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.

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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.