k\in \mathbb{N},$$and hence $\left[ M^{\left( n\right) }-M\right] _{\tau _{k}^{n}}\underset{n\rightarrow \infty }{\overset{\mathbb{P}}{\longrightarrow }}0$ for all $k\in \mathbb{N}.$

Finally, to prove that $\left[ M^{\left( n\right) }-M\right] _{\infty }\overset{\mathbb{P}}{\rightarrow }0$ we assume that it is not true, and then $\left[ M^{\left( n\right) }-M\right] _{\infty }\overset{\mathbb{P}}{\nrightarrow }0$ implies that there exist $\varepsilon >0$ and $\left\{
n_{k}\right\} _{k\in \mathbb{N}}\subset \mathbb{N}$ with $$d\left( \left[ M^{\left( n_{k}\right) }-M\right] _{\infty },0\right) \geq
\varepsilon$$for all $k\in \mathbb{N},$where $d\left( X,Y\right) :=\inf \left\{
\varepsilon >0:\mathbb{P}\left[ \left\vert X-Y\right\vert >\varepsilon \right] \leq \varepsilon \right\} $ is the Ky Fan metric. We shall denote the subsequence as the original sequence, trying to keep the notation as simple as possible. Using a diagonal argument, a subsequence $\left\{
n_{i}\right\} _{i\in \mathbb{N}}\subset \mathbb{N}$ can be chosen, with the property that $d\left( \left[ M^{\left( n_{i}\right) }-M\right] _{\tau
_{k}^{n_{i}}},0\right) <\frac{1}{k}$ for all $i\geq k.$ Since $$\lim_{k\rightarrow \infty }\left[ M^{\left( n_{i}\right) }-M\right] _{\tau
_{k}^{n_{i}}}=\left[ M^{\left( n_{i}\right) }-M\right] _{\infty }\quad
\mathbb{P}-a.s.,$$we can find some $k\left( n_{i}\right) \geq i$ such that $$d\left( \left[ M^{\left( n_{i}\right) }-M\right] _{\tau _{k\left(
n_{i}\right) }^{n_{i}}},\left[ M^{\left( n_{i}\right) }-M\right] _{\infty
}\right) <\frac{1}{k}.$$Then, using the estimation $$\mathbb{P}\left[ \left\vert \left[ M^{\left( n_{k}\right) }-M\right] _{\tau
_{k\left( n_{k}\right) }^{n_{k}}}-\left[ M^{\left( n_{k}\right) }-M\right]
_{\tau _{k}^{n_{k}}}\right\vert >\varepsilon \right] \leq \mathbb{P}\left[
\left\{ \sup\limits_{t\in \mathbb{R}_{+}}\left\vert M_{t}^{\left(
