Quadratic Variation of Brownian Motions Let $B$ be a Brownian motion. The following statement is well-known:

Let $(\pi_n)$ be a sequence of partitions of $[0, \infty)$ satisfying $\pi_n \subseteq \pi_{n+1}$ and $\text{mesh}(\pi_n) \rightarrow 0$ as $n \rightarrow \infty$. Then for any $t >0$, there exists a null set $N_t$ such that $$\lim_{n \to \infty}\sum_{s \in \pi_n}(B_{s'\wedge t}(\omega)-B_{s\wedge t}(\omega))^2=t$$ for every $\omega \in \Omega\setminus N_t$. 

Here $s'$ denotes the next element of the partition that is larger than $s$. 
My question is, is there any null set $N$ such that $$\lim_{n \to \infty}\sum_{s \in \pi_n}(B_{s'\wedge t}(\omega)-B_{s\wedge t}(\omega))^2=t$$ for every $t \in [0, \infty)$ and $\omega \in \Omega \setminus N$?

This statement is true. 
For any $q \in \mathbb{Q}$, there exists a null set $N_q$ such that $$\lim_{n \to \infty}\sum_{s \in \pi_n}(B_{s'\wedge t}(\omega)-B_{s\wedge t}(\omega))=t$$ for every $\omega \in \Omega \setminus N_q$. Define $$\Omega'=\bigcap_{q \in \mathbb{Q}}\Omega\setminus N_q.$$ We have $$P(\Omega')=1.$$
Let $\omega \in \Omega'$. Then for every $t \in \mathbb{R}$, there exists a rational sequence $(q_k)$ converging to $t$. Then $$t = \lim_{k \to \infty} q_k=\lim_{k \to \infty}\lim_{n \to \infty}\sum_{s \in \pi_n}(B_{s'\wedge q_k}(\omega)-B_{s\wedge q_k}(\omega))=\lim_{n \to \infty}\sum_{s \in \pi_n}(B_{s'\wedge t}(\omega)-B_{s\wedge t}(\omega)),$$ where the last equality follows from the continuity of Brownian motions.
