The $1-$dimensional Kolmogorov-Chentsov Theorem is as follows:
Suppose $(X_{t})_{t\in [0,1]}$ that satisfies $$\mathbb{E}|X_{t}-X_{s}|^{\alpha}\leq C|t-s|^{1+\beta},\ \text{for all}\ s,t\in[0,1]$$ and for some $\alpha,\beta, C>0$. Then there exists a continuous modification of $X$.
In the note by Amir Dembo, he gives a proof showing that the coefficients cannot be relaxed. That is, $\beta=0$ cannot work.
The example is as follows:
Consider the stochastic process $X_{t}(\omega)=\mathbb{1}_{\{\omega>t\}}$ for $t\in [0,1]$ and the uniform probability measure on $\Omega=(0,1]$. Then $$\mathbb{E}|X_{t}-X_{s}|=U\Big((s,t]\Big)=|t-s|\ \text{for all}\ 0<t<s\leq 1.$$ Thus, $\{X_{t}, t\in[0,1]\}$ satisfies the “insufficient inequality'' with $C=1$, $\beta=0$ and $\alpha=1$.
However, if $\{\tilde{X}_{t}\}$ is a modification of $\{X_{t}\}$ then a.s. $\tilde{X}_{t}(\omega)=X_{t}(\omega)$ at all $t\in(0,1]\cap\mathbb{Q}$, from which it follows that $s\mapsto \tilde{X}_{s}(\omega)$ is discontinuous at $s=\omega$.
I don't understand the last paragraph of his argument. Why does the existence of the continuous modification $\tilde{X}_{t}$ implies $\tilde{X}_{t}=X_{t}$ for $t\in [0,1]\cap\mathbb{Q}$ and Why does it follow that $\tilde{X}_{s}(\omega)$ is discontinuous at $s=\omega$?
I am really confused…
Thank you!
Best Answer
If I'm using the right definition of continuous modification (I'm rusty), then $\tilde X_t$ satisfies:
$\forall t \in [0,1], \mathbb P(\tilde X_t = X_t) = 1$
Therefore it follows that
$$\mathbb P( \tilde X_t \neq X_t, \text {for some }t \in \mathbb Q \cap [0,1]) \leq \sum_{t \in \mathbb Q} \mathbb P(\tilde X_t \neq X_t) = 0$$
by a union bound and noting that $\mathbb Q$ is countable, and so the event $\tilde X_t = X_t$ for all $t \in \mathbb Q$ happens almost surely.
Now, once we have that $\tilde X_t$ and $X_t$ agree on $\mathbb Q,$ we recall that we've fixed $X_t(w)= 1_{w>t}.$ So for any rational $r > w$ we have $\tilde X_r(w) = 0$ while for rationals $s < w$ we have $\tilde X_s(w) = 1.$ This doesn't seem very continuous to me - it's a jump discontinuity! (we're relying on the fact that $\mathbb Q$ is dense)