From: "G. A. Edgar" Subject: Re: Measurability of metric Date: Sat, 16 Oct 1999 13:23:26 -0400 Newsgroups: sci.math Keywords: Borel sigma algebra on ExE is the square of that on E? In article <7u9k3l$pk3$1@taliesin.netcom.net.uk>, Noel Vaillant wrote: > Let (E,d) be a metric space. Let B(E) be the borel sigma-algebra on E. > Let B(ExE) be the borel sigma-algebra on ExE. > > Whenever E is separable B(ExE) =B(E)xB(E), and the metric d is therefore > measurable with respect to the product sigma algebra B(E)xB(E). > > Do we know whether the 'separability' of E is a necessary condition > for this to be true ? If d were measurable, then the diagonal {(x,x):x in E}={(x,y):d(x,y)=0} would be a measurable set of E x E. But that is false if the cardinal of E is > c (the continuum). B(E)xB(E) is generated by rectangles A x B, and any one set in B(E)xB(E) belongs to the sigma-algebra generated by countably many rectangles. If E has cardinal > c, then coutably many sets canot separate points of E, so countably many rectangles cannot generate a sigma-algebra that contains the diagonal of E x E. > > If f and g are two measuble maps with values in E, do we need to have E > separable to ensure that d(f,g) is itself a measurable map ? In general, yes. Or you could use a more "sensible" definition of " f : X -> E is measurable", say f is the pointwise limit of a sequence of measurable functions with finitely many values. Or you could put a complete measure on X and use measurability with respect to that measure: Completeness will escape the counterexample I gave you above [at least if there are no measurable cardinals]. > > Regards. Noel. > > ------------------------------------------- > Dr Noel Vaillant > http://www.probability.net > vaillant@probability.net -- Gerald A. Edgar edgar@math.ohio-state.edu ============================================================================== From: "David C. Ullrich" Subject: Re: Measurability of metric Date: Sun, 17 Oct 1999 13:28:36 -0500 Newsgroups: sci.math Noel Vaillant wrote: > > > If d were measurable, then the diagonal {(x,x):x in E}={(x,y):d(x,y)=0} > > would be a measurable set of E x E. But that is false if the cardinal > > of E is > c (the continuum). B(E)xB(E) is generated by rectangles > > A x B, and any one set in B(E)xB(E) belongs to the sigma-algebra > > generated by countably many rectangles. If E has cardinal > c, > > then coutably many sets canot separate points of E, > > yep, so far so good. > > >so > > countably many rectangles cannot generate a sigma-algebra > > that contains the diagonal of E x E. > > > > I am still struggling to work out this last point, but i am sure I'll > crack it before the end of the week-end. :-) If not, I ll come back. Your question was one that I thought I should know the answer to but I didn't see what it was right away, so Edgar's reply was very interesting. Stop reading now if you don't want to see a more detailed version: Lemma: If S is a collection of sets then the sigma-algebra generated by S is the same as the union of the sigma-algebras generated by the countable subsets of S. (This is fairly clear if you think of generating the sigma-algebra by transfinite induction. If otoh you're taking the modern "intersection of all the sigma-algebras containing" as the definition of "the sigma-algebra generated by" then the key hint is to show that the union of the sigma-algebras generated by the countable subsets of S is a sigma-algebra.) Now back to your question. If d is measurable then the diagonal is in B(E)xB(E), which is by definition the sigma-algebra generated by the rectangles AxB with A and B elements of B(E). Hence there's some countable collection of rectangles C = {(A_n)x(B_n)} so that the diagonal is in the sigma-algebra generated by C. But that's impossible if the cardinality of E is larger than c: In that case there must exist x and y with x <> y, but such that the sets A_n and B_n can't tell them apart: If S = A_n for some n or S = B_n for some n then either [x and y are both in S] or [neither x nor y is in S]. It follows that a set in the sigma-algebra generated by the A_n x B_n cannot tell the difference between the point (x,x) and the point (x,y) either, in the same sense. (Hint: the class of all subsets of ExE which contain both or neither of the points (x,x) and (x,y) is a sigma-algebra...) Hence the diagonal is not in this sigma-algebra, since the diagonal contains (x,x) but not (x,y). > This is most useful. Thank you very much. > > Noel. > > ------------------------------------------- > Dr Noel Vaillant > http://www.probability.net > vaillant@probability.net ============================================================================== From: "Noel Vaillant" Subject: Re: Measurability of metric Date: Sun, 17 Oct 1999 22:00:39 +0100 Newsgroups: sci.math As a possible example of metric space (E,d) such that B(ExE)<>B(E)xB(E), someone suggested to look at cardE=c, with d the discrete metric d(x,y) = 0 if x=y and d(x,y)=1 if x<>y. The metric topology is the whole cardinal set 2^E The borel sigma-algebra is B(E)=2^E The product topology is the cardinal set 2^ExE So B(ExE)=2^ExE B(E)xB(E) is generated by rectangles AxB (E,d) is not seperable (as a metric space), but can be seperated by a countable set of subsets. I initially tried to prove that the diagonal was not in B(E)xB(E), but I am no longer sure whether this is true or not. Regards. Noel. ------------------------------------------- Dr Noel Vaillant http://www.probability.net vaillant@probability.net