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.