I've been studying a bit of probability theory lately and noticed that there seems to be a universal agreement that random variables should be defined as Borel measurable functions on the probability space rather than Lebesgue measurable functions. This is so in every textbook on probability theory which I consulted. In general, it seems to me that probability theory favors the Borel algebra more than the algebra of Lebesgue measurable sets. My question is: why?

In every course in measure theory, one is taught of the notion of a complete measure and completion of measures and I got the impression that a complete measure space is somewhat superior to a non-complete one (or at least that completeness makes life a bit easier on the technical level), so this preference of Borel sets puzzles me.

  • 7
    $\begingroup$ +1 for the question. I've always wondered why the opposite never happened! I've always wanted the composition of two measurable functions between two measure spaces to be measurable (and that was before I knew what a category was...). This does NOT hold with the definition I'd been given when studying measure theory as an undergrad. And I'm still unsure what you gain from completeness of the measure (but I seriously do not understand anything in analysis...). $\endgroup$
    – babubba
    Jul 12, 2010 at 19:45
  • 8
    $\begingroup$ Completion of a measure is always applicable, and this is the way I think of Lebesgue measure on R^n. $\endgroup$
    – Mark
    Jul 12, 2010 at 21:41
  • 1
    $\begingroup$ I would like to add a comment. 'The notion of probability does not enter into the definition of a random variable.' (Ref.: Page 43, V. K. Rohatgi and A. K. Md. E. Saleh, An Introduction to Probability and Statistics, Wiley Series in Probability and Statistics, John Wiley & Sons (Asia) Pte. Ltd. Singapore, 2001.) Indeed, measure theoretically, anything probabilistic is necessarily random, but not everything random are probabilistic. $\endgroup$ Feb 19, 2017 at 16:15
  • $\begingroup$ Related? Why do we consider Borel sets instead of measurable sets? $\endgroup$
    – BCLC
    Jul 24, 2021 at 7:43

7 Answers 7


One should be careful with the definitions here. Notation: Given measurable spaces $(X, \mathcal{B}_X), (Y, \mathcal{B}_Y)$, a measurable map $f : X \to Y$ is one such that $f^{-1}(A) \in \mathcal{B}_X$ for $A \in \mathcal{B}_Y$. To be explicit, I'll say $f$ is $(\mathcal{B}_X, \mathcal{B}_Y)$-measurable.

Let $\mathcal{B}$ be the Borel $\sigma$-algebra on $\mathbb{R}$, so the Lebesgue $\sigma$-algebra $\mathcal{L}$ is its completion with respect to Lebesgue measure $m$. Then for functions $f : \mathbb{R} \to \mathbb{R}$, "Borel measurable" means $(\mathcal{B}, \mathcal{B})$-measurable. "Lebesgue measurable" means $(\mathcal{L},\mathcal{B})$ measurable; note the asymmetry! Already this notion has some defects; for instance, if $f,g$ are Lebesgue measurable, $f \circ g$ need not be, even if $g$ is continuous. (See Exercise 2.9 in Folland's Real Analysis.)

$(\mathcal{L}, \mathcal{L})$-measurable functions are not so useful; for instance, a continuous function need not be $(\mathcal{L}, \mathcal{L})$-measurable. (The $g$ from the aforementioned exercise is an example.) $(\mathcal{B}, \mathcal{L})$ is even worse.

Given a probability space $(\Omega, \mathcal{F},P)$, our random variables are $(\mathcal{F}, \mathcal{B})$-measurable functions $X : \Omega \to \mathbb{R}$. The Lebesgue $\sigma$-algebra $\mathcal{L}$ does not appear. As mentioned, it would not be useful to consider $(\mathcal{F}, \mathcal{L})$-measurable functions; there simply may not be enough good ones, and they may not be preserved by composition with continuous functions. Anyway, the right analogue of "Lebesgue measurable" would be to use the completion of $\mathcal{F}$ with respect to $P$, and this is commonly done. Indeed, many theorems assume a priori that $\mathcal{F}$ is complete.

Note that, for similar reasons as above, we should expect $f(X)$ to be another random variable when $f$ is Borel measurable, but not when $f$ is Lebesgue measurable. Using $(\mathcal{F}, \mathcal{L})$ in our definition of "random variable" would not avoid this, either.

The moral is this: To get as many $(\mathcal{B}_X, \mathcal{B}_Y)$-measurable functions $f : X \to Y$ as possible, one wants $\mathcal{B}_X$ to be as large as possible, so it makes sense to use a complete $\sigma$-algebra there. (You already know some of the nice properties of this, e.g. an a.e. limit of measurable functions is measurable.) But one wants $\mathcal{B}_Y$ to be as small as possible. When $Y$ is a topological space, we usually want to be able to compose $f$ with continuous functions $g : Y \to Y$, so $\mathcal{B}_Y$ had better contain the open sets (and hence the Borel $\sigma$-algebra), but we should stop there.

