Constructing quantum field theories is a well-known problem. In Euclidean space, you want to define a certain measure on the space of distributions on R^n. The trickiness is that the detailed properties of the distributions that you get is sensitive to the dimension of the theory and the precise form of the action.

In classical mathematics, measures are hard to define, because one has to worry about somebody well-ordering your space of distributions, or finding a Hamel basis for it, or some other AC idiocy. I want to sidestep these issues, because they are stupid, they are annoying, and they are irrelevant.

Physicists know how to define these measures algorithmically in many cases, so that there is a computer program which will generate a random distribution with the right probability to be a pick from the measure (were it well defined for mathematicians). I find it galling that there is a construction which can be carried out on a computer, which will asymptotically converge to a uniquely defined random object, which then defines a random-picking notion of measure which is good enough to compute any correlation function or any other property of the measure, but which is not sufficient by itself to define a measure within the field of mathematics, only because of infantile Axiom of Choice absurdities.

So is the following physics construction mathematically rigorous?

Question: Given a randomized algorithm P which with certainty generates a distribution $\rho$, does P define a measure on any space of distributions which includes all possible outputs with certain probability?

This is a no-brainer in the Solovay universe, where every subset S of the unit interval [0,1] has a well defined Lebesgue measure. Given a randomized computation in Solovay-land which will produce an element of some arbitrary set U with certainty, there is the associated map from the infinite sequence of random bits, which can be thought of as a random element of [0,1], into U, and one can then define the measure of any subset S of U to be the Lebesgue measure of the inverse image of S under this map. Any randomized algorithm which converges to a unique element of U defines a measure on U.

Question: Is it trivial to de-Solovay this construction? Is there is a standard way of converting an arbitrary convergent random computation into a measure, that doesn't involve a detour into logic or forcing?

The same procedure should work for any random algorithm, or for any map, random or not.

EDIT: (in response to Andreas Blass) The question is how to translate the theorems one can prove when every subset of U gets an induced measure into the same theorems in standard set theory. You get stuck precisely in showing that the set of measurable subsets of U is sufficiently rich (even though we know from Solovay's construction that they might as well be assumed to be everything!)

The most boring standard example is the free scalar fields in a periodic box with all side length L. To generate a random field configuration, you pick every Fourier mode $\phi(k_1,...k_n)$ as a Gaussian with inverse variance $k^2/L^d$, then take the Fourier transform to define a distribution on the box. This defines a distribution, since the convolution with any smooth test function gives a sum in Fourier space which is convergent with certain probability. So in Solovay land, we are free to conclude that it defines a measure on the space of all distributions dual to smooth test functions.

But the random free field is constructed in recent papers of Sheffield and coworkers by a much more laborious route, using the exact same idea, but with a serious detour into functional analysis to show that the measure exists (see for instance theorem 2.3 in http://arxiv.org/PS_cache/math/pdf/0312/0312099v3.pdf). This kind of thing drives me up the wall, because in a Solovay universe, there is nothing to do--- the maps defined are automatically measurable. I want to know if there is a meta-theorem which guarantees that Sheffield stuff had to come out right without any work, just by knowing that the Solovay world is consistent.

In other words, is the construction: pick a random Gaussian free field by choosing each Fourier component as a random gaussian of appropriate width and fourier transforming considered a rigorous construction of measure without any further rigamarole?

EDIT IN RESPONSE TO COMMENTS: I realize that I did not specify what is required from a measure to define a quantum field theory, but this is well known in mathematical physics, and also explicitly spelled out in Sheffield's paper. I realize now that it was never clearly stated in the question I asked (and I apologize to Andreas Blass and others who made thoughtful comments below).

For a measure to define a quantum field theory (or a statistical field theory), you have to be able to compute reasonably arbitrary correlation functions over the space of random distributions. These correlation functions are averages of certain real valued functions on a randomly chosen distribution--- not necessarily polynomials, but for the usual examples, they always are. By "reasonably arbitrary" I actually mean "any real valued function except for some specially constructed axiom of choice nonsense counterexample". I don't know what these distribtions look like a-priory, so honestly, I don't know how to say anything at all about them. You only know what distributions you get out after you define the measure, generate some samples, and seeing what properties they have.

But in Solovay-land (a universe where every subset S of [0,1] is forced to have Lebesgue measure equal to the probability that a randomly chosen real number happens to be an element of S) you don't have to know anything. The moment you have a randomized algorithm that converges to an element of some set of distributions U, you can immediately define a measure, and the expectation value of any real valued function on U is equal to the integral of this function over U against that measure. This works for any function and any distribution space, without any topology or Borel Sets, without knowing anything at all, because there are no measurability issues--- all the subsets of [0,1] are measurable. Then once you have the measure, you can prove that the distributions are continuous functions, or have this or that singularity structure, or whatever, just by studying different correlation functions. For Sheffield, the goal was to show that the level sets of the distributions are well defined and given by a particular SLE in 2d, but whatever. I am not hung up on 2d, or SLE.

