The usual category of measure spaces consists of objects $(X, \mathcal{B}_X, \mu_X)$, where $X$ is a space, $\mathcal{B}_X$ is a $\sigma$-algebra on $X$, and $\mu_X$ is a measure on $X$, and measure preserving morphisms $\phi \colon (X, \mathcal{B}_X, \mu_X) \to (Y, \mathcal{B}_Y, \mu_Y)$ such that $\phi_\ast \mu_X(E) = \mu_X(\phi^{-1}(E)) = \mu_Y(E)$ for all $E \in \mathcal{B}_Y$.

The category of measurable spaces consists of objects $(X, \mathcal{B}_X)$ and measurable morphisms $\phi \colon (X, \mathcal{B}_X) \to (Y, \mathcal{B}_Y)$.

Products exist in the category of measurable spaces. They coincide with the standard product $(X \times Y, \mathcal{B}_X \times \mathcal{B}_Y)$, where $X \times Y$ is the Cartesian product of $X$ and $Y$ and $\mathcal{B}_X \times \mathcal{B}_Y$ is the coarsest $\sigma$-algebra on $X\times Y$ such that the canonical projections $\pi_X \colon X \times Y \to X$ and $\pi_Y \colon X \times Y \to Y$ are measurable. Equivalently, $\mathcal{B}_X \times \mathcal{B}_Y$ is the $\sigma$-algebra generated by the sets $E \times F$ where $E \in \mathcal{B}_X$ and $F \in \mathcal{B}_Y$.

However, in the category of measure spaces, products do not exist. The first obstacle is that the canonical projection $\pi_X \colon X \times Y \to X$ may not be measure preserving. A simple example is the product of $(\mathbf{R}, \mathcal{B}[\mathbf{R}], \mu)$ with itself, where $\mathcal{B}[\mathbf{R}]$ is the Borel $\sigma$-algebra on $\mathbf{R}$. In this case,

$(\pi_{\mathbf{R}})_\ast\mu\times \mu([0,1]) = \mu\times \mu(\pi_\mathbf{R}^{-1}([0,1])) = \mu \times \mu([0,1]\times \mathbf{R}) = \infty \neq 1 = \mu([0,1])$.

In addition, there may be multiple measures on $X\times Y$ whose pushforwards on $X$ and $Y$ are $\mu_X $ and $\mu_Y$. Terry Tao mentions that from the perspective of probability, this reflects that the distribution of random variables $X$ and $Y$ is not enough to determine the distribution of $(X, Y)$ because $X$ and $Y$ are not necessarily independent.

Given that the products in the usual category fail to exist, is it possible to define a new categorical structure on the class of measure spaces such that products do exist?

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    $\begingroup$ The answer to your question as stated is a somewhat vacuous "yes". For instance, I can certainly build a category whose collection of objects is the class of measured spaces, but with precisely one morphism between any two objects. Then this category certainly has products --- it's equivalent to the category with one object and no nontrivial morphisms. But this, I assume, is not what you want. $\endgroup$ Dec 14, 2010 at 20:13
  • $\begingroup$ One thing to point out is that (and I haven't checked this) you may have pullbacks but not products. This would arise from the failure of the existence of a terminal object. $\endgroup$
    – David Roberts
    Dec 14, 2010 at 20:14
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    $\begingroup$ So I think the answer to any reasonable revision is "no". I doubt that there is any way to non-violently modify what are the morphisms between measured spaces that accomplishes what you want. $\endgroup$ Dec 14, 2010 at 20:15
  • $\begingroup$ A standard, somewhat agressive (but I wouldn't say "violent") way to add products to a category is simply to complete it. The free completion of a category $C$ is the opposite category to the category of all functors $C \to \text{SET}$. Then $C$ embeds fully and faithfully in its completion. If you want to be less free, you can pick some collection of limits that do exist in your category (e.g. pullbacks if David Roberts is correct) and repeat the trick just considering those functors $C \to \text{SET}$ that preserve the limits you want. $\endgroup$ Dec 14, 2010 at 20:18
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    $\begingroup$ Somewhat contrary to what Theo has said, I think (although I may have misread Damek's question) that the view point advocated (evangelised?) by Dmitri Pavlov in several answers to somewhat related questions ought to be relevant. See mathoverflow.net/questions/20740/… for instance, although I think there are others. $\endgroup$
    – Yemon Choi
    Dec 14, 2010 at 20:42

2 Answers 2


To clarify Chris Heunen's answer, let me point out that most notions of measure theory have analogs in the category of smooth manifolds. For example, the analog of a measure space (X,M,μ), where X is a set, M is a σ-algebra of measurable subsets of X, and μ is a measure on (X,M), is a smooth manifold X equipped with a density μ. Likewise, the analog of a measure-preserving morphism is a volume-preserving smooth map.

