Explanation Borel set

Question

I'm getting totally crazy with the Borel set, I use this concept for maybe two years, but I still don't know what it really is. I would really appreciate if somebody could finally explain me what is a the Borel set. These are my question:

Consider a measurable set $(X,\mathscr M,\mu)$ where $\mathcal T$ the topology of $X$.

1) The Borel set is such that $\mathscr B\subset \mathscr M$ and it's the smallest $\sigma -$algebra generated by all the element of $\mathcal T$. What does it mean? I agree that all element of $\mathcal T$ are in $\mathscr B$, and also all element of $$\{Y\mid \exists U\in\mathcal T: Y=X\backslash U\}$$ by definition of a $\sigma-$algebra, but I know that there is a lot of other set, for exemple, let take the measurable set $(\mathbb R,\mathscr M,m)$ where $m$ is the Lebesgue measure and $\mathbb R$ provide with it's usual topology. Why $[1,2[$ is in the Borel set ?

2) A consequence of the first question, how to prove in general that a set is in the Borel set (excepted if it's an open or a close set, it's clearly obvious) ?

3) What is the fantastic side of this set ? Evry body talks about the Borel set like something gorgeous, where is the genius there ?

4) Does the Borel set of $(X,\mathscr M,\mu)$ depend of $\mu$ ? Or only of the Lebesgue measure ? And is a Lebesgue measurable set is always $\mu-$ measurable ?

5) By the way why a $\sigma -$algebra is called a $\sigma -$algebra ? What the $\sigma $ mean ?

I hope my questions are clear.

I'll probably complete this post in the future, and when I'll have more answers.

Answer 1

Lets look at this in a more intuitive way:

A collection of sets $\{E_k\}$ forms an algebra if the collection is closed under complements and finite unions. The collection of sets forms a $\sigma-$algebra if the collection is closed under complements and countable unions.

A 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. This includes any open set, any countable intersection of open sets (known as $G_\delta$ sets), any closed set, any countable union of closed sets (also known as $F_\sigma$ sets), $G_{\delta\sigma}$ sets, $F_{\sigma\delta}$ sets, etc...

Now the set $[1,2[$ is a Borel set because $$ [1,2[=\bigcup_{k=1}^\infty \left[1,2-\frac{1}{k}\right] $$ which makes $[1,2[$ a $F_\sigma$ set and hence a Borel set. In general, if a set is neither open nor closed, but Borel, you should be able to create either a $F_\sigma$ set or a $G_\delta$ exactly equal to the set in question, as I did above.

Let $X$ be a set and let $\mathcal{P}(X)$ be the power set of $X$, then the collection $\mathcal{B}\subseteq\mathcal{P}(X)$ that consists of all Borel sets from $\mathcal{P}(X)$ forms a $\sigma-$algebra of sets and is the smallest possible collection that can do so collection that forms a $\sigma-$algebra and contains all the open sets from $\mathcal{P}(X)$. In other words, every possible $\require{enclose} \enclose{horizontalstrike}{\sigma-}$algebra that can be formed from $\enclose{horizontalstrike}{\mathcal{P}(X)}$ must contain $\enclose{horizontalstrike}{\mathcal{B}}$ the intersection of all $\sigma-$ algebras that contain the open sets is the Borel $\sigma-$algebra. This is what makes Borel sets so special; they are the skeleton of any $\enclose{horizontalstrike}{\sigma-}$algebra, for lack of a better phrase., especially from a topological standpoint.

If you notice in the above discussion, nowhere does the concept of measure become involved. In the case of the Lebesgue measure on $\mathbb{R}$, it turns out that the collection of measurable sets forms a $\sigma-$algebra. But this is simply a consequence of restricting the outer measure to the collection of sets for which countable additivity holds.

Now since the collection of measurable sets forms a $\sigma-$algebra, then it must be that every Borel set in $\mathcal{P}(\mathbb{R})$ is measurable as well. And while it is not immediately obvious, there do exist non-Borel sets that are measurable. Thus not all measurable sets are Borel. This is an important distinction!

Answer 2

To add some informal comments to what was already mentioned here, the word "algebra" is somewhat unfortunate, as the algebraic structure (in the abstract algebra sense) comes from treating the symmetric difference of two member sets as addition, while intersection serves the role of multiplication; however, we are mostly concerned with unions instead of symmetric difference (not all the time-check Littlewood's First Principle!). In this sense, an algebra of sets is a formal Boolean algebra. This is because the intersection (or "multiplication") of any set with itself always returns the set you start with, i.e. $x*x=x$ for all $x$ in the collection.

