# Canonical measure on an infinite dimensional Euclidean space R^N

1. Jun 4, 2009

### jostpuur

I just encountered the Wikipedia page There is no infinite-dimensional Lebesgue measure, and I was left slightly confused by it. They say that a Lebesgue measure $m_n$ on $\mathbb{R}^n$ has the property that each point $x\in\mathbb{R}^n$ has an open environment with non-zero finite measure, and then go on to say that it would be natural to demand the same property for a measure on some infinite dimensional norm space. In fact it does not seem very natural to demand it, IMO. I shall now forget about norm spaces, and deal with the vector space $\mathbb{R}^{\mathbb{N}}$ only (possibly equipped with a non-proper Euclidean metric $d:\mathbb{R}^{\mathbb{N}}\times\mathbb{R}^{\mathbb{N}}\to [0,\infty]$, for the purpose of dealing with some balls). Considering the fact that a cube $[0,1]\times [0,1]\times\cdots\subset\mathbb{R}^{\mathbb{N}}$ is supposed to have a measure 1, and the fact that you can insert an infinite amount of disjoint balls of fixed radius into the corners of this cube, it is immediately clear that the measure of such small balls is supposed to be zero. So IMO the conclusion that "there is no infinite-dimensional Lebesgue measure" seems to be premature and exaggerated.

It is a fact that a very natural measure exists on the set $[0,1]^{\mathbb{N}}$, which generalizes the measures on $[0,1]^n$.

Suppose $X$ is some set, $\mathcal{F}\subset\mathcal{P}(X)$ is some collection of its subsets, that $\sigma(\mathcal{F})$ is the $\sigma$-algebra generated by $\mathcal{F}$, and that $\mu_1,\mu_2$ are two probability measures defined on $\sigma(\mathcal{F})$. In this situation we can ask that will $\mu_1$ and $\mu_2$ be the same, if $\mu_1(A)=\mu_2(A)$ for all $A\in\mathcal{F}$. It turns out that this will not necessarily be the case always, but $\mu_1$ and $\mu_2$ will be the same if $\mathcal{F}$ is a $\pi$-system, meaning that $A\cap B\in\mathcal{F}$ for all $A,B\in\mathcal{F}$.

The measure on $[0,1]^{\mathbb{N}}$ can be defined as follows. First set

$$\mathcal{F}=\big\{[a_1,b_1]\times [a_2,b_2]\times\cdots\times [a_n,b_n]\times [0,1]\times [0,1]\times\cdots\; |\; 1\leq n,\; 0\leq a_k\leq b_k\leq 1\;\textrm{for}\; 1\leq k\leq n\big\}$$

and

$$\mu:\mathcal{F}\to [0,1],\quad \mu([a_1,b_1]\times [a_2,b_2]\times\cdots\times [a_n,b_n]\times [0,1]\times [0,1]\times\cdots)=\prod_{k=1}^n (b_k-a_k)$$

Now $\mathcal{F}$ is a $\pi$-system, so $\mu$ extends uniquely to become a Borel measure (with respect to the product topology) on $[0,1]^{\mathbb{N}}$.

This was not yet the same thing as a measure on $\mathbb{R}^{\mathbb{N}}$, but IMO it seems that we can easily reduce the problem measuring sets in $\mathbb{R}^{\mathbb{N}}$ to the measures on small cubes $[0,1]^{\mathbb{N}}$, since we can write $\mathbb{R}^{\mathbb{N}}$ as a union of this kind of identical cubes. This is an uncountable union, since the indexes of the cubes lie in $\mathbb{Z}^{\mathbb{N}}$, and

$$\textrm{card}(\mathbb{R}) = \textrm{card}(\{0,1\}^{\mathbb{N}}) \leq \textrm{card}(\mathbb{Z}^{\mathbb{N}}),$$

but I don't think that uncountability is a serious problem. Suppose $X\subset\mathbb{R}^{\mathbb{N}}$ is an arbitrary set. We then restrict the attention to the intersection of $X$ with the cubes. If the intersection is non-measurable in some of the cubes, then we say that $X$ is not measurable. If this does not happen, then we divide $\mathbb{Z}^{\mathbb{N}}$ into two subsets $I,J$ so that the measures of the intersection of $X$ with cubes $I$ are non-zero, and the measures of the intersection of $X$ with the cubes $J$ are zero. If $I$ is uncountable, then we set the measure of $X$ to be infinite, and if $I$ is not uncountable, then we define the measure of $X$ by summing up its measures in the cubes $I$.