The category of smooth manifolds equipped with a density together with volume preserving maps as morphisms does not have good categorical properties. The problem comes from the fact that preservation of volume is too strong a condition to allow for good categorical properties.

If one drops the data of a density and the property of volume preservation, then the resulting category of smooth manifolds has relatively good categorical properties, such as existence of finite products and, more generally, existence of all finite transversal limits.

The same is true for the category of measure spaces. However, in this case we cannot simply drop the data of a measure from the definition and expect to get a category with good properties. The reason for this is that the data of a measure in fact combines two independent pieces of data. The first piece of data tells us which sets have measure zero and which ones don't. The second piece of data tells us the actual values of measure on sets of non-zero measure.

The analog of dropping the data of a density for measure theory is dropping the second piece of data described above, but not the first one. This can already be seen for smooth manifolds: If we have a smooth manifold, we don't need a density on it to say which sets have measure 0.

Thus one is naturally led to the notion of a measurable space that knows which measurable sets have measure 0. I describe it in more detail in [1], therefore here I offer only a brief summary of the main definitions. A measurable space is a triple (X,M,N), where X is a set, M is a σ-algebra of measurable subsets of X, and N⊂M is a σ-ideal of measure 0 sets. A morphism of measurable spaces f: (X,M,N)→(Y,P,Q) is an equivalence class of maps of sets g: X→Y such that the preimage of every element of P is an element of M and the preimage of every element of Q is an element of N. Two maps g and h are equivalent if they differ on a set of measure 0. Here for the sake of simplicity I assume that both measurable spaces are complete (any subset of a measure 0 subset is again a measure 0 subset). Every measurable space is equivalent to its completion [2], hence we do not lose anything by restricting ourselves to complete measurable spaces. In general, one has to modify the above definition to account for incompleteness, as explained in the link above.

Finally, one has to require that measurable spaces are localizable. One way to express this property is to say that the Boolean algebra M/N of equivalence classes of measurable sets is complete (i.e., it has arbitrary suprema and infima). Many basic theorems of measure theory fail without this property. In fact, as explained in the link above, theorems such as Radon-Nikodym theorem and Riesz representation theorem are equivalent to the property of localizability. Remarkably, the category of localizable measurable spaces is equivalent to the opposite category of the category of commutative von Neumann algebras [7]. This statement can be seen as another justification for the property of localizability.

Henceforth I assume that all measurable spaces are localizable.

Being in possession of a good category of measurable spaces we can prove that it admits finite products, and, more generally, arbitrary finite limits.

Let me point out that at this point measure theory diverges from smooth manifolds: The functor that sends a smooth manifold to its underlying measurable space is colax monoidal but not strong monoidal with respect to the monoidal structures given by the categorical products.

More precisely, the category of measurable spaces admits two natural product-like monoidal structures. One is given by the categorical product mentioned above, and the other one is the spatial product, which is often simply called the product in many textbooks on measure theory. By the universal property of product there is a canonical map from the spatial product to the categorical product, which is a monomorphism but not an isomorphism unless one of the spaces is atomic, i.e., a disjoint union of points.

The forgetful functor F from the category of smooth manifolds to the category of measurable spaces is strong monoidal with respect to the categorical product on the category of smooth manifolds and the spatial product on the category of measurable spaces. Therefore it is also colax monoidal with respect to the categorical product on measurable spaces, but not strong monoidal because F is essentially surjective on objects and the map from the spatial product to the categorical product is not always an isomorphism.

Here is an instructive example for the last statement. Let Z=F(R), where R is the real line considered as a smooth manifold. Consider the spatial product of Z and Z, which is canonically isomorphic to F(R×R). The spatial product maps monomorphically to the categorical product Y=Z×Z. However, Y is much bigger than F(R×R). For example, the diagonal map Z→Y=Z×Z is disjoint from F(R×R) in Y. The space Y also has a lot of other subspaces whose existence is guaranteed by the universal property. Note that set-theoretically F(R×R) also has a diagonal subset. However, this subset has measure 0 and therefore is invisible in this formalism.

