Rough path theory- Verify that $\mathbb{X}_{s,t}=\int_s^t X_{s,r} \otimes dX_r$ This is exercise 7.7 from Martin Hairer's Rough Path notes. 

Verify that $\mathbb{X}_{s,t}=\int_s^t X_{s,r} \otimes dX_r$ where the integral is to be interpreted in the sense of (4.22) (I'll define this later), taking $(Y,Y') $ to be $(X,I)$. In fact, check that this holds not only in the limit $|P|\to 0$ but in fact for every fixed $|P|$, i.e. $\Bbb{X}_{s,t}=\int_P \Xi$

Originally, $\mathbb{X}_{s,t} \colon=\int_s^t X_{s,r} \otimes dX_r$ is a definition for a rough path. However if $Y$ with Gubinelli derivative $Y'$ is a controlled rough path, controlled by $X$, we define:
$$\mathbb{Y}_{s,t} \colon=\int_s^t Y_{s,r} \otimes dX_r\colon = \lim\limits_{|P|\to 0} \int_{p} \Xi$$
where:
$$\Xi_{u,v}= Y_u \otimes Y_{u,v}+Y'_u\otimes Y'_u\mathbb{X}_{u,v} $$
We define the integral:
$$\lim\limits_{|P|\to 0} \int_{p} \Xi\colon = \lim\limits_{|P| \to 0} \sum\limits_{[u,v]\in P} \Xi_{u,v}$$
So we compute:
$$\sum\limits_{[u,v]\in P} Y_u \otimes Y_{u,v}+Y'_u\otimes Y'_u\mathbb{X}_{u,v}=\sum\limits_{[u,v]\in P} X_u \otimes X_{u,v}+I\otimes I\mathbb{X}_{u,v}$$
$$\sum\limits_{[u,v]\in P} Y_u \otimes Y_{u,v}+Y'_u\otimes Y'_u\mathbb{X}_{u,v}=\sum\limits_{[u,v]\in P} X_u \otimes X_{u,v}+\mathbb{X}_{u,v}$$
$$\sum\limits_{[u,v]\in P} Y_u \otimes Y_{u,v}+Y'_u\otimes Y'_u\mathbb{X}_{u,v}=\sum\limits_{[u,v]\in P} X_u \otimes X_{u,v}+\sum\limits_{[u,v]\in P}\mathbb{X}_{u,v}$$
$$\sum\limits_{[u,v]\in P} Y_u \otimes Y_{u,v}+Y'_u\otimes Y'_u\mathbb{X}_{u,v}=\left(\sum\limits_{[u,v]\in P} X_u \otimes X_{u,v}\right)+\mathbb{X}_{s,t}$$
The last equality is because of telescoping series. I am stuck with the first term though. I need to show that $\sum\limits_{[u,v]\in P} X_u \otimes X_{u,v}=0$ and I feel like it's something trivial, but I can't see it.
So my question is, how do we establish $\sum\limits_{[u,v]\in P} X_u \otimes X_{u,v}=0$?
Edit, I made a rather elementary mistake, the last step is incorrect. I will post an answer soon.
 A: Actually, I found my mistake. Once we get to:
$$\sum\limits_{[u,v]\in P} Y_u \otimes Y_{u,v}+Y'_u\otimes Y'_u\mathbb{X}_{u,v}=\sum\limits_{[u,v]\in P} X_u \otimes X_{u,v}+\mathbb{X}_{u,v}$$
Without loss of generality let $X_s=0$, so:
$$X_{u} \otimes X_{u,v}+\mathbb{X}_{u,v}=X_{s,u} \otimes X_{u,v}+\mathbb{X}_{u,v}$$
Then note Chen's relation:
$$X_{u} \otimes X_{u,v}+\mathbb{X}_{u,v}=\Bbb{X}_{s,v}-\Bbb{X}_{s,u}-\Bbb{X}_{u,v}+\mathbb{X}_{u,v}=\Bbb{X}_{s,v}-\Bbb{X}_{s,u}$$
Note this is a telescoping series
$$\sum\limits_{[u,v]\in P} Y_u \otimes Y_{u,v}+Y'_u\otimes Y'_u\mathbb{X}_{u,v}=\sum\limits_{[u,v]\in P}\Bbb{X}_{s,v}-\Bbb{X}_{s,u}=\Bbb{X}_{s,t}-\Bbb{X}_{s,s}=\Bbb{X}_{s,t}$$
Can someone confirm this is okay?
A: I would rather set
$\Xi_{u,v}=X_{u,s}\otimes X_{u,v}+\mathbb{X}_{u,v}$. From this we have
$$
\Xi_{u,v}+\Xi_{v,w}=X_{u,s}\otimes X_{u,v}+\mathbb{X}_{u,v}+X_{v,s}\otimes X_{v,w}+\mathbb{X}_{v,w}
$$
[Chen's relation]
$$
= X_{u,s}\otimes X_{u,v}+X_{v,s}\otimes X_{v,w}+\mathbb{X}_{u,w}+X_{u,v}\otimes X_{v,w}
$$
$$
= X_{u,s}\otimes X_{u,w}+\mathbb{X}_{u,w}=\Xi_{u,w}
$$
from which the result follows.