Isn't this now a well defined and the most natural measure on $\mathbb{R}^{\mathbb{N}}$?

2. Jun 4, 2009

### jostpuur

I was first thinking that we could obtain a natural measure for example on $\ell^2\subset \mathbb{R}^{\mathbb{N}}$ by restricting the previous measure. However, we cannot obtain a nontrivial measure like this because the subset $\ell^2$ itself has zero measure. The Wikipedia article also makes more sense when one considers this. Anyway, I still insist that the conclusion "there is no infinite-dimensional Lebesgue measure" is not justified.

3. Jun 5, 2009

### Hurkyl

Staff Emeritus
(Assuming what you wrote is well-defined):

Here's one pathological property of your measure:

$$\mu [0, 1)^{\mathbb{N}} = 0$$

Because it's the union of $\mu [0, 1 - 1/n)^{\mathbb{N}}$ as n ranges over the positive integers, and each of those sets have zero measure.
These are open subsets of the cube

4. Jun 5, 2009

### jostpuur

There is something that doesn't make sense in this. Your remark leads to a following contradiction. Define

$$I_1 := \{1\}\times [0,1]\times [0,1]\times [0,1]\times\cdots$$
$$I_2 := [0,1]\times \{1\}\times [0,1]\times [0,1]\times\cdots$$
$$I_3 := [0,1]\times [0,1]\times \{1\}\times [0,1]\times\cdots$$
$$\quad\vdots$$

This is a countable sequence of sets with zero measure, but

$$[0,1]^{\mathbb{N}} = [0,1[^{\mathbb{N}}\;\cup\; I_1\;\cup \;I_2\;\cup\cdots$$

so thats like 1=0 then.

5. Jun 5, 2009

### jostpuur

Okey. I have two comments now. One to my own posting, and other one to Hurkyl's posting.

This was true, to my current knowledge.

This was not properly justified according to my current knowledge.

If I understood my lecture notes correctly, it is possible that a measure cannot be extended out from some $\pi$-system at all. If the measure can be extended, then the extension will be unique, anyway.

To prove the existence of an extension, one can use Caratheodory-Hahn-extension theorem. It says that more structure for $\mathcal{F}$ is needed. Besides being a $\pi$-system, $\mathcal{F}$ must contain $\emptyset$, and $\mathcal{F}$ must contain complements of its members. Then we must have a mapping $\mu:\mathcal{F}\to[0,\infty]$, which satisfies the condition

$$\mu\Big(\bigcup_{k=1}^{\infty} A_k\Big) = \sum_{k=1}^{\infty}\mu(A_k)$$

when $A_1,A_2,\ldots\in\mathcal{F}$ are disjoint, and $\mu(\emptyset)=0$. When these conditions are met, then $\mu$ will extend to $\sigma(\mathcal{F})$ and become a measure.

The $\mathcal{F}$ I defined earlier did not satisfy these conditions. However, this does not seem to be a big problem, because the previous construction can be corrected. Instead set

$$\mathcal{F}=\big\{ A\times [0,1]\times [0,1]\times\cdots\;|\; 1\leq n,\; A\subset [0,1]^n\;\textrm{is a Borel subset}\big\}$$

Now $\mathcal{F}$ satisfies the needed conditions, and we can set

$$\mu(X)=m_n(A),\quad\quad\textrm{when}\; X=A\times [0,1]\times [0,1]\times\cdots.$$

So I believe that the measure for $[0,1]^{\mathbb{N}}$, which I described earlier, does exist, even though I did not justify it properly in the original post.

I'm a afraid that this is the case:

$$\bigcup_{n=1}^{\infty} [0,1-\frac{1}{n}]^{\mathbb{N}} \subsetneq [0,1[^{\mathbb{N}}$$

Choose

$$x=\big(0,\frac{1}{2},\frac{2}{3},\frac{3}{4},\ldots\big)\in [0,1[^{\mathbb{N}}.$$

Now

$$x\notin [0,1-\frac{1}{n}]^{\mathbb{N}}$$

for all $n=1,2,3,\ldots$.

6. Jun 5, 2009

### Hurkyl

Staff Emeritus
Good catch. A colimit of a limit is not the limit of a colimit. I should know better than that! (In the sense of category theory, a nested union is a kind of colimit, and the Cartesian product is a kind of limit)

7. Jun 5, 2009

### Hurkyl

Staff Emeritus
Anyways, assuming this works out, it seems the question is to test your construction on innocuous sets that could be problematic.