If one were to suggest that this is the proper way to do field theory, and by "one" I mean "me", then one would get laughed out of town. So one must make sure that there isn't some simple way to de-Solovay such a construction for a general picking algorithm. This is my question.

EDIT (in response to a comment by Qiaochu Yuan): In my view, operator algebras are not a good substitute for measure theory for defining general Euclidean quantum fields. For Euclidean fields, statistical fields really, you are interested any question one can ask about typical picks from a statistical distribution, for example "What is the SLE structure of the level sets in 2d"(Sheffield's problem), "What is the structure of the discontinuity set"? "Which nonlinear functions of a given smeared-by-a-test-function-field are certainly bounded?" etc, etc. The answer to all these questions (probably even just the solution to all the moment problems) contains all the interesting information in the measure, so if you have some non-measure substitute, you should be able to reconstruct the measure from it, and vice-versa. Why hide the measure? The only reason would be to prevent someone from bring up set-theoretic AC constructions.

For the quantities which can be computed by a stochastic computation, it is traditional to ignore all issues of measurability. This is completely justified in a Solovay universe where there are no issues of measurability. I think that any reluctance to use the language of measure theory is due solely to the old paradoxes.

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    $\begingroup$ If you really want to sidestep all the set-theoretic issues, why are you using measure theory as a conceptual framework at all? $\endgroup$ Jun 26, 2011 at 2:14
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    $\begingroup$ One can effectively replace a measure space by a suitable algebra of random variables on it (see for example en.wikipedia.org/wiki/Abelian_von_Neumann_algebra and terrytao.wordpress.com/2010/02/10/245a-notes-5-free-probability), and it is possible that a suitable generalization of this construction may produce a "generalized measure theory" suitable for QFT. I have no idea if it's expected that this works, but my point is the assumption that measure theory is a reasonable framework for QFT seems to be the assumption you should be, but aren't, challenging. $\endgroup$ Jun 26, 2011 at 2:39
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    $\begingroup$ It's unclear to me why you want a measure defined on all subsets of $U$. As Andreas explained, your randomized algorithm does define a probability measure on some $\sigma$-algebra of subsets of $U$. Is there any reason to believe this $\sigma$-algebra is insufficient? $\endgroup$ Jun 26, 2011 at 20:16
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    $\begingroup$ Sufficient for what? Please add some substance to your questions... $\endgroup$ Jun 27, 2011 at 4:42
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    $\begingroup$ Dear Ron Maimon: if you cannot refrain from using language like "dinky" or "infantile" or "this answer is too trivial", then it might be best if you found another place to ask your questions. $\endgroup$
    – S. Carnahan
    Jun 27, 2011 at 15:31

3 Answers 3


I don't know anything about the space of all distributions dual to smooth test functions, but do know a fair bit about computable measure theory (from a certain perspective).

First, you mention that you have a computable algorithm which generates a probability distribution. I believe you are saying that you have a computable algorithm from $[0,1]$ (or technically the space of infinite binary sequences) to some set $U$ where $U$ is the space of distributions of some type.

Say your map is $f$. How are you describing the element $f(x) \in U$? In computable analysis, there is a standard way to talk about these things. We can describe each element of $U$ with an infinite code (although each element has more than one code). Then $f$ works as follows: It reads the bits of $x$; from those bits, it starts to write out the code for the $f(x)$. The more bits of $x$ known, the more bits of the code for $f(x)$ known.

(Note, not every space has such a nice encoding. If the space isn't separable, there isn't a good way to describe each object while still preserving the important properties, namely the topology. Is say, in your example above, the space of distributions that are dual to smooth test functions, is it a separable space--maybe in a weak topology? Does the encoding you use for elements of $U$ generate the same topology?)

The important property of such a computable map is that it must be continuous (in the topology generated by the encoding, but these usually coincide with the topology of the space). Since $f$ is continuous, we know we can induce a Borel measure on $U$ as follows. If $S$ is an open set then $f^{-1}(S)$ is open and $\mu(f^{-1}(S))$ is known. Similarly, with any Borel sets, hence you have a Borel measure.

Borel measures are sufficient for most applications I can think of (you can integrate continuous functions and from them, define and integrate the L^p functions), but once again, I don't know anything about your applications.

