I've edited, just skip the first attempt and go to the second one.


**THE FRAMEWORK:** let us consider a real topological vector space $V$.


We denote with $\mathscr C\_k(V)$ the set of all continous functions $f:[0,T]^k\to V$ such that $f\_{t\_1\cdots t\_k}=0$ whenever $t\_i=t\_{i+1}$ for some $0\le i\le k-1$.


We define the operator $\delta\_k:\mathscr C\_k(V)\to\mathscr C\_{k+1}(V)$ as follows:
$$
(\delta\_kf)\_{t\_1\cdots t\_{k+1}}:=\sum\_{j=1}^{k+1}(-1)^jf\_{t\_1\cdots \widehat t\_{j}\cdots t\_{k+1}}
$$
where the hat $\widehat{\cdot}$ means that argument is omitted.


So if $f\in\mathscr C\_1(V)$ then $(\delta\_1f)\_{ts}=f\_t-f\_s$ and if $g\in\mathscr C\_2(V)$ then $(\delta\_2g)\_{tus}=-g\_{us}+g\_{ts}-g\_{tu}$.


Next we define the following norms: if $g\in\mathscr C\_2(V)$ then, for $\mu>1$ we set
$$
\|g\|\_{\mu}:=\sup\_{0\le s<t\le T}\frac{|g\_{st}|}{|t-s|^{\mu}}
$$
while for $h\in\mathscr C\_3(V)$ we first set
$$
\|h\|\_{\mu,\rho}:=\sup\_{0\le s<t\le T}\frac{|h\_{tus}|}{|t-u|^{\rho}|u-s|^{\mu-\rho}}
$$
(here $0<\rho<\mu$) and then define the norm:
$$
\|h\|\_{\mu}:=\inf\left\{\sum\_j\|h\_j\|\_{\rho\_j,\mu-\rho\_j}:h=\sum\_jh\_j,\; h\_j\in\mathscr C\_3,0<\rho\_j<\mu\right\}\;.
$$


Finally let us denote for $k=2,3$
$$
\mathscr C\_k^{\mu}:=\{h\in\mathscr C\_k:\|h\|\_{\mu}<+\infty\}.
$$
and accept that 
$$
\ker\delta\_k=\operatorname{Im}\delta\_{k-1}
$$
(this holds for every $k$).


**THE PROBLEM:** let us take $h\in\mathscr C\_3^{\mu}$ such that $\delta\_3h=0$. Then it is not difficult to prove that there exists $B\in\mathscr C\_2$ such that $\delta\_2B=h$. Now fix $0\le s<t\le T$ and consider on $[s,t]$ a sequence of partitions $\{\pi\_n\}\_n$ whose mash tends to zero.


To fix ideas we write
$$
\pi\_n=\{s=r\_0^n<r\_1^n<\cdots<r\_{k\_n}^n<r\_{k\_n+1}^n=t\}
$$


Define then
$$
M\_{ts}^{\pi\_n}:=B\_{ts}-\sum\_{l=0}^{k\_n}B\_{r\_{l+1}^nr\_l^n}\;.
$$
Now accept this last one converges (up to passing to a subsequence): how can we show that the limit does not depend on the particular sequence of partitions chosen?


**MY ATTEMPT:** let us take another sequence of partitions whose mash tends to zero, say
$$
\sigma\_n=\{s=u\_0^n<u\_1^n<\cdots<u\_{h\_n}^n<u\_{h\_n+1}^n=t\}
$$


and prove that
$$
|M\_{ts}^{\pi\_n}-M\_{ts}^{\sigma\_n}|\to0\;\;\;n\to+\infty.
$$


Now
\begin{align\*}
M\_{ts}^{\pi\_n}-M\_{ts}^{\sigma\_n}
&=\sum\_{l=0}^{h\_n}B\_{u\_{l+1}^nr\_l^n}
-\sum\_{l=0}^{k\_n}B\_{r\_{l+1}^nr\_l^n}\\
&=\sum\_{l=0}^{k\_n}
\left(B\_{t\_{4l+4}^nt\_{4l+2}^n}-B\_{t\_{4l+3}^nt\_{4l+2}^n}-
B\_{t\_{4l+4}^nt\_{4l+3}^n}\right)-
\left(B\_{t\_{4l+3}^nt\_{4l+3}^n}-B\_{t\_{4l+2}^nt\_{4l+1}^n}-
B\_{t\_{4l+3}^nt\_{4l+2}^n}\right)\\
&=\sum\_{l=0}^{k\_n}
(\delta\_2B)\_{t\_{4l+4}^nt\_{4l+3}^nt\_{4l+2}^n}-
(\delta\_2B)\_{t\_{4l+3}^nt\_{4l+2}^nt\_{4l+1}^n}\\
&=\sum\_{l=0}^{k\_n}
h\_{t\_{4l+4}^nt\_{4l+3}^nt\_{4l+2}^n}-
h\_{t\_{4l+3}^nt\_{4l+2}^nt\_{4l+1}^n}\\
\end{align\*}
supposing wlog that $h\_n\le k\_n$ and setting, for $l\le h\_n$
\begin{align\*}
t\_{4l+1}^n&:=r\_l^n\\
t\_{4l+2}^n&:=u\_l^n\\
t\_{4l+3}^n&:=r\_{l+1}^n\\
t\_{4l+4}^n&:=u\_{l+1}^n\\
\end{align\*}
while, for $l>h\_n$ the definition of the odd terms stay the same, while $t\_{4l+2}^n=t\_{4l+4}^n:=t$.


