currently I am reviewing some exercises about Brownian motion and the following exercise gives me some trouble: Let $B$ be standard Brownian motion then one can prove that $$ \sup_{t \in [0,1]} |B_t| \leq \sum_{n \in \mathbb N} \sup_{i=0,1,\dots,2^{n}-1} |B_{2^{-n}(i+1)}-B_{2^{-n}i}|. $$
Why this makes sense: In easy words, the above states that we can bound the absolute value of the record time of Brownian motion by an infinite sum of its increments. Here we really need infinitly many since otherwise only taking the difference at two points might not capture that a very high record time was achieved in between.
Update: I answer the question in full detial below.
Found solution: Instead of working with the supremum it turns out that it sufficies totally to first work with some finite $2$-adic representation of $t$ then finding a sufficient bound and applying continuity of paths.
Proof: Let $t \in [0,1] \cap 2^{-N} \mathbb N_0$ then observe that we have the following $2$-adic representation given by $$ t = \sum^{N}_{n=1} a_k 2^{-n}, $$ where $a_k \in \{0,1\}$, now we bound the increment between time $0$ and time $t$, via the absolute values of a $2^{-n}$ difference in the time, i.e. $$ |B_t - B_0| \leq \sum^{N}_{n=1} |B_{a_1*2^{-1}+\dots+a_n 2^{-n}}-B_{a_1*2^{-1}+\dots+a_{n-1} 2^{-(n+1)}}|, $$ where we bound the increment by a $a_n 2^{-n}$-time difference. But now we done, since $$ |B_t - B_0| \leq \sum^{N}_{n=1} |B_{a_1*2^{-1}+\dots+a_n 2^{-n}}-B_{a_1*2^{-1}+\dots+a_{n-1} 2^{-(n+1)}}| \leq \sum^{N}_{n=1} \sup_{i=0,\dots,2^{n}-1} |B_{(i+1)2^{-n}}-B_{i*2^{-n}}|, $$ and since $\cup_{n \geq 0} [0,1] \cap 2^{-n} \mathbb N_0$ is dense in $[0,1]$ we are done.