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I am reading DeGroot's book titled 'Optimal Statistical Decisions' in which he says the following:

If $S$ is the $n$-dimensional space $\mathbb{R}^n$, then the $\sigma$-field will be taken to be the $\sigma$-field of Borel sets, i.e. , the smallest $\sigma$-field containing all $n$-dimensional intervals.

and

If the function $g$ is measurable with respect to a $\sigma$-field and $B$ is any Borel set on the real line, then the subset $g^{-1}(B)$ of $S$, defined by the relation $g^{-1}(B)=\{s:g(s)\in B\}$, also belongs to the $\sigma$-field.


I have no prior exposure to measure theory and topology, hence I find these two statements difficult to comprehend. However, I do understand what a $\sigma$-field is and the three properties that a collection of subsets of sample space $S$ must fulfill in order to become a $\sigma$-field.

I hope someone can provide a simple and concise explanation of what a Borel set is, so that I can develop a deeper understanding of these two statements above. I only want to learn from a probabilistic standpoint right now and would appreciate it if the explanation would leave out measure theory and topology altogether. Thanks.

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  • $\begingroup$ On $\mathbb{R}$, if I recall correctly the Borel sets are the sigma-algebra generated by the open intervals, or alternatively, the sigma-algebra generated by the closed intervals. In saying this, I still have no intuition whatsoever despite using them occasionally. +1 $\endgroup$ Commented Jun 5, 2019 at 7:59
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    $\begingroup$ There is no simple characterization of Borel sets. For a given set it may not be easy to say whether it is a Borel set or not. So you just have to work with the definition of Borel sigma algebra. $\endgroup$ Commented Jun 5, 2019 at 8:01
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    $\begingroup$ A "constructive" point of view (just to confirm the last comment). $\endgroup$
    – metamorphy
    Commented Jun 5, 2019 at 8:16
  • $\begingroup$ The fact that Borel sets are hard to characterize also comes from the fact that without the axiom of choice, it is consistent that all subsets of $\mathbb R$ are Borel. $\endgroup$ Commented Jun 5, 2019 at 8:44
  • $\begingroup$ @KaviRamaMurthy I understand your argument. Could you possibly explain Borel sigma algebra in that case? $\endgroup$ Commented Jun 5, 2019 at 8:58

2 Answers 2

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A Borel set is actually a simple concept. Any set that you can form from open sets or their complements (i.e., closed sets) using a countable number of intersections or unions is a Borel set. It really is nothing more than that.


Why we need this concept is a much more interesting discussion though and unless you venture into measure theory it will be hard to understand that. But I will attempt an exaplantion. Let's start with some motivation (courtesy Shreve Vol. II, Ch 1):

Consider a trial where you randomly choose a uniformly distributed random number $x$ over $[0,1]$. You define a function $f$ as follows: $$ f(x) = \begin{cases} 1 \qquad & \text{if } x \in \mathbb{R} \setminus \mathbb{Q} \\ 0 \qquad & \text{if } x \in \mathbb{Q} \end{cases} $$ What is $E(f(x))$ when $x$ is drawn in the way described?

Well, expectation is traditionally defined as follows: $$ E(f(x)) = \sum_{i = 1}^\infty f(x_i) P(x_i) \label{def_exp} \tag{1} $$

where $x_i \in [0,1]$ represent all possible numbers one can draw. A very naïve way to sum this up would be to say $P(x_i) = 0 \, \forall x_i \in \mathbb{R}$ and then conclude that the expected value of $f(x)$ is zero. However, that is not true because probabilities can only be summed up when you have a countable set of events (see Kolmogorov's axioms please) and the set of all reals from $[0,1]$ is not countable at all.

That's where Borel sets come in. What you do is map the events from a sample space $\Omega$ onto the Borel subsets of $\mathbb{R}$. Specifically we construct a probability measure $\mu$ as follows, $$ \mu(B) = P\{\omega : X(\omega) \in B \}, \qquad \omega \in \Omega; B \in \mathcal{B}(\mathbb{R}) $$

where $\mathcal{B}(\mathbb{R})$ is the Borel $\sigma$-field on $\mathbb{R}$. Because this definition satisfies the Kolmogorov axioms it qualifies as a valid probability measure. If furthermore, we are able to come up with a valid probability measure that gives a uniform distribution to $x$ on $[0,1]$ then we can use it in our expectation value calculation above.