The first one I thought of was

$$\left( [0, 1/2] \times [0, 2] \right)^\mathbb{N}$$

or possibly even to consider

$$\prod_{n \in \mathbb{N}} [0, a_n]$$

where $\prod_n a_n = 1$ or something like that.

If there are infinitely many terms larger than 1, I think your construction says that the measure is automatically infinity.

This made me realize there is another pathological set:

$$\mu \left\{ 0, 2 \right\}^{\mathbb{N}} = +\infty$$

(Yes, the repeated factor is the set with two elements, not an interval)

Oh, and yet another bad one:

$$\mu [-1/2, 1/2]^{\mathbb{N}} = +\infty$$

8. Jun 6, 2009

### jostpuur

I didn't look very carefully all of those examples, but it seems that you made a mistake, which was that a measure of a set would be infinite always, when the set intersects an uncountable amount of cubes. But that will not yet make the measure infinite. The measure is guaranteed to be infinite if the set's intersections with an uncountable amount of cubes have non-zero measure.

I took a closer look at the following subsets.

$$T_0 = [0,1]\times [0,1]\times [0,1]\times\cdots$$
$$T_1 = [-\frac{1}{2},\frac{1}{2}]\times [0,1]\times [0,1]\times\cdots$$
$$T_2 = [-\frac{1}{2},\frac{1}{2}]\times[-\frac{1}{2},\frac{1}{2}]\times [0,1]\times\cdots$$

$$T_{\infty} = [-\frac{1}{2},\frac{1}{2}]\times [-\frac{1}{2},\frac{1}{2}] \times[-\frac{1}{2},\frac{1}{2}]\times\cdots$$

The set $T_0$ intersects only the cube $(0,0,0,\ldots)\in \mathbb{Z}^{\mathbb{N}}$.

The set $T_1$ intersects the cubes $(0,0,0,\ldots)$ and $(-1,0,0,\ldots)$.

In general the set $T_n$ intersects the cubes $(k_1,k_2,\ldots ,k_n,0,0,\ldots)$ where $k_i\in \{0,-1\}$. There are $2^n$ cubes, and the measure of the restriction of $T_n$ in each of them is $1/2^n$. So $\mu(T_n)=1$ for all $n=0,1,2,3,\ldots$.

The set $T_{\infty}$ intersects an uncountable amount of cubes, but its measure in each one them is zero.

This is a same kind of situation like when a zero-function is integrated over a set with infinite measure. Usually if $f:X\to\mathbb{R}$ is a constant so that $f(x)=c$ for all $x\in X$, then

$$\int\limits_{X} f(x) d\mu(x) = c\mu(X),$$

but if $c=0$ and if $\mu(X)=\infty$, then

$$\int\limits_{X} f(x) d\mu(x) = 0.$$

In the same spirit, I decided in the original post, that a measure of a set like $T_{\infty}$ will be zero, because it is the only reasonable unique choice for its measure.

Is this an unacceptable property, since it still looks like that $T_{\infty}$ should have a measure 1?

Here's a possible philosophy to solve the problem:

The results

$$\int\limits_{-\infty}^{\infty} \chi_{[n,n+1]}(x) dx = 1,\quad\quad\forall n\in\mathbb{Z}$$

and

$$\int\limits_{-\infty}^{\infty}\Big(\lim_{n\to\infty} \chi_{[n,n+1]}(x)\Big)dx = 0$$

can be interpreted so that when a cube is "translated by an infinite amount", then its measure goes to zero, because it "goes outside the domain of Lebesgue measure". The cube $T_n$ is basically a cube $T_0$ which is translated a distance $\sqrt{n}$ in some direction. So $T_{\infty}$ is translated by an infinite distance. It could be a bad sign, if $T_{\infty}$ still had a measure 1, because then a cube's measure would vanish when it is translated by infinity to some directions, but would remain 1 in some other directions, indicating a violation in rotational symmetry. So the fact that the measure vanishes in an infinite translation also in direction $(-1,-1,-1,\ldots)$ could be a good sign too.

A following problem arises, however. Define

$$T'_1 = [-1,0]\times [0,1]\times [0,1]\times\cdots$$
$$T'_2 = [-1,0]\times [-1,0]\times [0,1]\times\cdots$$

$$T'_{\infty} = [-1,0]\times [-1,0]\times [-1,0]\times\cdots$$

Now suddenly a translation by an even greater amount results in the measure arising back above the zero, which is not quite compatible with the previous philosophy.

