Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Given a set Ω and a σ-algebra F of subsets, is there some natural way to assign something like a "uniform" measure on the space of all measurable functions on this space? (I suppose first it would be useful to know if there's even a natural σ-algebra to use on this space.)

The reason I'm asking is because I'd like to do the following. Let Ω be the (2-dimensional) surface of a sphere, with the uniform probability distribution. Let F be the Borel σ-algebra, and let G be the sub-algebra consisting of all measurable sets composed of lines of longitude. (That is, S is in G iff S is measurable and for all x in S, S contains all points with the same longitude as x.) Let A be the set of all points with latitude 60 degrees north or higher (a disc around the north pole).

Let f be a G-measurable function defined on Ω such that the integral of f over any G-measurable set B equals the measure of (A\cap B). (This is a standard tool in defining the conditional probability of A given G-measurable sets.) It's not hard to show that for any such function f, for almost-all x, f(x) will equal the unconditional measure of A.

What I'd like to be able to say is that for any x, for almost-all such functions f, f(x) will equal the unconditional measure of A. However, I can't say "almost-all" on the functions unless I have some measure on the space of functions.

Clearly I can do this by concentrating all the measure on the single constant function in this set. But I'd like to be able to pick out this most "generic" such function even in cases where A isn't so nice and symmetric.

Maybe there's some other, simpler question I should be asking first?

share|improve this question

4 Answers 4

up vote 20 down vote accepted

Let I be the unit interval with the Borel $\sigma$-algebra. There is no $\sigma$-algebra on the set of measurable functions from I to I such that the evaluation functional $e:I^I\times I\to I$ given by $e(f,x)=f(x)$ is measurable, as shown by Robert Aumann here, so even finding useful $\sigma$-algebras is a problem.

However, t is possible to talk about "almost all" functions in a function space even when it is not possible to have an appropriate measure. The trick is to find a characterization of a set having full (or zero) measure that can be applied to function spaces. There is a generalization of Lebesgue measure zero, independently found by various authors and knowyn as Haar measure zero or shyness that should be applicable to your problem. A nice survey of the theory and some of its extensions can be found here.

share|improve this answer
I wonder if the result cited can be beefed up to show that the category of measurable spaces is not cartesian closed. This was a question whose answer I have been looking for a while; last time I checked, nobody seemed to know. –  G. Rodrigues Jun 14 '10 at 16:06
The book of Benyamini and Lindenstrauss has a nice discussion of several kinds of null subsets of infinite dimensional Banach spaces and the relations between them. Even more sophisticated null sets are defined in recent papers of Lindenstrauss and Preiss. The notions are important for proving e.g. that Lipschitz functions from a space have a point of differentiability; this is also discussed in the book mentioned above. –  Bill Johnson Jun 16 '10 at 7:05
@Michael Greinecker, the survey no longer seems to be posted at William Ott's website. Would you please post a copy on your website? I doubt that Professor Ott would mind. –  Tom LaGatta Feb 4 '13 at 6:11
@Tom LaGatta The paper is open acess of the Bulletin of the AMS, so I included the official link. Thank you for telling me! –  Michael Greinecker Feb 4 '13 at 9:06
@G Rodrigues: It looks like there is now even a formal proof that the category of measurable spaces is not Cartesian closed! I can't really follow it, but it looks like they use similar arguments. See isabelle.in.tum.de/library/HOL/HOL-Probability/… –  Chris Heunen 2 days ago

The closest thing I know is the Effros borel structure. Let X be a Polish space. Let F(X) be the collection of closed subsets of X. The Effros borel structure on F(X) is the coarsest borel structure for which all sets of the form

{ E \in F(X) : E \subseteq A }

are measurable for all A \in F(X) . See: J.P.R. Christensen, TOPOLOGY AND BOREL STRUCTURE, North-Holland, 1974

share|improve this answer

Sorry for the necromancy. Here's an attempt at constructing a $\sigma$-algebra using the tensor product of $\sigma$-algebras. This should likely not result in a Borel structure (i.e., a $\sigma$-algebra generated as the Borel $\sigma$-algebra of a topological space), so I don't think it contradicts Aumann's work.