The members of an algebra must include finite unions and complements of members you know are there, as well as the parent set itself. The word "sigma" means that countably infinite unions of members must also themselves be members. Set algebras not specified as "sigma" algebras can exclude countable unions while still remaining algebras. Think of finite subsets of the natural numbers and their respective complements - you can take finite unions all day and your results will be members of this collection, but you would never get the set of all odd natural numbers, so the set of odd natural numbers is not a member of this algebra; but, the set of all odd natural numbers IS a countable union of elements in this set.

Since complements are included in algebras, the intersection of any two members of an algebra must be in the algebra, since $A\cap B=\left(A^c \cup B^c\right)^c$. Likewise, countable intersections of members of sigma-algebras are always themselves members of the sigma algebra (in other words, a sigma-algebra is "closed" under countable intersection).

With the Borel sigma algebra, the definition "the smallest sigma algebra containing the open intervals" (or open sets; recall, every open set is a countable union of open intervals, though) is somewhat difficult to understand at first glance, and in my opinion is given because it is more formal than defining the Borel sets by activity - i.e. "generating" by any combination of intersection, union, and complements of open sets. It is important to keep in mind, by the way, that Borel sets are more than just countable unions and intersections of open and closed sets. There are countable unions, intersections, complements, etc. of the sets you've mentioned that are also Borel; since a sigma-algebra includes a countable union of members as itself a member of the collection, and likewise with countable intersections of members, then we can take e.g. a countable intersection of a countable union of members and return a Borel set. The "Borel hierarchy" article on Wikipedia is a good exposition of this.

Back to the "smallest sigma algebra" definition. Think of it this way: if a sigma algebra includes all open intervals, it must include all complements, countable unions, and countable intersections of open intervals applied in any combination you choose, i.e. EVERY sigma algebra including the open intervals must include such sets, because of the fact that we are considering a sigma algebra in general in the first place (no matter what other members not having to do with open sets happen to be lurking around).

On the other hand, if a certain set is included in every single sigma-algebra that includes the open sets, it must be some cut-and-paste operation of unions, intersections and complements on the open sets, because if it is not such a set, it fails to belong to every sigma algebra that includes the open sets; namely, by our very assumption, it fails at least to belong to the sigma-algebra arising only from operating with unions, intersections and compliments on open sets! Hopefully, this clears up the "smallest sigma algebra" characterization for you.

As another poster mentioned, the Lebesgue measurable sets include members that are not Borel sets, i.e. cannot be "made" by operating on open sets via complements, and countable unions and intersections. Additionally, there are sets that are not Lebesgue measurable at all (the Vitali set, for instance), in fact every Lebesgue measurable set of positive measure includes a non-measurable subset!

Non-Borel measurable sets can be reached in this way; the Cantor set, a set of zero Lebesgue measure, can be mapped via a measurable function to a set of positive measure. Such a set, as mentioned before, contains a nonmeasurable set. Consider the inverse of our measurable function operating on this nonmeasurable subset, and you will get a subset of the measure zero Cantor set, hence you will get a (hopefully) measurable subset (of measure zero). This means a measurable function can, counterintuitively, map a measurable set to a nonmeasurable set. But it cannot map a Borel set to a nonmeasurable set. So we are stuck with a Lebesgue measurable set (of measure zero!) that is not Borel.

This is, in fact, why Lebesgue measurable sets are needed rather than just strictly sticking to Borel sets only. If we did the latter, we should hope that all subsets of a set of zero measure are themselves measurable with zero measure, but if we only consider the "easy" Borel sets as the measurable ones, we are hit with the inevitable problem of dealing with the fact that there are zero-measure sets with subsets that aren't even measurable (in that they are not Borel) at all. Lucky for us, Caratheodory verified that a "complete" extension of the standard Borel measure (i.e. the measure that reports the length of intervals) exists that can return a zero measure for every subset of a zero measure set, while still preserving measurements we expect on Borel sets. This is, as you may guess, exactly the Lebesgue measure.

