Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm writing notes for my upcoming class in Game Theory and I realized some time ago that I only need three properties of the Lebesgue measure $\lambda$ on $\mathbb{R^n}$.

  1. It is a non-negative measure on Borel subset of $\mathbb{R}^n$ that is strictly positive on open non-empty sets.
  2. Every proper hyperplane in $\mathbb{R}^n$ has measure zero.
  3. There is a polynomial function $P: (\mathbb{R}^n)^{n+1}\to \mathbb{R}$ such that: For all $(x_1,x_2,\ldots,x_{n+1})\in (\mathbb{R}^n)^{n+1}$ there is a one to one onto function $\ell: \{1,2,\ldots, n+1\}\to \{1,2,\ldots, n+1\}$ satisfying $$ \lambda(conv\{x_1,x_2,\ldots,x_{n+1}\}) =P(x_{\ell(1)},x_{\ell(2)},\ldots,x_{\ell(n+1)}) $$ where $conv$ stands for convex hull.

The Lebesgue measure $\lambda$ satisfies (3) by identifying the volume of a positively oriented simplex with determinant of vertices times a constant.

Q1. I am only interested however in the existence of a measure satisfying (1), (2), (3). Do I have to fully construct the cumbersome Lebesgue measure to do this?

Q2. Can I construct a measure satisfying (1) and (2), which measures bounded sets finitely, in a manner that is less cumbersome than the Lebesgue measure? (Thanks @5PM: Added this question in Edit)

Q3. Suppose instead of polynomial P I require that P be real analytic. Will a measure satisfying (1) (2) (3) always be Lebesgue absolutely continuous with an analytic density function (like the normal distribution on $\mathbb{R^n}$)? In other words, if I show the existence of a measure satisfying (1) (2) (3) have I simply developed the Lebesgue measure.

share|improve this question
1  
There are other measures that satisfy (1)(2)(3); for example $d\mu = q(x)\,dx$ for any positive polynomial $q:\mathbb R\to \mathbb R$. However I do not think that any of them will be less cumbersome than the Lebesgue measure; the opposite is more likely. Even forgetting property (3), do you know a measure that satisfies (1)(2) and is less cumbersome than the Lebesgue measure? –  user53153 Jan 23 '13 at 3:17
    
@5Pm yes, surely. Thanks. I don't know the answer to your last question regarding (1) (2). As you know, I want to develop these without the Lebesgue measure or constructing it. I want an elementary but full development that goes directly to those three properties without going down the Lebesgue measure road. –  Rabee Tourky Jan 23 '13 at 3:19
    
I'm not sure what exactly you will do with it, but I guess you want to make some genericity argument. In that case, it might be easier to forget about measure completely and use being in the complement of a lower-dimesnional manifold as your notion of genericity. Those students who know Lebesgue measure, can then easily make the connection. –  Michael Greinecker Jan 23 '13 at 8:35
    
@Michael Greinecker. the motivation arises from a proof that I have that two person finite games have an equilibrium without appealing to a fixed point theorem. It simply uses the fact that if $\sigma_1$ and $\sigma_2$ are two simplexes in $\mathbb{R}$ $f:\mathbb{R^n} \to \mathbb{R}^n$ is a polynomial function –  Rabee Tourky Jan 23 '13 at 8:41
    
@RabeeTourky That sounds fascinating. Is there some place where one can see that proof? –  Michael Greinecker Jan 23 '13 at 8:43

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.