There is no infinite dimensional Lebesgue measure

Imagine that there was a countably additive and translation invariant measure, {m}, on a Hilbert Space {\mathcal{H}} such that every ball {B(v, \varepsilon)} has positive measure, finite measure. Because {\mathcal{H}} is infinite dimensional, there exists {\{e_{i} | i = 1, \cdots, \infty\}} such that {\langle e_{i}, e_{j}\rangle = \delta_{ij}}. Note that the points {e_{i}} and {e_{j}} are far apart: {\|e_{i} - e_{j}\|^2 = \|e_{i}\|^2 + \|e_{j}\|^2 = 2}, if {i \neq j}. Now, the ball, {B(0, 2)}, contains the countable collection of disjoint balls: {\{B(e_{i}, \frac{1}{2}) | i = 1,\cdots, \infty\}} (you can see where this is going). Thus, because all balls, {B(e_{i}, \frac{1}{2})} have the same measure (by translation invariance), and {m} is countable additive, it follows that {\infty = \sum_{i=1}^\infty m(B(e_{i}, \frac{1}{2})) < B(0, 2) < \infty}. Thus, we’ve reached a contradiction.

In some cases, it’s desirable to have an analogue of a measure on a Hilbert space {\mathcal{H}}. The most common application is to make statements such as “Property X is true for a.e. function.” One way to do make these statements is through the concept of Prevalence. Another is the definition of a Gaussian like measure called the Abstract Wiener Space.

Note that I wrote this article before realizing that Wikipedia has a similar article.


Is the Product of Measurable Spaces the Categorical Product?

This post requires some knowledge of measure theory.

Today I’m going to show that the product of two measurable spaces {(X, \mathcal{B}_X)} and {(Y,  \mathcal{B}_Y)}, is actually the product in the category of measurable spaces. See Product (category theory).

The category of measurable spaces, {\mathbf{Measble}}, is the collection of objects {(X,\mathcal{B}_X)}, where {X} is a set and {\mathcal{B}_X} is a {\sigma}-algebra on {X}, and the collection of morphisms {\phi : (X, \mathcal{B}_X) \rightarrow  (Y, \mathcal{B}_Y)} such that

  1. {\phi : X \rightarrow Y};
  2. {\phi^{-1}(E) \in \mathcal{B}_X} for all {E \in \mathcal{B}_Y}.

Such a function {\phi} is called a measurable morphism, and {(X, \mathcal{B}_X)} is called a measurable space.

Given two measurable spaces {(X, \mathcal{B}_X)} and {(Y, \mathcal{B}_Y)} we can define their 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 {\sigma}-algebra generated by the sets of the form {E \times Y} and {X \times F} with {E \in \mathcal{B}_X} and {F \in  \mathcal{B}_Y}. We will need the following definition: A {\sigma}-algebra {\mathcal{B}} on a set {Z} is said to be coarser than a {\sigma}-algebra {\mathcal{B}'} on {Z} if {\mathcal{B} \subseteq \mathcal{B}'}. In exercise 18 of Terry Tao’s notes on product measures, it is shown that {\mathcal{B}_X \times  \mathcal{B}_Y} is the coarsest {\sigma}-algebra on {X\times Y} such that the projection maps {\pi_X} and {\pi_Y} are both measurable morphisms.

Now, we are finally ready to show that {(X\times Y,  \mathcal{B}_X\times \mathcal{B}_Y)} is the categorical product of the measurable spaces {(X,  \mathcal{B}_X)} and {(Y,  \mathcal{B}_Y)}. If {\phi_X : (Z, \mathcal{B}_Z)  \rightarrow (X, \mathcal{B}_X)} and {\phi_Y : (Z,  \mathcal{B}_Z) \rightarrow (Y, \mathcal{B}_Y)} are measurable morphisms, then we need to show that there exists a unique measurable morphism {\phi_{X\times Y} : (Z, \mathcal{B}_Z)  \rightarrow (X\times Y, \mathcal{B}_X \times  \mathcal{B}_Y)} such that {\phi_X = \pi_X \circ  \phi_{X\times Y}} and {\phi_Y = \pi_Y \circ  \phi_{X\times Y}}. Because the cartesian product is the product in the category of sets, we only have one choice for such a map. Indeed, {\phi_{X \times Y} = (\phi_X, \phi_Y)}.

We claim that {\phi_{X \times Y}} is measurable. Indeed, because the pullback {\phi_{X \times Y}^{-1} : 2^{X\times  Y} \rightarrow 2^Z} respects arbitrary unions and complements, and {\phi_{X\times Y}(\emptyset) =  \emptyset}, we only need to show that {\phi_{X\times Y}^{-1}(E \times Y) \in \mathcal{B}_Z} and {\phi_{X\times Y}^{-1}(X \times F) \in  \mathcal{B}_Z} for all {E \in  \mathcal{B}_X} and {F \in  \mathcal{B}_Y} (see remark 4 of Terry Tao’s notes on abstract measure spaces). This is easy to show:

\displaystyle  \begin{array}{rcl}  \phi^{-1}_{X\times Y} (E\times Y)  &=& (\pi_X \circ \phi_{X\times Y})^{-1}(E) \\ &=&  \phi_X^{-1}(E) \\ \phi^{-1}_{X\times Y} (X \times F) &=& (\pi_Y  \circ \phi_{X\times Y})^{-1}(F) \\ &=& \phi_Y^{-1}(F)  \end{array}

are both {\mathcal{B}_Z} measurable because {E \in \mathcal{B}_X} and {F \in  \mathcal{B}_Y}.

Thus, we have shown that {(X \times Y, \mathcal{B}_X\times  \mathcal{B}_Y)} is actually the product of the measurable spaces {(X, \mathcal{B}_X)} and {(Y,  \mathcal{B}_Y)} in the category of measurable spaces. This is reassuring, otherwise the term “product space” would be misleading.

I would like to know if there is a category of measure spaces with objects {(X, \mathcal{B}_X, \mu_X)}, where {X} is a set, {\mathcal{B}_X} is a {\sigma}-algebra on {X}, and {\mu_X : \mathcal{B}_X \rightarrow [0, +\infty]} is a measure. There would need to be some extra condition on morphisms in this category, otherwise we couldn’t distinguish between triples {(X, \mathcal{B}_X, \mu_X)} and {(X,  \mathcal{B}_X, \mu_X')} where {\mu_X \neq  \mu_X'}. If this was a category, I don’t believe products could exist. Indeed, if the measure spaces {(X, \mathcal{B}_X,  \mu_X)} and {(Y, \mathcal{B}_Y,  \mu_Y)} are not {\sigma}-finite, then there can be multiple measures on {(X \times Y,  \mathcal{B}_X \times \mathcal{B}_Y)}: see Terry Tao’s notes on product measures.

Another question is whether equalizers, coproducts, coequalizer, etc. exist in the category of measurable spaces. If a sufficient number of these properties exist, then we can take categorical (co)limits of measurable spaces. These might be of some interest already, but I have not looked into it.