The only solution that I can come up to this is the following modification. The cubes whose centers' distance from the center of $[0,1]^{\mathbb{N}}$ must be excluded from the measure computation. This means that we define

$$(\mathbb{Z}^{\mathbb{N}})_0 := \big\{ (k_1,k_2,k_3,\ldots)\in\mathbb{Z}^{\mathbb{N}}\;|\; k_n=0\;\textrm{when}\; n>N,\quad\textrm{for some}\; N\in\mathbb{N}\big\}$$

and then consider only the intersections of a set $X\subset\mathbb{R}^{\mathbb{N}}$ with the cubes corresponding to $(\mathbb{Z}^{\mathbb{N}})_0$.

It could be that this is getting farther from the original goal, but isn't this where that attempt to find the canonical measure leads to?

Last edited: Jun 6, 2009
9. Jun 6, 2009

### jostpuur

The Lebesgue measure $dm_1$ on $\mathbb{R}$ is an example of a measure which is invariant under translations $x\mapsto x+R$, and as result the measure is not concentrated close to the origo.

A Gaussian measure $d\mu(x) = e^{-Ax^2} dm_1(x)$ is an example of a measure which is concentrated close to the origo, and as result is not invariant under translations $x\mapsto x+R$.

In the previous posts I have attempted to described such measure on $\mathbb{R}^{\mathbb{N}}$, which is invariant under finite translations, but which still is concentrated only to close to the origo. This might appear paradoxical, but one has to recall that most of the Euclidean distances between points in $\mathbb{R}^{\mathbb{N}}$ are infinite. So this vector space is so great in size, that even a (finite-)translation invariant measure can vanish when one goes sufficiently far away from origo.

As a heuristic remark, suppose we have substituted infinity to $x=\infty$, and then take the limit $A\to 0^+$ in $e^{-Ax^2}$. The limit will remain zero.

Last edited: Jun 6, 2009
10. Jun 6, 2009

### Hurkyl

Staff Emeritus
Ah, I misread the opening post -- I thought that's how you defined the extension to all of RN. But still, you see how they are pathological and you went in the opposite direction: to make my example sets have measure 0.

Which confirms the statement originally under consideration -- that there is no infinite dimensional analog of the Lesbegue measure. The measures that you can define on RN are not going to behave like finite dimensional Lesbegue measures.

11. Jun 6, 2009

### jostpuur

If I want to make the vector space $\mathbb{R}^{\mathbb{N}}$ to become a norm space, I would have to restrict the attention to some subspace, like $\ell^2\subset\mathbb{R}^{\mathbb{N}}$, and some vectors far away from origo would have to be abandoned.

If it turns out that making the vector space to become a measure space $(\mathbb{R}^{\mathbb{N}},dm_{\mathbb{N}})$ will have the effect of making a large volume far away from origo to become a null set, it would seem acceptable to me.

The claim "there is no infinite dimensional Lebesgue measure" sounds similar as a claim "there is no infinite dimensional norm space".

12. Jun 6, 2009

### Hurkyl

Staff Emeritus
If you pick any (Hamel) basis and use the standard dot product... *presto* instant inner product space.

13. Jun 7, 2009

### jostpuur

Okey, what I said was not true, but I believe that I explained already sufficiently why the Wikipedia article's explanation is not satisfying.

One should not approach the measure in an infinite dimensional vector space like this: "Let's first demand all kind of properties from the measure." ... and then... "oh, it turned out that the measure does not exist."

A better approach would be this: "Let's demand that we are going to get a measure, and then see what properties it must have so that it's going to exist."

14. Jun 7, 2009

### Hurkyl

Staff Emeritus
That's because you seem to want it to say something other than what it's actually saying.

One should if one actually wants those properties.

Oh, by the way, the theorem:
{mu is a measure on RN and mu is translation invariant and there exists an open set U such that mu(U) is finite and positive} is a contradiction​

is equivalent to the theorem

{mu is a measure on RN} implies at least one of {mu is not translation invariant} and {for every open set U, mu(U) = 0 or +infinity}​

Last edited: Jun 7, 2009
15. Jun 8, 2009

### jostpuur

Your statement is logically true, but that is a big "if". The Wikipedia article contains some suspicious property demands about properties that I dont' really see any reason to demand.

