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Set of points of strict increase is Borel

  1. Sep 3, 2012 #1
    Hello all,

    I'm thinking about the following exercise from Intro to stoch. analysis:

    Let [itex]V[/itex] be a continuous, nondecreasing function on [itex]\mathbb{R}[/itex] and [itex]\Lambda[/itex] its Lebesgue-Stieltjes measure. Say [itex]t[/itex] is a point of strict increase for [itex]V[/itex] if [itex] V(s) < V(t) < V(u)[/itex] for all [itex]s<t[/itex] and all [itex]u>t[/itex]. Let [itex]I[/itex] be the set of such points. Show that [itex]I[/itex] is a Borel set and [itex]\Lambda(I^C) = 0[/itex].

    My attempt to this exercise:
    By definition, for any rational point [itex]t \in I[/itex] there exists an [itex]\epsilon[/itex]-neighbourhood of [itex]t[/itex] containing [itex]s<t[/itex] and [itex]u>t[/itex] where [itex]s,u \in I[/itex]. The neighbourhood, forming an open set, is Borel. Countable (due to rationality of [itex]t[/itex]s) union of these neighbourhoods forms [itex]I[/itex] which is Borel too.

    Complement of [itex]I[/itex] is hence a countable union of connected sets (say [itex]J_i, i=1,...,n[/itex]) where, due to the nondecreasing property [itex]V(x) = V(y)[/itex] for all [itex]x,y[/itex] in particular [itex]J_i[/itex]. Since [itex]\Lambda(J_i) = |V(x)-V(y)| = 0[/itex] for any connected set [itex]J_i[/itex], hence [itex]\Lambda(I^C) = \Lambda(\cup J_i) = 0[/itex].

    Intuitively, I feel that my proof misses or skips something important...
     
  2. jcsd
  3. Sep 3, 2012 #2
    I have very little experience with measure theory, but from what I do know, this seems fine to me. If I might ask, where do you feel your proof is lacking? If you elaborate on this, I may be able to offer some input.
     
  4. Sep 4, 2012 #3
    Thank you for your response, Christoff. My uncertainty followed from my considering only rational [itex]t[/itex]s and I was unsure whether my approach doesn't neglect some sets.
     
  5. Sep 4, 2012 #4
    Hmm, good point. I hadn't considered that. I don't think, however, that it is too difficult to show that any irrational point of strict increase is contained in one of your [itex]\epsilon[/itex]-neighbourhoods of a rational point.

    eg. Let [itex]q\in I[/itex] be an irrational point of strict increase. Then there exists a largest open interval [itex](s,u)[/itex] containing q with [itex](s,u)\subset I[/itex]. Since the rationals are dense, there exists a rational [itex]t\in (s,u)[/itex]. Since t is rational, there exists a maximal open interval [itex](s',u')[/itex] containing t, such that [itex](s',u')\subset I[/itex]. By maximality of [itex] (s',u')[/itex], we must have [itex](s,u)\subset (s',u')[/itex].

    In conclusion, you don't miss any irrational points. I think that should do it.
     
  6. Sep 4, 2012 #5
    Thank you Christoff :-)

    Probably for the irrational points it follows directly from their property of being accumulation points of sequences of rationals.
     
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