  • 1
    $\begingroup$ Fair enough. My question arose out of some textbook examples of probability spaces and random variables (e.g the interval $[0,1]$ with the Borel algebra and Lebesgue measure) in which the underlying space had some familiar topology and the $\sigma$-algebra was chosen to be the Borel algebra rather than its completion. Maybe this is more of a phenomenon of examples rather than of theorems in probability theory. Can you mention some basic theorem in probability requiring the space in question to be equipped with a complete probability measure? The CLT, for instance, does not, as far as I know. $\endgroup$
    – Mark
    Jul 12, 2010 at 20:25
  • 3
    $\begingroup$ I will mention, though, that despite the bad behavior of Lebesgue measurability under function composition, the class of Lebesgue measurable functions does have the following very nice property - it is closed under a.e. pointwise limits (The class of Borel measurable functions is not - an example is easy to find). This defect will typically occur whenever you don't have completeness in the domain. Considering that the notion of almost sure convergence of random variables is commonly used in probability theory, I would expect completeness to be a rather important property. $\endgroup$
    – Mark
    Jul 12, 2010 at 20:32
  • 5
    $\begingroup$ Mark: You ask "Can you mention some basic theorem in probability requiring the space in question to be equipped with a complete probability measure?" Pretty much every theorem about continuous-time stochastic processes has such an assumption, often via the phrase "the usual conditions". I don't offhand know a counterexample where something goes wrong in the absence of completeness, but I bet they exist. I think as much as anything it's a convenience, so that you can say "almost surely" almost everywhere, to paraphrase Friedrichs. $\endgroup$ Jul 12, 2010 at 20:51
  • 4
    $\begingroup$ I'll also point out mathoverflow.net/questions/19459/… , where it is mentioned that even a $C^\infty$ function may fail to be $(\mathcal{L}, \mathcal{L})$-measurable. $\endgroup$ Jul 12, 2010 at 21:01
  • 6
    $\begingroup$ Nate - completeness is required for the debut theorem. The first time at which a (jointly measurable) process hits a measurable set is measurable. Completeness is still required for the first time at which a right-continuous process hits a given value, although it's not required for continuous processes. $\endgroup$ Jul 13, 2010 at 18:50

One reason is that probabilists often consider more than one measure on the same space, and then a negligible set for one measure (added in a completion) might be not negligible for the other. The situation becomes more acute when you consider uncountably many different measures (such as the distributions of a Markov process with different starting points.)

Another reason is that probabilists often need to consider projections of events: Instead of asking if Brownian motion (say) has some property at time $t$, we would like to know if there exists a time where Brownian motion has that property. Projections of Borel sets in a Polish space are Analytic (also known as Suslin) sets, and these sets are universally measurable (i.e., measurable in the completion of any Borel measure); a good source for this is [1]. In contrast, projections of Lebesgue measurable sets might fail to be Lebesgue measurable which then hinders further analysis.

[1] Arveson, William. An invitation to C*-algebras. Vol. 39. Springer Science & Business Media, 2012.

  • 26
    $\begingroup$ This is also a reason why the Borel sigma algebra on the domain is often preferred in ergodic theory. (A closely related reason is because of the connection between ergodic theory and topological dynamics; a topological dynamical system has a canonical Borel sigma algebra but not a canonical Lebesgue sigma algebra.) On the other hand, a significant portion of ergodic theory is also concerned with almost everywhere convergence (wrt some reference invariant measure, of course), and then it becomes useful for the domain sigma algebra to be complete... $\endgroup$
    – Terry Tao
    Jul 17, 2010 at 19:43

Some reasons can be found here. Borel measurable functions are much nicer to deal with. Every continuous function is Borel measurable, but the inverse of a Lebesgue measurable set may not be Lebesgue measurable. Moreover, Borel measurable functions are very well behaved when it comes to conditioning. If $f:(X,\Sigma)\to\mathbb{R}$ is Borel measurable, then a function $g:X\to\mathbb{R}$ is measurable with respect to $(X,\sigma(f))$ if and only if there exists a Borel measurable function $h:\mathbb{R}\to\mathbb{R}$ such that $g=h\circ f$.

On a more conceptual note, the less measurable sets you have in your codomain, the easier it is for a function to be measurable. And if a random variable should represent a random quantity, then all empirically interesting questions can be formulated in terms of simple intervals and their combinations. For, say, statistical applications there is no empirical difference between Borel sets and a Borel set modified by a null set. The distributions (on the reals) commonly applied can usually be given by a cumulative distribution function and such a function essentially determines the probability of intervals.


To fix notations, let's call $(\Omega,\mathcal{F},\mathbb{P})$ our probability space and $X: \Omega\to \mathbb{R}$ our random variable.

The Lebesgue measure on the range of $X$ has no role to play at all: what we are interested in is the law of $X$, which can very well be discrete. Asking that $X$ is Borel measurable is what is needed to ensure that $X\in A$ is a well defined event as soon as $A$ is open, which is the least one would want. But to know the probability of $X\in A$ for some Lebesgue-measurable set has less meaning (as when $A$ is measurable with respect to the completion of the Borel algebra with respect to another measure).

