# 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.