I made this answer community wiki, so feel free to edit it. If it's wrong, please correct it.

I figured I'd answer the question to provide a quick reference for the future.

Let $(X,\Sigma_X)$ and $(Y, \Sigma_Y)$ be two measurable spaces, and let $H = \operatorname{Hom}(X,Y)$ be the set of measurable functions from $X$ to $Y$. Define the evaluation map $\operatorname{eval} : H \times X \to Y$ by $$\operatorname{eval}(h,x) = h(x).$$

Now, simply define $\Sigma_{H}$ to be the minimal $\sigma$-algebra on $H$ so that the evaluation map $\operatorname{eval} : H \times X \to Y$ is measurable, where $H \times X$ is equipped with the tensor product $\sigma$-algebra $\Sigma_H \otimes \Sigma_X$.

I think that $\Sigma_H$ should be well-defined, even though it's unlikely to be Borel in most interesting situations. There should always be some minimal solution, even if it's the whole power set $2^H$.

Here are some general thoughts on why it is important that the evaluation function is measurable, and why this is good enough for most interesting applications, e.g., applied analysis, physics or computation. This means that f $B \in \Sigma_Y$ is any measurable event in $Y$, then $$\operatorname{eval}^{-1}(B) = \big\{ (h,x) : h(x) \in B \big\} \in \Sigma_H \otimes \Sigma_X.$$

For example, this always describes solution-sets to equations, since $$\{ h(x) = y \} = \operatorname{eval}^{-1}({y}).$$

When $Y$ is a measurable hierarchy (i.e., a pre-ordered measurable space), then this also includes inequalities, e.g., $$\{ h(x) \le y \} = \operatorname{eval}^{-1}(\downarrow{y}),$$ where $\downarrow{y} = \{ y' \le y \}$ denotes the down-set of $y \in Y$. Basically, $$\mbox{if you can write it down, it's probably measurable.}$$

This is very useful computationally, since the hom-set $\operatorname{Hom}(H \times X, Y)$ is adjoint to $\operatorname{Hom}(H,Y^X)$ via the process of currying. The adjoint to the evaluation map is called function application, and in computer science is known as Apply. Ultimately, this means that anything you work out computationally is measurable, which means no more appendices full of nasty measurability proofs by hand.

Note that $Y^X$ is a measurable space when equipped with the tensor-product $\sigma$-algebra, and in most cases of interest its $\sigma$-algebra is not generated by a topology (reference Jochen Wengenroth's answer to this question).

Furthermore, this should be useful in measure theory, and may lead toward an answer to Kenny Easwaran's question. If you can see a way to answer it, go ahead and edit this answer.

share|improve this answer

Generally there is no natural sigma-algebra on the function space.

I can give a counterexample to the previous answer: Let's assume that $P$ is the power set over the real numbers, $C$ is the union of all countable and cocountable sets over the real numbers, and $B$ is the discrete measurable space on booleans.

Now the predicate $D\ a\ b = (a = b)$ is $P$-$B^C$-measurable, but the corresponding predicate $D'\ (a, b) = (a = b)$ is not $(P \times C)$-$B$-measurable.

Why? A function $f$ if $P$-$B^C$-measurable iff for all $a \in \mathbb{R}$ the function $f\ a$ is $C$-$B$-measurable, as all sets are measurable in $P$, we only need to check that $f\ x$ maps into the carrier of $B^C$. As all singleton sets are measurable in $C$, the function $\lambda b.\ b = a$ is measurable for all $a$. Hence $D$ is $P$-$B^C$-measurable.

Now the property we want to have about the $B^C$ space is that a $P$-$B^C$-measurable function is also $(P \times C)$-$B$-measurable! Hence, when the measurable function space exists, then $D'$ is $(P \times C)$-$B$-measurable.

But the diagonal set $\{(x, x) \mid x \in \mathbb{R} \}$ is not in $P \times C$.


When working with the usual topologies (i.e. Euclidean spaces) you may take a look into the compact-open topology: http://en.wikipedia.org/wiki/Compact-open_topology

You could use the Borel-space of the compact-open topology as measurable function space.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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