Integrability along a Hilbert space?

[LieToMe]

Suppose we have an infinite dimensional Hilbert-like space but that is incomplete, such as if a subspace isomorphic to $\mathbb{R}$ had countably many discontinuities and we extended it to an isomorphism of $\mathbb{R}^{\infty}$. Is there a measure of integrating along any closed subset of such a space? How does one distinguish between when the gaps it may have defies integrability versus when they resemble isolated points that can be integrated over with respect to the topology of the space?

 

[Stephen Tashi]

such as if a subspace isomorphic to $\mathbb{R}$ had countably many discontinuities

What does it mean for a subspace to be "continuous"? - or not continuous ? Are you using "continuous" to mean "closed"?

 

[LieToMe]

What does it mean for a subspace to be "continuous"? - or not continuous ? Are you using "continuous" to mean "closed"?

I'm thinking of a more open interpretation where at every point in the domain space, there is an open ball in the inverse image of the codomain that contains that limit point.

 

[Stephen Tashi]

What is a "domain space"? The terms "domain" and "co-domain" apply to a function. The definition of "subspace" and "function" are different.

 

[LieToMe]

What is a "domain space"? The terms "domain" and "co-domain" apply to a function. The definition of "subspace" and "function" are different.

Right but I'm talking about integrating a function over a Hilbert space.

 

[Stephen Tashi]

Right but I'm talking about integrating a function over a Hilbert space.

You didn't use the term "function" in your original post, but you did mention "subspace". I suggest you state your question using standard terminology. The domain of a function can be a subset of a Hilbert Space that is not a subspace of that Hilbert Space.

 

[LieToMe]

You didn't use the term "function" in your original post, but you did mention "subspace". I suggest you state your question using standard terminology. The domain of a function can be a subset of a Hilbert Space that is not a subspace of that Hilbert Space.

I said "along" a subset of such a space, which is pretty conventional. What else are you going to integrate if not a function over that space?

The domain issue you pointed it is the entire point of the post. If there are countably many points missing from any finite dimensional subspace that is extended to space otherwise isomorphic to infinite extension of a Euclidean space, then is the space still guaranteed to be integrable over any closed subset?
I suspect one needs to impose a restriction of measurability which reduces the structure to countably many dimensions, but this is also sensical for extending Euclidean space.

 

[fresh_42]

I said "along" a subset of such a space, which is pretty conventional. What else are you going to integrate if not a function over that space?

You say it yourself: over a subset not along! And of course is your question even with "over" far too broad to be answered in a meaningful way. What do we know about the domain and the function? Next, a Hilbert space is complete by definition, incomplete inner product spaces are called pre-Hilbert space. Anyway, the key is measure theory. You need a measurable subset to perform an integration, the entire space is not.

 

[LieToMe]

be answered in a meaningful way

But I'm looking precisely for a broad answer to cover bases to research. Does a direct product of measurable Euclidean spaces to a Hilbert-like space preserve measurability? Maybe, maybe not. If one has only countably many dimensions, seems like there is a good chance. I imagine in uncountable cases there could be a subset that is not measurable, but in such a case, what additional constraints are needed?

 

[S.G. Janssens]

But I'm looking precisely for a broad answer to cover bases to research. Does a direct product of measurable Euclidean spaces to a Hilbert-like space preserve measurability?

See e.g. Aliprantis and Border, Infinite Dimensional Analysis, p. 520 in the third edition, in the section on the Kolmogorov Extension Theorem. (Keyword: infinite product $\sigma$-algebra. In the definition, note the analogy with the general definition of the product topology. Also, note that measurability as such does not require the existence of a measure.)

If one has only countably many dimensions, seems like there is a good chance. I imagine in uncountable cases there could be a subset that is not measurable, but in such a case, what additional constraints are needed?

This is puzzling. Already for finite products there may be non-measurable subsets. Do you mean something more specific?

 

[member 587159]

Even $\mathbb{R}$ with the standard Borel algebra has subsets that are nog Borel-measurable (or even Lebesgue-measurable, cfr axiom of choice).

Anyway, the book "Real and functional analysis" by Lang has a very good section on the integral of functions with values in a Banach space. Worth checking out imo.

 

[S.G. Janssens]

Anyway, the book "Real and functional analysis" by Lang has a very good section on the integral of functions with values in a Banach space.

It is my impression that the questions in the OP were more about the domain side than about the co-domain side.

What I mean is that questions of measurability and integration of functions from subsets of $\mathbb{R}$ into a Banach space may be quite different from the same questions for functions from a Banach space into $\mathbb{R}$, although there are also relations.

(I am aware that you yourself know that.)

 

[member 587159]

s

It is my impression that the questions in the OP were more about the domain side than about the co-domain side.

What I mean is that questions of measurability and integration of functions from subsets of $\mathbb{R}$ into a Banach space may be quite different from the same questions for functions from a Banach space into $\mathbb{R}$, although there are also relations.

(I am aware that you yourself know that.)

Thanks for the clarification. I missed that.

I must admit that it is not entirely clear to me what the OP is asking. All I can add at the moment is that every locally compact Hausdorff group (for example, the additive structure of a normed vector space) admits a canonical Borel $\sigma$-algebra on which we can define a unique left (right) translation-invariant regular Borel measure (left Haar measure).

Thus in the case of the OP, every normed space has a canonical choice of $\sigma$-algebra and measure. In particular, completeness is not relevant here. What more does one want?

 

[martinbn]

s

Thanks for the clarification. I missed that.

I must admit that it is not entirely clear to me what the OP is asking. All I can add at the moment is that every locally compact Hausdorff group (for example, the additive structure of a normed vector space) admits a canonical Borel $\sigma$-algebra on which we can define a unique left (right) translation-invariant regular Borel measure (left Haar measure).

Thus in the case of the OP, every normed space has a canonical choice of $\sigma$-algebra and measure. In particular, completeness is not relevant here. What more does one want?

In the infinite dimensional case, you don't have a locally compact space. There are no translation invariant measures. You have to settle for something else.

The original question is not clear at all. It is way too broad. Give us some motivation and more specifics.

 

[member 587159]

In the infinite dimensional case, you don't have a locally compact space. There are no translation invariant measures. You have to settle for something else.

The original question is not clear at all. It is way too broad. Give us some motivation and more specifics.

Yes, of course. I really should get some sleep. Thanks for the correction.

 

[Stephen Tashi]

Perhaps relevant: https://en.wikipedia.org/wiki/Infinite-dimensional_Lebesgue_measure

 

[WWGD]

All I can think of is that product of uncountable copies of the Reals is not metrizable. For Riemann integration you can allow even uncountable discontinuities as long as the measure of the set is 0.