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I am new to measure theory. I have good grasp on statistics and probability concepts, and not I am trying to learn it from the measure theory perspective.

I encountered Borel Sets. The definition (taken from wikipedia) is as follows,

Borel set is any set in a topological space that can be formed from open sets (or, equivalently, from closed sets) through the operations of countable union, countable intersection, and relative complement.

Now, what I don't understand is, why we have a different definition, instead of just calling these sets as subsets of $\mathbb{R}$. For example, Say I have two sets, $(- \infty,b] , [b,\infty)$. Then by $(-\infty,b]\cap [b,\infty)$ we have ${b}$. Then by, $(-\infty,b]\cap \{b\}^c$ we have $(-\infty,b)$ and so on.

The point I am making here is, with enough open interval and closed interval subsets, we can achieve any subsets on the Real Numbers. So why even bother with the definition? Is there any advantage of having Borel sets be defined in such a way, instead of just calling those subsets of $\mathbb{R}$ ?

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    $\begingroup$ When I was at your level of understanding in measure theory I would probably not have understood any of the highly sophisticated answers you got. So feel free to ask for clarifications if needed! $\endgroup$
    – Ruy
    Commented Jul 18, 2022 at 17:32
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    $\begingroup$ Ruy , thanks for that. But I think I understood most of the answers. Basically I was also thinking the same that, we are using borel sets because sone subsets of R is not Borel. And that is Vitali sets. So that clears up $\endgroup$ Commented Jul 19, 2022 at 5:35

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You write:

"The point I am making here is, with enough open interval and closed interval subsets, we can achieve any subsets on the Real Numbers."

This is wrong! It is true that "nicely-describable" sets of reals tend to be Borel, but it is definitely not true that all sets of reals are Borel. The easiest way to prove this is just by a counting argument: there are $2^{\aleph_0}$-many Borel sets but $2^{2^{\aleph_0}}$-many sets of real numbers in general, so "most" sets are non-Borel. With more care we can produce concrete (i.e. non-Vitali) examples of non-Borel sets, such as Lusin's original example or a combinatorial one due to Schmerl after some easy rephrasing to "put it in $\mathbb{R}$," but I recommend starting by understanding the counting argument mentioned above.


There's an interesting technical subtlety here. It's consistent with $\mathsf{ZF}$ (= set theory without the axiom of choice) that $\mathbb{R}$ is a countable union of countable sets - while of course $\mathsf{ZF}$ proves that $\mathbb{R}$ is uncountable, $\mathsf{ZF}$ does not prove that a countable union of countable sets is countable! - and it follows easily from this that the smallest $\sigma$-algebra containing the open sets is $\mathcal{P}(\mathbb{R})$. But above, I said there were examples of concrete non-Borel sets, and concrete examples shouldn't require the axiom of choice. So, what gives?

The issue is that if we drop choice, we have to be very careful about what "Borel set" means. The "from-above" definition ("element of the smallest $\sigma$-algebra containing all open sets") and "from-below" definition ("having a Borel code") no longer coincide. See this old post of mine for some details.

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  • $\begingroup$ "It's consistent with ZF [...] that $\mathbb{R}$ is a countable union of countable sets" So $\mathbb{R}$ is countable under ZF alone? I don't really know what the Zermelo Frenkel axioms are because my understanding of set theory is rather rudimentary, so this is weird. $\endgroup$
    – Caio Lins
    Commented Jul 18, 2022 at 13:48
  • $\begingroup$ @CaioLins While $\mathbb{R}$ is provably uncountable in $\mathsf{ZF}$, what $\mathsf{ZF}$ cannot prove is that a countable union of countable sets is countable! See e.g. the discussion here. $\endgroup$ Commented Jul 18, 2022 at 13:52
  • $\begingroup$ Well that's very surprising! Thanks for the answers. $\endgroup$
    – Caio Lins
    Commented Jul 18, 2022 at 13:53
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Take a look at Vitali Sets. In short, we can't consistently define the Lebesgue measure (which assings intervals to their lengths, i.e. $(a, b] \mapsto b - a$) on all the subsets of $\mathbb{R}$, so we restrain ourselves to a class of subsets where it can be done. This class, the Borel sets, is not the biggest one where this is possible (take a look at this extension theorem to see how to extend this notion of length to a larger class of subsets of $\mathbb{R}$), but it is big enough to contain all sets that are of interest, usually.

And this definition of Borel sets that you presented, as far as I know, is not the standard one. The class of (real) Borel sets is usually defined as the smallest $\sigma$-algebra which contains all open sets of $\mathbb{R}$.