Let me finish by mentioning that locales arguably provide much better formalism for measure theory, which in particular does not suffer from problems with sets of measure 0, e.g., we don't need to pass to equivalence classes or even mention the words “almost everywhere”. The relevant functor sends a measurable space (X,M,N) to the locale M/N (as explained above M/N is a complete Boolean algebra, hence a locale). We obtain a faithful functor from the category of measurable spaces to the category of locales. Let's call its image the category of measurable locales. Then measurable morphisms of measurable locales correspond bijectively to equivalence classes of measurable maps. Note that we don't need to pass to equivalence classes to define a measurable morphism of measurable locales. Every measurable space (or locale) can be uniquely decomposed into its atomic and diffuse part. The atomic part is a disjoint union of points and the diffuse part does not have any isolated points. An example of a diffuse space is given by F(M) where M is a smooth manifold of non-zero dimension. (In fact all these spaces are isomorphic as measurable spaces if the number of connected components of M is countable.) Here is the punchline: If Z is a diffuse measurable locale, then it does not have any points, in particular it is non-spatial. This can serve as an explanation of why we cannot construct a reasonable concrete category of measurable spaces and why we always have to use equivalence classes if we want to stay in the point-set measure theory.

References that advocate (but do not evangelize) the viewpoint described in this answer:

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    $\begingroup$ Very thorough. I was hoping you would turn up on this question to provide an answer! $\endgroup$
    – Yemon Choi
    Dec 15, 2010 at 19:09
  • $\begingroup$ While this all sounds interesting, I'm having trouble digesting this. Just to check if I'm on the right track: Up to isomorphism there seem to be only very few objects in this category (under suitable smallness conditions like 'separability' and $\sigma$-finiteness), in particular there is only one non-atomic space, right? $\endgroup$ Dec 16, 2010 at 6:05
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    $\begingroup$ @Theo: Yes. Every measurable space can be canonically decomposed as a coproduct (disjoint union) of its atomic and diffuse part. Atomic spaces are coproducts of points (cardinality A), diffuse spaces are coproducts of real lines (cardinality D). Thus the pair of cardinal numbers (A,D), where D is either zero or infinite (finite coproducts of real lines are isomorphic to countable coproducts of real lines) completely characterizes isomorphism classes of measurable spaces. $\endgroup$ Dec 16, 2010 at 19:07
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    $\begingroup$ (In the above, measurable spaces are assumed to be of countable Maharam type, otherwise the classification is slightly more complicated, as given by Maharam's theorem.) $\endgroup$ May 25, 2013 at 1:31

If preserving measure is too strong a choice of morphisms to guarantee products, one could weaken that a bit. Of course, taking measurable functions as the morphisms would give a category equivalent to that of measurable spaces, but that completely ignores the measures. How about taking as morphisms measurable functions $\phi \colon X \to Y$ such that $\phi_* \mu_X \ll \mu_Y$? The latter means that $\phi_*\mu_X$ is absolutely continuous with respect to $\mu_Y$, i.e. that $\mu_X(\phi^{-1}(E))=0$ whenever $\mu_Y(E)=0$. Then certainly the projection $\pi \colon X \times Y \to X$ is a well-defined morphism, and if I did my doodles right, so is the tuple inherited from the product structure on the category of measurable spaces.

UPDATE: The diagonal map $X \to X \times X$ is not necessarily a morphism as defined above, and hence the category of measure spaces with these morphisms does not have products. Do we really need an even weaker choice of morphisms?

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    $\begingroup$ You've disqualified the diagonal map $\mathbb R\to \mathbb R\times \mathbb R$! $\endgroup$ Dec 14, 2010 at 23:01
  • $\begingroup$ Why? Let $A$ and $B$ be measurable subsets of $X=\mathbb{R}$, put $E=A \times B$, and suppose $\mu_{X \times X}(E)=0$. Denote the diagonal map by $\Delta \colon X \to X \times X$. Then $0=\mu_{X \times X}(A \times B)=\mu_X(A)\mu_X(B)$, so $\min(\mu_X(A),\mu_X(B))=0$. Hence $\mu_X(\Delta^{-1}(E))=\mu_X\{x \mid (x,x) \in A \times B\}=\mu_X(A \cap B) \leq \min(\mu_X(A),\mu_X(B))=0$. By construction of the product $\sigma$-algebra and measure, this extends to all measurable E, and so $\Delta_* \mu_X \ll \mu_{X \times X}$. Am I missing something stupid? $\endgroup$ Dec 15, 2010 at 0:29
  • $\begingroup$ What about the line $\{y=x\}$ in $\mathbf{R}^2$ for $x \in [0, 1]$. This has Borel measure zero, but $\mu_R(\Delta^{-1}(\{y=1\})) = \mu_R([0,1]) = 1$ $\endgroup$ Dec 15, 2010 at 1:00
  • $\begingroup$ Whoops! I meant to say that $\Delta^{-1}(\{y=x\}) = [0,1]$. This has measure $1$. $\endgroup$ Dec 15, 2010 at 1:01
  • $\begingroup$ Ah, I see. My reasoning above only holds for $E$ of the form $A \times B$; the flaw is that it doesn't extend to all measurable $E$. I'll edit the answer. Or should I delete it? $\endgroup$ Dec 15, 2010 at 6:37

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