Additivity of quadratic variation Let $t<u$. $[X,Y]_t$ and $[X,Y]_u$ are the quadratic cross variations of the processes $X$ and $Y$ at times $t$ and $u$. By that I mean
$$\lim_{\text{mesh}(\pi(t))\rightarrow 0}\sum_{i=0}^{n-1} (X_{s_{i+1}}-X_{s_{i}})(Y_{s_{i+1}}-Y_{s_{i}}) \rightarrow [X,Y]_\tau \quad \text{in probability}$$
where $\pi(t) = \{s_i\}$ and $0 = s_0 < s_1 < \ldots < s_n = \tau$. $\text{mesh}(\pi(t))$ is the length of the largest of the intervals $(s_i,s_{i+1})$.
The question is to show that
$$\lim_{\text{mesh}(\pi(t))\rightarrow 0}\sum_{i=0}^{n-1} (X_{s_{i+1}}-X_{s_{i}})(Y_{s_{i+1}}-Y_{s_{i}}) \rightarrow [X,Y]_u - [X,Y]_t\quad \text{in probability}$$
where $t = s_0 < s_1 < \ldots < s_n = u$.
My attempt is as follows. Since $[X,Y]_t$ and $[X,Y]_u$ both converge to the first sum above for $\tau = t$ and $\tau = u$ in probability, respectively, their difference converges to the difference of these sums in probability. (Here I used a result that I had seen in a book on SDE's). For $[X,Y]_u$ it is very tempting to choose a mesh such that $0 = s_0 < s_1 < \ldots < s_n = u$ where $s_m = t$ for some $m \leq n$. However I haven't been able to justify why this would be a valid move.
 A: Your argumentation is absolutely correct, but since you are still not convinced, let's have a closer look. For simplicity of notation, I'll denote by $\Pi[a,b]$ a partition of the form $\{a=s_0<\ldots < s_n=b\}$ for any $a \leq b$.
By assumption,
$$\lim_{\text{mesh}(\Pi[0,t])\rightarrow 0}\sum_{s_i \in \Pi[0,t]} (X_{s_{i+1}}-X_{s_{i}})(Y_{s_{i+1}}-Y_{s_{i}}) \to [X,Y]_t \qquad \text{in probability.}$$
This means that for any $\epsilon>0$, $r>0$ we can find $\delta=\delta(t)>0$ such that
$$\mathbb{P} \left( \left| \sum_{s_i \in \Pi[0,t]} (X_{s_{i+1}}-X_{s_{i}})(Y_{s_{i+1}}-Y_{s_{i}})- [X,Y]_t \right| >r \right) < \epsilon$$
for any partition $\Pi[0,t]$ with $\text{mesh}(\Pi[0,t]) < \delta$.
Now let $\Pi[t,u]$ be an arbitrary partition of the interval $[u,t]$ such that $\text{mesh}(\Pi[t,u]) < \delta:= \min\{\delta(t),\delta(u)\}$. Denote by $\Sigma[0,t]$ an equidistant partition of $[0,t]$, i.e. $$s_k := \frac{k}{n} t, \qquad k=0,\ldots,n$$ where $n \in \mathbb{N}$ is chosen sufficiently large such that $\frac{1}{n} < \delta$. Then
$$\mathbb{P} \left( \left| \sum_{s_i \in \Sigma[0,t]} (X_{s_{i+1}}-X_{s_{i}})(Y_{s_{i+1}}-Y_{s_{i}})- [X,Y]_t \right| >r \right) < \epsilon \tag{1}$$
and
$$\mathbb{P} \left( \left| \sum_{s_i \in \Sigma[0,t]+\Pi[t,u]} (X_{s_{i+1}}-X_{s_{i}})(Y_{s_{i+1}}-Y_{s_{i}})- [X,Y]_u \right| >r \right) < \epsilon. \tag{2}$$
($\Sigma[0,t]+\Pi[t,u]$ denotes the partition obtained by "glueing" both partitions together. Note that $\text{mesh}(\Sigma[0,t]+\Pi[t,u])<\delta$.)  Now it follows from the triangle inequality that
$$\begin{align*}&\mathbb{P} \left( \left| \sum_{s_i \in \Pi[t,u]} (X_{s_{i+1}}-X_{s_{i}})(Y_{s_{i+1}}-Y_{s_{i}})- ([X,Y]_u-[X,Y]_t) \right| >2r \right) \\ &\leq \mathbb{P} \left( \left| \sum_{s_i \in \Sigma[0,t]} (X_{s_{i+1}}-X_{s_{i}})(Y_{s_{i+1}}-Y_{s_{i}})- [X,Y]_t \right| >r \right) \\ &\quad + \mathbb{P} \left( \left| \sum_{s_i \in \Sigma[0,t]+\Pi[t,u]} (X_{s_{i+1}}-X_{s_{i}})(Y_{s_{i+1}}-Y_{s_{i}})- [X,Y]_u \right| >r \right)\\ &< 2\epsilon \end{align*}$$
Since this holds for any partion $\Pi[t,u]$ with mesh size $<\delta$ and $r, \epsilon$ are arbitrary, this shows that
$$\lim_{\text{mesh}(\Pi[t,u])\rightarrow 0}\sum_{i=0}^{n-1} (X_{s_{i+1}}-X_{s_{i}})(Y_{s_{i+1}}-Y_{s_{i}}) \rightarrow [X,Y]_u - [X,Y]_t\quad \text{in probability}.$$