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  • $\begingroup$ Note that we don't even need Vitali sets to get non-Borel sets - there are concrete examples of non-Borel sets! (See my answer.) $\endgroup$ Commented Jul 18, 2022 at 13:03
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A Borel set is any set that can be produced by countable unions or intersections of open sets or complements of such sets! But it is not the class of all subsets of real numbers but a much smaller set. It can also be defined as the smallest $\sigma$ algebra which includes all open and all closed sets! For instance the set of Lesbegue measurable sets is bigger than Borel sets but even this does not coincide with the set of all subsets, because there are examples (very hard to produce) of non-measurable sets! See Royden's : "Real Analysis" for an excellent introduction!! :

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    $\begingroup$ While the conclusion "it is not the class of all subsets of real numbers but a much smaller set" is correct, your specific claim is wrong: you can use closed sets since complementation is included in the operations used to build the Borel sets. The Borel $\sigma$-algebra is equivalently the smallest $\sigma$-algebra containing all the closed intervals. $\endgroup$ Commented Jul 18, 2022 at 12:54
  • $\begingroup$ You are right, I had forgotten the correct definition! Thanks! $\endgroup$
    – user1054388
    Commented Jul 18, 2022 at 12:59
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Let $(X, \tau) $ be a topological space. Then sigma algebra generated by open sets is called borel sigma algebra and a member of Borel sigma algebra is called borel set.

A borel set can be obtained by countable union, intersection, set complementation of open sets.

You are almost correct. It is very difficult to find a set which is not borel . But believe me there are many such sets!

The cardinality of collection of all borel sets is $\mathfrak{c}$ i.e there are as many borel sets as reals.

But collection of all subsets of $\Bbb{R}$ has cardinality $2^{\mathfrak{c}}$ .

$\mathfrak{c} < 2^{\mathfrak{c}}$ by Cantor's theorem.

So there are many sets (many!) in $\Bbb{R}$ which are not borel.


Let $\mathcal{C}$ denote the famous Cantor set ( my favorite).

Then $\mathcal{C}$ has $2^{\mathcal{c}}$ many subsets. But Lebesgue measure of Cantor set is $0$ , hence every subset is measurable. Cantor set has a subset which is not Borel but still Lebesgue measurable.


Every borel set is Lebesgue measurable. Hence a set which is not Lebesgue measurable can be borel.

Vitali set, Bernstein set are non borel subsets of $\Bbb{R}$ ( here we need some choice!)

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Not every subset of $\mathbb{R}$ is Borel. Wikipedia gives an example of a non-Borel set, citing Kechris, Descriptive Set Theory; the construction is sufficiently complicated that I too will defer to Kechris here.

Now, you might say: given a set $A$, can't we just write $$A=\bigcup_{a\in A}{\{a\}}$$ and conclude that, since each $\{a\}$ is Borel, so is $A$? No, because the union defining $A$ need not be countable. There is a term in the union for each element of $A$, and $A$ can be quite large indeed. In the example Wikipedia cites, $A$ is equinumerous with $\mathbb{R}$.

Now, you might say: OK, but can't we write down some sort of list of Borel sets, rather than starting with the open sets and then taking closure under some operations? Actually, one can. But it's not as interesting as you might hope.

The key is to recognize that the axioms of a $\sigma$-algebra in fact encode logical formulas. The atomic formulas are any proposition of the form "$x\leq c$" for some constant $c$. "And" and "or" correspond to intersection and union; negation to set complements, and we can quantify over $\mathbb{Q}$ since the universal quantifier is a countable intersection. To be precise, let $\mathscr{L}=\{\leq\}\sqcup\{r:r\in\mathbb{R}\}$ be the language with one relation and $|\mathbb{R}|$-many constant symbols. Thus $S$ is Borel iff $S=\{x:\phi(x)^{\mathbb{R}}\}$, where $\phi$ is an infinitary ($\mathcal{L}_{\omega_1,\omega}$) logical formula in $\mathscr{L}$ that quantifies only over $\mathbb{Q}$, and where the superscript is to indicate that the symbols in $\mathscr{L}$ are interpreted in the obvious way. (In fact, since the continuous preimage of an open set is open, you can use continuous functions in $\phi$ too.)

That logical representation also underlies the counting argument Noah Schweber mentioned. There are only $|\mathbb{R}|\times|\mathbb{N}|=|\mathbb{R}|$-many such formulas, so there are only as many Borel sets as there are real numbers. But there are definitely more subsets of $\mathbb{R}$ than that.

(I screwed up describing the logic for formulas characterizing Borel sets; h/t Noah Schweber for the correct name.)

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  • $\begingroup$ This is wrong - you need to use infinitary ($\mathcal{L}_{\omega_1,\omega}$) formulas, not just first-order formulas, in your characterization. Note that an $n$-quantifier first-order $\mathscr{L}$-sentence corresponds to a Boolean combination of ${\bf \Sigma^0_n}$ sets, so any Borel set which is not of finite Borel rank gives a counterexample. $\endgroup$ Commented Jul 18, 2022 at 16:54
  • $\begingroup$ @NoahSchweber: Corrected, thanks! $\endgroup$ Commented Jul 18, 2022 at 22:01

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