One assumption in the Wikipedia article, which is suspicious, is that the space is assumed to be separable. If one considers an example $\ell^2\subset\mathbb{R}^{\mathbb{N}}$, which is separable, or some other separable subspaces of $c_0\subset\mathbb{R}^{\mathbb{N}}$ (the space of sequences that converge towards zero), which will have zero measure (if the measure is going to exist), and on the other hand considers a space $\ell^{\infty}\subset\mathbb{R}^{\mathbb{N}}$, which is not separable, and which will have non-zero measure (if the measure is going to exist) because $[0,1]^{\mathbb{N}}\subset\ell^{\infty}$, it starts to look intuitively plausible, that if a non-trivial Lebesgue-like measure will exist on some norm space, this space will not be separable.

The Wikipedia article comments like this:
A very quick examination reveals that at least sufficiently small $\ell^2$-balls are going to have zero measure. It is immediately clear that even though open sets have non-zero measure in finite dimensions, this is not a kind of property that one should insist in infinite dimensions too. An insistence that open sets should have non-zero measure also in infinite dimensions, is another example of a suspicious demand.

The big problem with property demands is that one cannot know in advance what the suitable properties are going to be. They must be found during the study/research. A one good example about this emerged in this thread already. It could be that I was originally keeping up hopes that this could be true

$$m_{\mathbb{N}}(x+X) = m_{\mathbb{N}}(X),\quad\quad\forall \;x\in\mathbb{R}^{\mathbb{N}},$$

but I have now replaced this hope with a new one, which is this:

$$m_{\mathbb{N}}(x+X) = m_{\mathbb{N}}(X),\quad\quad\forall \;x\in\ell^2\subset \mathbb{R}^{\mathbb{N}}$$

Even though the vector space where the measure is intended to exist is larger than $\ell^2$. One cannot know in advance what the translation invariance is going to mean.

16. Jun 8, 2009

### Hurkyl

Staff Emeritus
The reasons seemed fairly clear to me.

We insist that the measure is translation invariant, because translations are an isomorphism of (the affine space corresponding to) RN.

We insist the measure is nonzero, beacuse otherwise it can't tell us anything at all.

We insist that there exists a nonempty open set of finite measure, because otherwise the measure cannot tell us anything interesting about open sets, which play a central role in topology and analysis.

And these three qualities are enough to arrive at the contradiction.

Your goal seems to be to find some sort of "natural" measure on RN, and there's nothing wrong with trying to do that. But I really and truly don't understand your fixation on the adjective "Lesbegue-like".

How does that change anything? Restricted to $\ell^2$, you have a translation-invariant measure, and you're right back to square 1.

17. Jun 8, 2009

### jostpuur

I am mostly interested in finding the natural measure to the vector space $\mathbb{R}^{\mathbb{N}}$, and there is no immediate reason to equip it with any particular topology. So the search for the measure cannot begin with properties of open sets. It could be good idea to ask what kind topology would make the measure $dm_{\mathbb{N}}$ a Borel measure, after the measure is found.

Do you notice that there is no objective mathematical claim that we would disagree about? In particular, I am not contradicting any theorems mentioned in the Wikipedia article.

Perhaps I should start proving stuff, and then post the theory to the independent research forum? It looks like the brainstorming is done.

The last change I would make to the construction is that the cube $[0,1]^{\mathbb{N}}\subset\mathbb{R}^{\mathbb{N}}$ would be replaced with a cube $[-\frac{1}{2},\frac{1}{2}]^{\mathbb{N}}\subset\mathbb{R}^{\mathbb{N}}$ so that the origo gets to the center of this all.

Last edited: Jun 8, 2009
18. Jun 8, 2009

### zetafunction

Interesting problem

physicists solve it using a Gaussian Measure or a Delta (infinite) measure,

the problem of 'infinite measure' is important because if solved it could provide a tool to compute Functional integrals and hence propagators and physical quantitites

19. Jun 8, 2009

### gel

I think if you find a translation invariant measure $\mu$ on RN which is finite on some open set then $\mu_\lamba(S)\equiv\mu(\lambda S)$ will give you an uncountable collection of non-equivalent measures (over $\lambda>0$). Also, you can add $\mu(LS)$ for linear operators L. So, if there's one then there's a lot of them.

20. Jun 8, 2009

### gel

Rather than just construct some fairly arbitrary measure, it would be better if you could show that it is the unique measure satisfying some useful or interesting properties. Or, if not unique, then classify all measures satisfying whatever property you come up with.