Set of points of strict increase is Borel

[camillio]

Hello all,

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

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

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

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

Intuitively, I feel that my proof misses or skips something important...

 

[christoff]

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.

 

[camillio]

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.

Thank you for your response, Christoff. My uncertainty followed from my considering only rational ts and I was unsure whether my approach doesn't neglect some sets.

 

[christoff]

Thank you for your response, Christoff. My uncertainty followed from my considering only rational ts and I was unsure whether my approach doesn't neglect some sets.

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 \epsilon-neighbourhoods of a rational point.

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

In conclusion, you don't miss any irrational points. I think that should do it.

 

[camillio]

Thank you Christoff :-)

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