And we can very easily come up with such a measure for $[0,1]$. For any open interval $(a,b) \subset [0,1]$ we choose $\mu((a,b)) = b - a$ as our probability measure. Now what is this measure for our set of rationals? Consider the rationals $q_1 = 0, q_2, q_3 \cdots$ to be the sequence of all rational numbers in $[0,1]$. For any $\epsilon > 0$ and $i$, define the set:

$$ Q_i = \left(q_i - \epsilon/2^i, q_i + \epsilon/2^i\right) $$

Then, $$ Q_{[0,1]} := Q \cap \left[0,1\right] \subseteq \bigcup_{i = 1}^\infty Q_i \\ \implies \mu \left( Q_{[0,1]} \right) \leq \sum_{i=1}^\infty \epsilon/2^{i-1} = 2\epsilon $$

But $\epsilon$ is arbitrary which means $\mu\left(Q_{[0,1]}\right) = 0$ (actually there are a few more steps before we can conclude this but I will omit them to keep the scope of this answer limited to probability theory).

But if $\mu(Q_{[0,1]} = 0$ then $\mu(Q^c_{[0,1]}) = 1$. As a last step realise that $\mu$ is the $P$ for \ref{def_exp} because it is a valid probability measure for a uniform distribution on $[0,1]$. Therefore, $E(f(x)) = 1$.

And that's one reason why we need a measure theoretic viewpoint to properly appreciate probability. More advanced applications in financial mathematics cannot work properly unless one takes the subtleties of measures into account.

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  • $\begingroup$ Thank you for this write-up, my naive interpretaion of $E(f(x))$ as follows: if a random experiment consists of drawing real numbers from $[0,1]$, "on an average, do you draw an irrational number or a rational number?" Would that be correct? I am self-learning probability theory and basic mathematical finance; from Schreve's book (Vol I currently). I'm an autodidact mostly. $\endgroup$
    – Quasar
    Commented Jan 5, 2021 at 23:07
  • $\begingroup$ @Quasar The term "on average" is not very precise. Do you mean "draw more irrational numbers than you draw rational ones"? $\endgroup$
    – Apoorv
    Commented Jan 5, 2021 at 23:15
  • $\begingroup$ I believe, I should interprete simply as the likelihood of drawing an irrational number, as $E[I_A] = P(A)$ for an event $A$. $\endgroup$
    – Quasar
    Commented Jan 5, 2021 at 23:24
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    $\begingroup$ @Quasar A better way would be to say that if you get £1 every time you pull an irrational number and £0 if you pull a rational number then how much money do you expect to have after $n$ rounds? And the answer is $n$ pounds. $\endgroup$
    – Apoorv
    Commented Jan 5, 2021 at 23:34
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    $\begingroup$ thumbs up for the answer showing the necessity of measure theory. There seems to be a minor typo in the last equation by the way (just to make it perfect), $\epsilon/2^{1-i}$ looks like should be $\episilon/2^{i-1}$? $\endgroup$
    – chichi
    Commented Feb 21, 2023 at 4:15
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There are several different ways to define Borel sets, you named one of them. What the definitions don’t tell you is what Borel sets mean, and why we care about them.

When measure theory was developed, it was intended as a set theoretic formalization of measuring “things”. Before that, geometry gave those answers (such as splitting some 2 dimensional form into many triangles and then summing up their area).

But with the advent of analytic geometry, geometry got reinterpreted as the study of subsets of $\mathbb{R}^n$. So what people wanted is a nice measure theory that doesn’t contradict prior intuition about how measuring areas work (e.g. moving something around doesn’t change it’s size) and it should be able to measure any subset of $\mathbb{R}^n$.

But shortly after, this was proven impossible, so people settled for the next best thing, which is not measuring all subsets of $\mathbb{R}^n$.

So you can view the Borel sets (or their completion, the Lebesgue sets), as subsets of $\mathbb{R}^n$ that you can measure, without your math breaking (i.e. without getting contradictions). See also: Vitali sets

That also means there are events that you cannot assign probability to! (Albeit these are very pathological sets)

I recommend looking into Terence Tao’s Introduction to Measure Theory for a more detailed introduction.

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