Inequality regarding a probability measure

Question

First of all, I am sorry for the ''not clear title' for this question but I cannot find a better way to describe this seemingly very simple and standard inequality,

So.. I am reading a paper 'Two-dimensional Navier-Stokes Equation Driven by a space time white noise' by Daprato and Debussche. And I came across an inequality regarding a probability measure.(It is in the proof of main theorem p.198) It seems very strandard but I cannot see why it is.

$$\mathbb{P}\left[ \sup_{t\in [0,T]} \left| u_N(t,u_0) \right|_{\mathcal{B}^\sigma_{p,\rho}} \geq M \right]\leq \sum_{k=0}^{[T/t^*_M]} \mathbb{P}\left[\sup_{t\in [kt^*_M,(k+1)t^*_M]} \left| u_N(t,u_0) \right|_{\mathcal{B}^\sigma_{p,\rho}}\geq M \right]$$

Here, $\mathcal{B}^\sigma_{p,\rho}$ is a Besov space and $\mathbb{P}$ is a probability measure.

I initially thought it is a typo and the $M$ in the right hand side should be replaced with $M/[T/t^*_M]$ but I realized that there is a possibility that I am missing something. So I wanted to hear from someone else.

I thank in advance for any help with this.

Answer 1

It looks fine to me.

If we let $I_k = [k t^\ast_M, (k+1) t^\ast_M]$ be the relevant subintervals of $[0,T]$, then the supremum of $|u_n|$ over $[0,T]$ must be almost attained along some sequence of points, and by pigeonhole infinitely many of them must be in one of the $I_k$, call it $I_{k_0}$, so that $\sup_{I_{k_0}} |u_n| = \sup_{[0,T]} |u_n|$. So if the sup over $[0,T]$ is at least $M$, then the sup over some $I_k$ must also be at least $M$ (and conversely). In other words, the event $\{\sup_{[0,T]} |u_n| \ge M\}$ equals the union of the events $A_k = \{\sup_{I_k} |u_n| \ge M\}$.

Now we just use the union bound.