Also, if the function $f$ doesn't always converge to a point in $U$, but only does so almost everywhere, the function $f$ is not continuous, but it is still fairly nice and I believe stuff can be said about the measure, although I need to think about it.

Update: If $f$ converges with probability one, then the set of input points that $f$ converges on is a measure one $G_{\delta}$ set, in particular it is Borel. The function remains continuous on that domain (in the restricted topology). Hence there is still an induced Borel measure on the target space. (Take a Borel set; map it back. It is Borel on the restricted domain, and hence Borel on [0,1]).

Update: Also, I am assuming that your algorithm directly computes the output from the input. I will give an example what I mean. Say one want to compute a real number. To compute it directly, I should be able to ask the algorithm to give me that number within $n$ decimal places with an error bound of $1/10^n$. An indirect algorithm works as follows: The computer just gives me a sequence of approximations that converge to the number. The computer may say $0,0,0,...$ so I think it converges to 0, but at some point it starts to change to $1,1,1,...$. I can never be sure if my approximation is close to the final answer. Even if your algorithm is of the indirect type, it doesn't matter for your applications. It will still generate a Borel map, albeit a more complex one than continuous, and hence it will generate a Borel measure on the target space. (The almost everywhere concerns are similar; they also go up in complexity, but are still Borel.) Without knowing more about your application it is difficult for me to say much specific to your case.

Am I correct in my understanding of your construction, especially the computable side of it? For example, is this the way you describe the computable map from $[0,1]$ to $U$?

On a more general note, much of measure theory has been developed in a set theoretic framework. This isn't very helpful with computable concerns. But using various other definitions of measures, one is able to once again talk about measure theory with an eye to what can and cannot be computed.

I hope this helps, and that I didn't just trivialize your question.

  • $\begingroup$ Yes on everything. The only topology I was thinking about is what you call the topology generated by the encoding. The only sticking point left is the you mention about almost everywhere convergence. $\endgroup$
    – Ron Maimon
    Jun 28, 2011 at 5:37
  • $\begingroup$ I made some edits to the post addressing the almost everywhere convergence. (They are marked Update.) $\endgroup$
    – Jason Rute
    Jun 28, 2011 at 18:36
  • $\begingroup$ Thank you for explaining this constructive measure theory business. Although your answer is self-contained, I was wondering if you can add a literature pointer, just for my own edification. For the specific questions: I didn't think about topology on the space of distributions, because Solovay guarantees that the measure will be defined on all subsets without worrying about topology. But, as you pointed out, there is the implicit topology in the statement that the random-picking algorithm converges. This allows you to easily make a countable dense set in the support. $\endgroup$
    – Ron Maimon
    Jul 1, 2011 at 1:49
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    $\begingroup$ Ron, there are two books: Computability in Analysis and Physics by Pour-El and Richards and Computable Analysis by Klaus Weihrauch. The first has less material, but might be easier to read. The second is heavy on notation. The first also has a section on Fourier transforms which may be of interest to you. $\endgroup$
    – Jason Rute
    Jul 4, 2011 at 15:21

The question is not very clear, but the paragraph following it suggests that you might mean the following. Suppose we have an operation $P$ that takes as input an infinite sequence of binary digits (or, almost equivalently, a number in $[0,1]$) and always produces an output in some set $U$ (of distributions). Does this induce a measure defined on all subsets of $U$? In general (with or without Solovay, and regardless of what the elements of $U$ are), such a $P$ induces a measure on some subsets of $U$, namely those whose inverse-image under $P$ is Lebesgue measurable. In Solovay's model where all sets of reals are Lebesgue measurable, the induced measure is thus defined on all subsets of $U$. In a universe where not all sets of reals are Lebesgue measurable, the natural induced measure on $U$ will not, in general, be defined on all subsets of $U$. For example, $P$ might be the identity map of $[0,1]$. Or, if you insist on $U$ being a set of distributions, $P$ could send each $x\in[0,1]$ to the Dirac delta-distribution concentrated at $x$.

Instead of asking whether the natural induced measure on $U$ is defined on all subsets, one could ask (and maybe you meant to ask) whether there is a reasonable extension of this measure to all subsets of$U$. In that case, I'd like to know what would count as reasonable.

