can find a set that is not the union of these sets and thus it's complement intersected with a set in the collection and itself intersected with a set are nonempty and disjoint.
– asd
Oct 4, 2014 at 15:43
• But even if we have an infinite collection of sets, a finite number of them might cover the union of all of them. While a $k$ cardinality family of disjoint subsets form only a finite number of unions, and we can pick a subset distinct from these, that doesn't show the one we pick will not be contained in their union. Oct 4, 2014 at 15:51
• I think I might understand your concern. The idea the picture is meant to convey is not quite "adding" but "carving" instead. If some set A is contained in their union but not equal to their union for some subcollection then it follows that $A^\complement \cap B_n$ is non-empty which is in the σ algebra and disjoint with $A \cap B_n$ also being in the $\sigma$ algebra. Now of course we have to replace the sets containing some part of A with their respective intersection with the complement of A, and intersection with A.
– asd
Oct 4, 2014 at 16:23

Show that if a $\sigma$-algebra is infinite, that it contains a countably infinite collection of disjoint subsets. An immediate consequence is that the $\sigma$-algebra is uncountable.

• Can you explain that a little more please? Feb 24, 2014 at 1:40
• What is the cardinality of the power set of a countably infinite set? Sep 12, 2014 at 13:19
• $2^{\mathbb{N}}$? Sep 20, 2014 at 14:46
• It's not easy to show the existence of a sequence of disjoint members in the sigma-algebra.
– Vim
May 8, 2016 at 15:16

I elaborate ncmathsadist's answer a little bit:

We define the notion of atom: An atom is a non-empty minimal measurable set. That is, a non-empty measurable set $A$ is an atom iff for any measurable set $B\subseteq A$, either $B=\emptyset$ or $B=A$. Note that any two distinct atoms are disjoint. We have the following cases:

Case 1: There are finitely many atoms $A_{1},A_{2},\ldots,A_{n}$ and $\bigcup_{i}A_{i}=X$. In this case, the $\sigma$-algebra is finite and every measurable set is a union of atoms.

Case 2: There are finitely many atoms $A_{1},A_{2}\ldots,A_{n}$ but $\bigcup_{i}A_{i}\neq X$. (This includes the case that there is no atom) In this case, let $Y=X\setminus\bigcup_{i}A_{i}$. We can construct inductively a sequence of measurable set $\{B_{n}\}$ such that $Y=B_{1}\supsetneq B_{2}\supsetneq B_{2}\supsetneq\ldots$. Define $C_{1}=B_{1}\setminus B_{2}$, $C_{2}=B_{2}\setminus B_{3}$ etc, then $\{C_{n}\}$ is a sequence of pairwisely disjoint non-empty measurable sets.

Case 3: There are infinitely many atoms. Choose a countably infinite sub-family of atoms, then they are pairwisely disjoint.

The proof above is terse. I've filled out the missing bits, any error corrections are greatly appreciated.

What we want to prove: A sigma algebra $$\mathcal{F}$$ of set $$X$$ is either finite or uncountable.

Proof: First suppose $$X$$ is finite. Then, the largest sigma algebra of $$X$$, $$\mathcal{P}(X)$$, is finite. Now, suppose $$X$$ is infinite. We assume that $$\mathcal{F}$$ is countably infinite, and derive a contradiction.

As $$\mathcal{F}$$ is countably infinite, denote $$\mathcal{F} = \{A_i : i \in \mathbb{N} \}$$. For $$\forall x \in X$$, define $$B_x := \cap_{x \in A_i} A_i$$. As $$\mathcal{F}$$ is countable, note that $$B_x$$ is a countable intersection of elements of $$\mathcal{F}$$, so is an element of $$\mathcal{F}$$. Therefore, for $$\forall x, y,z \in X$$, $$B_x \cap B_y \in \mathcal{F}$$, and $$B_x \backslash B_z \in \mathcal{F}$$.

I claim that $$B_x \cap B_y \neq \emptyset \implies B_x = B_y$$. As $$B_x \cap B_y \in \mathcal{F}$$, $$\exists i_0$$ such that $$B_x \cap B_y = A_{i_0}$$. Let $$z \in B_x \cap B_y$$. Then, $$B_z = \cap_{z \in A_i} A_i = \left( \cap_{\substack{z \in A_i \\ i\neq i_0}} A_i\right) \cap A_{i_0} \subset A_{i_0} = B_x \cap B_y.$$ Suppose that $$x \notin B_z$$. Then, $$x \in B_x \backslash B_z \in \mathcal{F}$$. This is a contradiction because as $$z \in B_x, B_z$$, we know $$B_x \backslash B_z$$ is strictly smaller than $$B_x$$. However, $$B_x$$ is defined to be the intersection of all sets in $$\mathcal{F}$$ containing $$x$$, so is the `smallest' set in $$\mathcal{F}$$ containing $$x$$ which is also a subset of all other sets in $$\mathcal{F}$$ containing $$x$$. Therefore, we conclude $$x \in B_z$$. As $$B_z$$ also contains $$x$$, we obtain $$B_x \subset B_z$$. Thus, we have $$B_x \subset B_z \subset B_x \cap B_y \subset B_x.$$ We can conclude $$B_x = B_z = B_x \cap B_y$$. By symmetry, $$B_y = B_z = B_x \cap B_y$$. Thus, $$B_x = B_y$$, and our claim is proven.

