Lebesgue measure on Frechet space?

It is well known that there are no Lebesgue measures on infinite-dimensional Banach spaces (see e.g. http://en.wikipedia.org/wiki/There_is_no_infinite-dimensional_Lebesgue_measure). However, I couldn't find anything about Lebesgue measures on infinite-dimensional Frechet spaces. The question seems very natural in the context of creating a mathematically rigorous definition of the path integral of quantum field theory.

So: 1. Can an infinite-dimensional Frechet space have a measure which is locally finite, strictly positive and translation-invariant? 2. Can a separable infinite-dimensional Frechet space have a measure which is locally finite, non-zero and translation-invariant?

The motivation for the formulations of 1+2 are that the analogous statements for Banach spaces are false. 2 is more important for me since all interesting examples as separable, as far as I can tell.

If such measures exist, I'd be glad to get some references on whatever is known about them.

Thx!

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No. The proof on the Wikipedia page you cite can be adapted to the Frechet setting. – Ryan Budney Nov 28 '10 at 18:51
If you're interested in fresh approaches to defining the path integral, I recommend Kevin Costello's in-progress book on the topic: math.northwestern.edu/~costello/renormalization – j.c. Nov 28 '10 at 21:41
In an appendix to Weil's book on integration on topological groups he proves a theorem to the effect that a topological group with a measure that resembles Haar measure (there are precise conditions, which I don't recall right now) has to be locally compact. There is a theorem of Riesz that any locally compact Hausdorff topological real vector space is finite-dimensional. (Well, Riesz gave the proof only for normed real vector spaces, but in Weil's Basic Number Theory I think he gives a proof for top. vector spaces.) – KConrad Nov 29 '10 at 5:01
The same argument works for any infinite dimensional separable TVS $X$. A non-negative and non-zero measure $\mu$ on $X$ vanishes on no non-empty open set, because countably many translates of the latter cover $X$. On the other hand, any non-empty open set of $X$ does contains infinitely many disjoint translates of some non-empty open set, thus it has infinite $\mu$ measure. – Pietro Majer May 31 '11 at 20:39