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.



Inequality regarding a probability measure
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.


