From: chernoff@math.berkeley.edu (Paul R. Chernoff) Subject: Re: Invariant measure on R^d Date: 20 Jan 2001 23:35:18 GMT Newsgroups: sci.math Summary: Translation-invariant measures on R^d are scaled Lebesgue measures In article , Prosper Abutbul wrote: >If m is a translation invariant measure on R^d and m(B) = 1 when >B is the standard unit cube in R^D then must m equal the standard >Lebesgue measure on R^d ? > >Thank you, >Prosper The answer is yes, and the proof is elementary. For simplicity, we do the 1-dimensional case. Since m is translation-invariant, obviously m({point}) = 0 since B contains any finite number of points. all of which have the same measure, by translation invariance. So m([0.1]) = 1 = m([0,1)) for example. Now the half-open interval [0,1) is the union of n half-open intervals of length 1/n, hence if I is an interval of length 1/n, m(I) = 1/n . Next, if r > 0 is rational (m/n), then by a similar argument m([0,r)) = m m([0,1/n)) (sorry for the double use of the symbol m!) = m/n = r. By countable additivity, m{[0,x)) = x for all real x > 0, hence m([a,b)) = b -a = \mu([a.b)) where \mu is Lebesgue measure. From this it follows that m(E) = \mu(E) for each Lebesgue-measureable set E. -- # Paul R. Chernoff chernoff@math.berkeley.edu # # Department of Mathematics 3840 # # University of California "Against stupidity, the gods themselves # # Berkeley, CA 94720-3840 struggle in vain." -- Schiller #