Let's say we have Borel $\sigma$-algebra for the interval $(0,1]$. Borel $\sigma$- algebra will have all singleton sets. Which implies it will contain all rationals and irrationals. But i doubt that there will not be any set in Borel $\sigma$-algebra which will contain all irrationals. Because we only have countable additivity axiom, and all irrationals are uncountable, and hence cant be build from taking unions of all singleton sets of irrationals. But since there will be a set which will contain all rationals, because its just union of all countable singleton sets of rationals. But if such a set exist then, our Borel $\sigma$; algebra must have its complement(a set with all its elements irrationals). Is such set exists(Containing all irrationals)??

Why such a question?? I am learning Probability theory where i found that professor assigned the measure to all rationals $=0$, and said that since irrationals are the complement of that, assigned measure $=1$ to them.So i got confused and posted a question. He is right in his way, because there is a set which includes all rationals, so its complement must be there.

  • 1
    $\begingroup$ Yes, the set of all rationals, and the set of all irrationals, are there. Together with some other sets, such as "all rationals plus countably many irrationals" and "all irrationals but countably many". $\endgroup$
    – user491874
    Dec 20, 2017 at 13:52
  • $\begingroup$ Please actually ask a well posed question if you want any useful feedback. $\endgroup$ Dec 20, 2017 at 14:10
  • $\begingroup$ What is your question? $\endgroup$
    – Jack M
    Dec 20, 2017 at 14:18

3 Answers 3


It seems you are confused because on the one hand you have that $\Bbb Q$ (the set of all rationals) is contained in your $\sigma$-Algebra $\mathcal A$, and there is a rule which says

If $X\in\mathcal A$, then the complement $\Bbb R -X\in\mathcal A$.

But on the other hand you are not able to construct $\Bbb I=\Bbb R-\Bbb Q$ (set of all irrational numbers) as union of countably many singletons.

And this is okay. Just because the rules of the $\sigma$-Algebra state that anything which you can construct from countably many singleton is in $\mathcal A$, does not mean that everything in $\mathcal A$ must be constructable in this way. You actually found the counterexample $\Bbb I$.


Your question (which I have understood to be: does the $\sigma$-algebra contain the set of irrational numbers?) mentions the Borel $\sigma$-algebra by name, but your discussion seems to be mostly about which sets can be reached by countable unions of singletons, while the Borel $\sigma$-algebra is the sets which can be reached by operations on open intervals, not singletons. Singletons are not open intervals in the usual topology on $(0,1]$. The Borel $\sigma$-algebra contains the singletons, but it is not restricted to be only those sets reached as countable unions of singletons. It contains the open intervals.

Which $\sigma$-algebra are you interested in here? One containing only those sets generated by singletons, or the one generated by open intervals? I will try to address both.

For any set $X$, you may consider the $\sigma$-algebra generated by the singletons in $X$. The $\sigma$-algebra will contain all countable subsets of the space, because of countable additivity. However it will not necessarily contain all uncountable subsets. But please remember that $\sigma$ algebras are also closed under complementation and intersection. If $\{x\}$ is in the $\sigma$-algebra, so is $X\setminus\{x\}.$ In fact the complement of any countable set is also in the $\sigma$-algebra. To address your example, the irrational numbers are in the $\sigma$-algebra generated by the singletons, because their complement is countable.

But consider a subset which is uncountable, but whose complement is also uncountable. For example if $X=(0,1]$, consider the subset $(0,1/2].$ This is not in the $\sigma$-algebra generated by singletons.

So what about the Borel $\sigma$-algebra? This is the $\sigma$-algebra generated by open intervals like $(1/3,1/2)$. Since $\sigma$-algebras are closed under complementation, it also includes the closed sets, including the singletons. The Borel $\sigma$-algebra is a refinement of the $\sigma$-algebra generated by singletons. And because $\sigma$-algebras are closed under countable union, it includes the rationals. And because it's closed under complementation, it includes the complement of the rationals. So yes, the irrationals are in the Borel $\sigma$-algebra.


Remember that one of the operations that gets used in building a $\sigma$-algebra is complementation: a $\sigma$-algebra must be closed under taking complements.

So we can see that the set of irrationals is in the Borel $\sigma$-algebra $\mathcal{B}$ - which is defined to be the smallest $\sigma$-algebra containing every open interval - as follows:

  • Every singleton $\{a\}$ can be written as a countable intersection of open intervals: $$\{a\}=\bigcap_{n\in\mathbb{N}}(a-{1\over n}, a+{1\over n}).$$ So since $\sigma$-algebras are closed under countable intersections, we know that every singleton is in $\mathcal{B}$.

  • The set $\mathbb{Q}$ of rational numbers is countable, hence a countable union of singletons. So since $\sigma$-algebras are closed under countable unions, we know that $\mathbb{Q}$ is in $\mathcal{B}$.

  • Finally, the set of irrationals is the complement of $\mathbb{Q}$; since $\sigma$-algebras are closed under complements, we know that the set of irrationals is in $\mathcal{B}$.

Note that this argument in fact shows something stronger: every countable set of reals, and every co-countable set of reals (= set of reals whose complement is countable), is in $\mathcal{B}$. So for example both the set of algebraic numbers and the set of transcendental numbers are in $\mathcal{B}$, and so on.

One point worth mentioning is the following. There are three common definitions of $\sigma$-algebra of sets of reals:

  • A collection of sets containing $\mathbb{R}$ and closed under complements, countable unions, and countable intersections. This is the definition I used above.

  • A collection of sets containing $\mathbb{R}$ and closed under complements and countable unions.

  • A collection of sets containing $\mathbb{R}$ and closed under complements and countable intersections.

It turns out that, despite looking different, these three definitions are equivalent. This is a good exercise. HINT: think about how complementation turns intersections into unions and vice versa.


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