# Set of points of strict increase is Borel

1. Sep 3, 2012

### 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 $t$s) 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...

2. Sep 3, 2012

### 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.

3. Sep 4, 2012

### camillio

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

4. Sep 4, 2012

### christoff

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.

5. Sep 4, 2012

### camillio

Thank you Christoff :-)

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