Let $$B = \{ B_x\}_{x \in X}$$. If $$|B| = n < \infty$$, then we have that $$\mathcal{F}$$ is finite. This is because for any $$A \in \mathcal{F}$$, $$A$$ can be expressed $$A = \cup_{a \in A} B_a$$ (as $$B_a \subset A,$$ $$\forall a \in A$$, and $$B_x$$ exists and is non-empty for $$\forall x \in X$$ as $$x \in X \in \mathcal{F}$$). So, there can exist at most $$2^n$$ elements of $$\mathcal{F}$$, a contradiction to the assumption that $$\mathcal{F}$$ is countably infinite. Now assume that $$B$$ is uncountably infinite. As we established $$B_x \in \mathcal{F}, \forall x \in X$$ (so $$B \subset \mathcal{F}$$), we have that $$\mathcal{F}$$ is uncountably infinite, a contradiction to the assumption that $$\mathcal{F}$$ is countably infinite.

Finally, assume that $$B$$ is countably infinite. Using the claim, we can let $$B = \{ C_i \}_{i \in \mathbb{N}} \cup C_0$$ where $$C_0 = \emptyset$$, and $$C_i$$ are disjoint non-empty sets such that $$C_i = C_j \implies i = j$$, $$\forall i, j \in \mathbb{Z}_{\ge 1}$$. Consider the set $$C = \{\cup_{i \in I} C_i: I \subset \mathbb{N}\}$$. Note that $$I \neq J \implies \cup_{i \in I} C_i \neq \cup_{i \in J} C_i$$. As $$I$$ runs over all elements of the power set of $$\mathbb{N}$$, $$C$$ has uncountably infinitely many elements. However, as each element of $$C$$ is composed of a countable union of elements of $$\mathcal{F}$$, $$C \subset \mathcal{F}$$. This is a contradiction to the assumption that $$\mathcal{F}$$ is countably infinite.

• Great post. But I find that the $B_z$ only obfuscates the whole idea (also in the original proof), which is this: If $B_x$ is the smallest measurable set containing $x$, and $B_y$ is the smallest measurable set containing $y$, then if $x\in B_y, y\in B_x$, we have $B_x=B_y$ by definition. Otherwise, assume without loss of generality that $x\not\in B_y$. Then $B_x\setminus B_y$ is also a measurable set containing $x$, but since $B_x$ is the smallest, it follows that $B_x\cap B_y=\emptyset$. Apr 4 at 22:06

Warning: the following proof is incorrect!

As ncmathsadist mentioned before, there is also a direct prove. For those who want to know how to proceed:

Let be $\mathcal{S}$ an infinite $\sigma$-Algebra. Take any pairwise distinct sets $A_1,A_2,\ldots\in\mathcal{S}$. Then \begin{align*}G_0&=A_0^\complement\cap A_1\cap A_2\cap A_3\cap\cdots,\\ G_1&=A_0\cap A_1^\complement\cap A_2\cap A_3\cap\cdots,\\ G_2&=A_0\cap A_1\cap A_2^\complement\cap A_3\cap\cdots,\\ \vdots\end{align*} all lie in $\mathcal{S}$, because the countable intersection is closed in $\sigma$-Algebras, and it is easy to see, that they are pairwise disjoint. But also the countable and finite union of any collection of the $G_k’s$ must lie in $\mathcal{S}$.

You will recognise, that for distinct collections, you’ll get different unions (because the $G_k$’s are pairwise disjoint). Wich Collection you choose, you can encode with a binary sequence $a_0,a_1,a_2,\ldots$ by setting $a_k=1$ for taking $G_k$ into the collection and $a_k=0$ otherwise. So you get an injection from $\{0,1\}^\mathbb{N}$ to $\mathcal{S}$. But you already know, that $\{0,1\}^\mathbb{N}$ is an uncountable set. And that’s why $\mathcal{S}$ must also be uncountable.

• Thanks. I like this direct proof. May 25, 2014 at 21:20
• Joseph, can you point out, why all but finitely many Gi s are not empty? Aug 11, 2014 at 17:01
• They aren't, if $A_0$ and $A_1$ happen to to be disjoint, then $G_3,G_4,...$ are all empty. Sep 17, 2014 at 3:11
• @bringingdownthegauss You are right, my proof is incomplete. But I see no way to correct it. It seems, that there is no other way to proof it as the questioner did. What should I do with an incorrect answer in stackexchange? Sep 24, 2014 at 17:38
• @Josef: Below, I have indicated a variant of your argument, which yields a correct proof. Oct 17, 2014 at 22:05

Sigma algebras are just Venn diagrams. (with some caveats because of all the "countable union" business)

A sigma field $$\mathcal{F}$$ on $$X$$ defines an equivalence relation on $$X$$ where $$x\sim y$$ iff $$\forall E\in \mathcal{F},x\in E\iff y\in E$$. This partition is just the partition defined by the Venn diagram -- by the little intersection regions. The important point is that there is a bijection $$\mathcal{F}\leftrightarrow \mathcal{P}(X/\sim)$$ -- this should also be obvious with the Venn diagrams.

So what are the possible values for the cardinality of a power set?