The Borel sets are special for the reasons I've already mentioned - they are, intuitively, "elementary" in construction, while non-Borel sets (measurable or no) are typically described in a roundabout way in the context of Lebesgue measure. One can, however, represent every Borel set, at base, as some conglomeration of unions, intersections, and compliments, of open intervals. They may not "look" elementary, e.g. this apparently nonsensical set: $$\bigcap_{k=1}^\infty \bigcup_{n=1}^\infty \bigcap_{i=n}^\infty \bigcap_{j=n}^\infty \{x:\left|f_i(x)-f_j(x)\right|<\frac{1}{k}\}$$

While daunting at first appearance, this is just symbology for the set of points where the sequence of functions converges in the Cauchy sense, with intersection and union standing in for "all" and "there exists (at least one)", respectively: for all $k$ there exists an $n$ such that for all $i,j\geq n$, the distance between $f_i$ and $f_j$ at this point is less than the given "epsilon" bound $1/k$. This set itself is Borel (hence measurable) as it is a porridge of intersections and unions of Borel sets.

Answer 3

Let me try to help here. First of all you need to understand what is the definition of a $\sigma$-algebra generated by a set. So let $S \subset X$ be an arbitrary subset of $X$, then we say that $\sigma(S)$ is $\sigma$-algebra generated by $S$ and is defined by $$ \sigma(S)= \bigcap\left\{ \mathcal{F}\subset \mathcal{P}(X) : \mathcal{F} \ \text{is $\sigma$-algebra and } S \subset \mathcal{F}\right\} $$ it is clearly seen that $\sigma(S)$ is the smallest $\sigma$-algebra such that $S \subset \sigma(S)$. Then you can use $S$ as the usual topology $\mathcal{T}$ of $X$ to generate the borel $\sigma$-algebra. I think that with this information you can answer yourself the first 3 questions.

For the fourth one, I suggest you to read about the Caratheodory and Hahn extension theorems, you will see that if $\mathcal{B}$ are the borel sets, and $\mathcal{L}$ the lebesgue ones then $$ \mathcal{B} \subset \mathcal{L} \subset \mathcal{P}(X) $$ this last line shows that there exists sets who are not lebesgue measurable, see for example the Vitalli set. Then of course if you are working with the outer measure $\mu^{*}$ then all the sets in $\mathcal{L}$ are measurable with $\mu^{*}$, again this is consecuence of caratheodory extension theorem.

Finally the $\sigma$ is due to the property that if a countable (this is what $\sigma$ represents, remember also that capital $\Sigma$ is used to numerable sums) unions of sets on the $\sigma$-algebra is an element of the $\sigma$-algebra. Again the $\sigma$-aditivity property of a measure is that $$ \mu \left( \bigcup_{n=1}^{\infty} A_i \right) = \sum_{n=1}^{\infty} \mu(A_i) $$ for a disjoint countable sequence $\left\{ A_i \right\}$ in the $\sigma-algebra$.

I hope you find this helpful.

Answer 4

Point One

The measures usually considered are constrcuted by specifying their action on a few basic sets like half open intervals. Then one wants to lift them up onto a sigma-algebra. Sigma-algebras constitute a natural domain for measures. That is a place where measures feel home. ;)

Now comes an interesting point:

As the smallest sigma-algebra that contains all half open intervals are really just the Borel sets,
this will be the common domain for all measures constructed this way.

The only difference lies in how much further they can be extended.
Here is the first time the behavior of a measure under consideration becomes involved.

So for example the Lebesgue measure lives up to the Lebesgue measurable sets (that is slightly more sets than the Borel sets) while the Dirac point measure can be pushed up to even all sets. But both live on the common measure independent domain of Borel sets.

Point Two

Abstractly, one proves the Caratheodory measurability.

In practice, one rather checks wether outer measure agrees with inner measure both being finite. That unveals already alot of sets though not all of them. But one must note that this is not a characterization but only some sort of criterion...

Point Three

Yes, that's true: Too many people talk as if the Borel sets are so fantastatic even as if they're the only ones in the world. Ask them why! ;)

Hopefully they will tell you because they allow to consider integration of continuous functions. Unfortunately, many people draw this conclusion from the analogy between measurability and continuity. That is really bad!! The point is rather that continuous functions are almost piecewise constant a.k.a. simple. And that is the crucial ingredient of Lebesgue integration.

After all, one can also consider exotic measure spaces; they're just less useful regarding the above...

Point Five

Sigma stands for sum. The countability restriction comes from the point that sums involve at most countably many nonvanishing terms. Then it becomes reasonable but not exclusive to consider a priori only countable families. Note that there are branches also considering measures additive under uncountable families as set theory!!!