  • $\begingroup$ The question is as follows: you can define the random free field (for example) by picking all Fourier components as Gaussian random with width $k^2/L^2$. This is a complete definition in Solovay universe. Is it a complete definition in the standard universe? $\endgroup$
    – Ron Maimon
    Jun 26, 2011 at 14:26
  • $\begingroup$ It seems very unlikely to me that you can extend the measure in the presence of choice to arbitrary subsets, precisely because the measure has certain translation properties which should allow a Vitali set (although I didn't do an explicit construction). $\endgroup$
    – Ron Maimon
    Jun 26, 2011 at 14:34
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    $\begingroup$ In answer to your first question, I think that what you propose is a complete definition in the sense that nothing more would need to be said to specify which measure you mean. The measure it defined would not, however, have as its domain the collection of all subsets of $U$. $\endgroup$ Jun 26, 2011 at 15:20
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    $\begingroup$ In answer to your second question: Your mention of translation invariance looks like at least the beginning of saying "what would count as reasonable". There are no translation-invariant extensions of Lebesgue measure to all sets (in the standard universe where AC holds), but there might be extensions that are not translation-invariant. This is one reason why I would want to know what sort of extensions you're willing to consider. $\endgroup$ Jun 26, 2011 at 15:22
  • $\begingroup$ The construction of theorem 2.3 in the Sheffield paper is what I want to avoid. This is a theorem of Gross which is cited, which constructs a probability measure which has the properties of the random picking measure. This theorem is trivial in Solovay-land, because the definition in the statement of the theorem automatically constructs the measure, but it is obviously considered nontrivial by Sheffield et. al. Is there a way to transfer the trivial proof to the usual universe, and avoid this Gross thing. $\endgroup$
    – Ron Maimon
    Jun 26, 2011 at 18:36

I don't know anything about the Solovay land, but I can say a little about the random functions and this may be related to what you're going for.

People have for a while been considering random functions which are generated in this way in the context of nonlinear dispersive equations. Probably the most interesting examples are the Gibbs measures associated to infinite dimensional Hamiltonian systems like nonlinear Schrödinger or wave equations. See for example these slides of Jim Colliander:


which has an outline of the technicalities to determine on which Banach space your measure will be supported. More can be found from this lecture of Gigliola Staffilani:


Now, if you want to put a measure on an infinite dimensional space (like the space of distributions), its support will necessarily be extremely thin even if you do get a dense subset. So for instance if you start with some $f \in L^2$ and randomize its Fourier coefficients to make a random function $f^\omega$, you get an increased integrability $f^\omega \in L^p$ for all $p < \infty$ almost surely. There are other measures you can use besides Gaussian measures where the same phenomenon will occur (random $\pm 1$'s will also do the trick by the well-known Khintchine's inequality).

Using Gibbs measures (or just ad hoc randomizations like randomized Fourier coefficients), people have been able to establish almost surely globally defined flows for random data which is "supercritical" when measured in a Sobolev space. The Gibbs measures just come with the nice feature of being invariant under the flow provided you can construct the flow. In physical space, the random data looks much better than a typical element in the Banach space. For this reason, you can show solutions to nonlinear evolution equations exist almost surely and even establish some kind of almost sure well-posedness when you deterministically would not have such a result. But there is a limit to what can be achieved with this freedom. In particular, if your variances not only fail to decay but even grow with the frequency, then it can be very difficult or simply impossible to construct a solution to the equation. The example you gave of variance like $k^2$, for instance, is likely to be too large at frequency infinity -- i.e. too irregular -- for any sensible solution to a familiar nonlinear evolution equation to exist, even though, of course, it will make sense as a measure on the space of distributions and have support in some negative Sobolev space you can explicitly compute. Since Fourier multipliers applied to the random Fourier series will also be random Fourier series, you will also be able to solve linear PDE with the random data with no difficulty, and these solutions will also possess higher integrability than you would ever sensibly ask for. But if the equation you have in mind is the cubic nonlinear Schrödinger and you're insistent about very large data, what you're asking for could likely be hopeless for reasons much more serious than this axiom of choice stuff. You have to pay attention to the space because even cubing -- let alone solving the equation -- may be impossible.

  • $\begingroup$ I inverted the variance by accident--- I meant inverse variance is k^2. This is the typical free bosonic quantum field variance, and it is the same as a Boltzmann distribution for an elastic sheet. The inverse k^2 variance is still irregular at high frequencies at high dimensions, it is only continuous (Brownian) in 1d, and somewhat regular in dimension 2. $\endgroup$
    – Ron Maimon
    Jun 28, 2011 at 12:50
  • $\begingroup$ I am aware of the issues with nonlinear functions of quantum fields. Those need to be dealt with by using the appropriate renormalization at the computational stage. I just wanted to make sure that given a solution to the computational problem (admittedly the more difficult one), that there are no further AC type difficulties in defining the theory. $\endgroup$
    – Ron Maimon
    Jun 28, 2011 at 12:56

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