Please answer this question with basic language if possible.

Please let the following notation apply $X : \Omega \longrightarrow \mathbb{R}$ is a random variable, where $(\Omega, \mathcal{F}, P)$ is a measure space.

Question 1: If $\Omega = \mathbb{R}$ then is the $\sigma$-algebra $\mathcal{F}$ necessarily $\mathcal{B}(\mathbb{R})$? If not, what determines the $\sigma$-algebra $\mathcal{F}$?

Question 2: Given a probability distribution with $X$ (above) as the underlying random variable then what is the domain of the probability distribution and importantly, why? Is it the Borel $\sigma$-algebra $\mathcal{B}(\mathbb{R})$?

  • 1
    $\begingroup$ Question1: No, it can be any $\sigma$-field over $\mathbb{R}$. A common choice is $\mathcal{B}(\mathbb{R})$ if $\Omega = \mathbb{R}$, though. Question2: I dont know what you mean by "domain of distribution". I assume you mean what you can measure with the pushforward measure. First of all you need a $\sigma$-field on $\mathbb{R}$,say $\mathcal{A}$ (e.g. the borel field). Then, per def. $P^X (A) = P(X^{-1}(A))$ for $A \in \mathcal{A}$ $\endgroup$ Sep 22, 2022 at 12:39

1 Answer 1


To answer your first question, it need not be the case in general, and you choose one. To answer your second question, yes, the $\sigma$-algebra associated to a probability distribution will be the Borel sets of $\mathbb{R}$. That being said, one could consider a similar measure over smaller $\sigma$-algebras or (in some cases) bigger $\sigma$-algebras that gives the same measure for common sets if one wishes. Let me elaborate, then.

As you consider different examples using probability theory you will convince yourself that Borel sets are a sufficiently big $\sigma$-algebra in the sense that most everyday properties can be expressed in terms of them. Try, for example, to define random variables and to express the probabilities of these events in the form $\{X\in B\}$ where $B$ is a Borel set and $X$ tracks a quantity in the result of an experiment: "The dice toss is greater than 4" or "The train arrives between 10:00 and 10:30". It can be useful to remember that some examples of Borel sets are: open intervals, closed intervals, and intervals in general, countable unions of intervals, and more general mathematically useful sets such as compact sets, G-$\delta$ sets or F-$\sigma$ sets.

Now, let's see how this relates to distributions. A random variable $X$ is a measurable function $X:(\Omega, \mathcal{F})\to (\mathbb{R}, \beta(\mathbb{R}))$. This means that $\{X\in B\}=X^{-1}(B)$ is $(\Omega, \mathcal{F})$-measurable for every Borel set $B$. This means that we will be able to calculate probabilities associated to the function taking values over sets in the $\sigma$-algebra in the codomain. Why were the Borel sets chosen for this? Well, I hope to have motivated (a little) that they are a big enough $\sigma$-algebra to guarantee that you can express most (all) everyday events as the random variable taking a value inside a Borel set. Now, the probability distribution associated to $X$ maps sets $B$ to the probability that $X$ lies in $B$. Because of the previous definition, we can guarantee that the sets $B$ for which $\{X\in B\}$ is an event are the Borel sets, and this is why it is natural to define $\mathbb{P}_X: \beta(\mathbb{R})\to [0,1]$.

After expressing it this way, it is clear that if one wished one could choose to restrict the probability distribution function to a smaller $\sigma$-algebra of sets (but I have never seen that done). Conversely, if the random variable behaves well enough then it could be possible to guarantee thar $\{X \in B\}$ are events for $B$ members of a bigger $\sigma$-algebra and thus extend the function to it, but this is not guaranteed for the general case.

Just as a curiosity: why do we not choose a bigger $\sigma$-algebra of sets when we define the notion of a random variable? Choosing a bigger $\sigma$-algebra of sets will restrict the functions that are random variables, as more and more sets will have to belong to $\mathcal{F}$ which will become increasingly difficult to satisfy. It turns out that other bigger $\sigma$-algebras like for example that of Lebesgue measurable sets $\mathcal{M}$ will restrict too much: continuous functions $f: (\mathbb{R}, \mathcal{M}) \to (\mathbb{R}, \mathcal{M})$ need not be measurable random variables in that case. This is non-trivial and is usually proved in a measure theory course.

To finish now, and to elaborate on the first question: is it useful and can you to consider bigger $\sigma$-algebras than the Borel sets for a probability (or measure) space over the reals? Yes! Here are some examples:

  • When you have only finite results for an experiment, for example only dice tosses, you can define an equiprobable space over those finite results. If one wanted to do it, this would induce a probability distribution over the reals, but one can choose the power set $\mathcal{P}(\mathbb{R})$ as its $\sigma$-algebra instead. The probability measure can be defined as: $\mathcal{P}(A) = \frac{\#(A \cap B)}{\#B}$ where $B$ is the set containing all finite results for the experiment.
  • A very useful space in the related area of measure theory is $(\mathbb{R}, \mathcal{M}, \mathcal{L})$ where $\mathcal{M}$ is the $\sigma$-algebra of Lebesgue-measurable sets of $\mathbb{R}$ and $\mathcal{L}$ is the Lebesgue measure. This will not be a probability measure, because the measure of the entire real line is infinite, but if one restricts the space to $[0,1]$ this gives us a probability space.

It turns out both $\sigma$-algebras are bigger than the Borel $\sigma$-algebra.

In general it would be easier not to have to worry about $\sigma$-algebras, but it turns out it is not possible to define measures satisfying all desirable properties over the power set $\mathcal{P}(\Omega)$. One interesting example of this can be found in Georgi's book for infinite sequences of coin tosses, where it is impossible to define a probability measure satisfying intuitive properties.


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