Note that when $\mathcal{F}$ is complete with respect to $\mathbb{P}$, then automatically $X$ is measurable with respect to the completion induced by the law of $X$ of the Borel algebra of $\mathbb{R}$.

Note also that a random variable can take its values in a measured space (a topological space say, since we are talking of Borel algebras) not having a singled out measure, in which case the question is empty.


A conceptual answer can be given in the framework of this answer.

The functor that sends a measurable space X to the set of random variables on X, i.e., equivalence classes of (unbounded) real or complex valued functions on X, sends colimits into limits and satisfies the solution set condition. Hence it is representable by the representable functor theorem.

The representing object is a (complete) measurable space Z such that morphisms of measurable spaces from X to Z are exactly random variables on X.

The measurable space Z has very interesting structure. For example, it contains a copy of real (or complex) numbers with the usual Lebesgue measurable structure. It also contains an atom corresponding to each real number, a copy of Cantor set with non-Lebesgue sets of measure 0 etc.

The underlying reason of these effects is that preimages of sets of measure 0 under morphisms of measurable spaces are again sets of measure 0. Thus a morphism from an atomless measurable space X to real numbers equipped with the usual Lebesgue structure cannot be constant, because a single-point set has measure 0 in the reals. Thus we have to add an atom for each real number and a lot of other stuff to get all random variables.

In fact, one can/should view the process described above as the canonical functor F from the appropriate category of topological spaces to the category of measurable spaces.

This functor should be contrasted with another canonical functor G from the category of smooth manifolds and submersions to the category of measurable spaces.

Every manifold is a topological space, however, G does not factor through F. One really needs the additional structure of smooth manifold to define G.

However, there is always a canonical map (actually, a monomorphism) G(M)→F(M) for every smooth manifold M.

Real (or complex) numbers form both a topological space and a smooth manifold, thus we get two measurable spaces F(R) and G(R) out of them by applying F and G respectively.

It is the former measurable space that should be used in the definition of random variables, not the latter one. In other words, we should think of R (or C) in the definition of random variable as a topological space, not as a smooth manifold.

Thus there is no reason to deal with non-complete spaces. Complete spaces are indeed technically superior to non-complete spaces.

Unfortunately, probabilists seem to be unfamiliar with this relatively easy construction and instead they have to phrase their definitions of random variable in a way that relies on a subtle difference between Borel and Lebesgue measurable sets.

  • 5
    $\begingroup$ Being slow and pedestrian, I'm having a bit of difficulty following how this interesting POV answers the original question: "In general, it seems to me that probability theory favors the Borel algebra more than the algebra of Lebesgue measurable sets. My question is: why?" $\endgroup$
    – Yemon Choi
    Jul 13, 2010 at 17:55
  • $\begingroup$ Well, in the answer I demonstrated how one can get rid of Borel algebra completely. Doesn't this answer the question? $\endgroup$ Jul 14, 2010 at 10:41
  • 1
    $\begingroup$ Dmitri: I think you are answering the question: "can we get rid of the Borel algebra and work throughout with complete algebras". The original question is, in my reading, more like "why do probabilists use Borel and not Lebesgue, seeing as some of them are really quite smart"? It's a question which invites those who use the Borel setup to justify why they have done so. $\endgroup$
    – Yemon Choi
    Jul 14, 2010 at 21:37
  • 7
    $\begingroup$ Analysts != probabilists. Besides, just how much of the categorical viewpoint have Doob, Malliavin, Meyer, Talagrand, Lyons et al. needed? And the first sentence of your penultimate paragraph does not quite ring true to me. In my inexpert opinion, the categorists' dictum that we should always embed our category with nice objects but bad structure into a category with potentially unfathomable objects but with good structure, is not in my view always the most appropriate one. $\endgroup$
    – Yemon Choi
    Jul 15, 2010 at 19:46
  • 1
    $\begingroup$ @Yemon: I edited the last paragraph to indicate my answer to the original question more clearly. $\endgroup$ Aug 13, 2010 at 6:33

On pages 209-210 of his book "Real Analysis, A Comprehensive Course in Analysis, Part I" Barry Simon gives an argument for sticking with Borel measurable sets and functions.


Barry Simon argues that Lebesgue measurable functions are not closed under composition, that it complicates arguments such as constructing product measures, requiring an extra completion set, and that nothing is gained since every Lebesgue measurable function is equal a.e. to a Borel function, and equivalence classes that matter.


Different measures on the set of all Borel-measurable sets have different completions. For some probability distributions on the set of Borel subsets of $\mathbb R$, the completion is something different from the set of Lebesgue-measurable sets. All subsets of the Cantor set are Lebesgue-measurable since the Lebesgue measure of the Cantor set is $0$. But consider the Cantor distribution, which decides whether the $n$th ternary digit is $0$ or $2$ based on the $n$th independent coin toss. It assigns positive probability to many subsets of the Cantor set, so not all of its subsets are in the completion.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.