Then we can write
\begin{align\*}
|M\_{ts}^{\pi\_n}-M\_{ts}^{\sigma\_n}|
&\le\sum\_{l=0}^{k\_n}
|h\_{t\_{4l+4}^nt\_{4l+3}^nt\_{4l+2}^n}|+
|h\_{t\_{4l+3}^nt\_{4l+2}^nt\_{4l+1}^n}|\\
&=\sum\_{l=0}^{k\_n}
\left|\sum\_jh\_{t\_{4l+4}^nt\_{4l+3}^nt\_{4l+2}^n}^j\right|+
\left|\sum\_ih\_{t\_{4l+3}^nt\_{4l+2}^nt\_{4l+1}^n}^i\right|\\
&\le\sum\_{l=0}^{k\_n}\left[
\sum\_j\left|h\_{t\_{4l+4}^nt\_{4l+3}^nt\_{4l+2}^n}^j\right|+
\sum\_i\left|h\_{t\_{4l+3}^nt\_{4l+2}^nt\_{4l+1}^n}^i\right|\right]\\
&=\sum\_{l=0}^{k\_n}\left[
\sum\_j\left|\frac{h\_{t\_{4l+4}^nt\_{4l+3}^nt\_{4l+2}^n}^j}{|t\_{4l+3}^n-t\_{4l+2}^n|^{\rho\_j}|t\_{4l+4}^n-t\_{4l+3}^n|^{\mu-\rho\_j}}\right|
|t\_{4l+3}^n-t\_{4l+2}^n|^{\rho\_j}|t\_{4l+4}^n-t\_{4l+3}^n|^{\mu-\rho\_j}\right]\\
+&
\left[
\sum\_i\left|\frac{h\_{t\_{4l+3}^nt\_{4l+2}^nt\_{4l+1}^n}^i}{|t\_{4l+2}^n-t\_{4l+1}^n|^{\rho\_i}|t\_{4l+2}^n-t\_{4l+2}^n|^{\mu-\rho\_i}}\right|
|t\_{4l+2}^n-t\_{4l+1}^n|^{\rho\_i}|t\_{4l+3}^n-t\_{4l+2}^n|^{\mu-\rho\_i}\right]\\
\end{align\*}
and since this is true for every $\{h^j\}\_j,\{h^i\}\_i\subset\mathscr C\_3(V)$ such that $\sum\_jh^j=\sum\_ih^i=h$ and $0<\rho\_j,\rho\_i<\mu$, passing to the $\inf$ on these parameters, we get 
\begin{align\*}
|M\_{ts}^{\pi\_n}-M\_{ts}^{\sigma\_n}|
\le
\|h\|\_{\mu}\left[\sum\_{l=0}^{k\_n}\left(\max\{|t\_{4l+3}^n-t\_{4l+2}^n|,|t\_{4l+4}^n-t\_{4l+3}^n|\}\right)^{\mu}\\
+\left(\max\{|t\_{4l+2}^n-t\_{4l+1}^n|,|t\_{4l+3}^n-t\_{4l+2}^n|\}\right)^{\mu}\right]\\
\le\|h\|\_{\mu}\left[\sum\_{l=0}^{k\_n}\left(|t\_{4l+3}^n-t\_{4l+2}^n|+|t\_{4l+4}^n-t\_{4l+3}^n|\right)^{\mu}\\
+\left(|t\_{4l+2}^n-t\_{4l+1}^n|+|t\_{4l+3}^n-t\_{4l+2}^n|\right)^{\mu}\right]
\end{align\*}


From this, I tried as follows: call $m\_0=1$ and take $\pi\_{m\_0}$; then since the mash tends to zero, there exists $m\_1>m\_0$ such that $m\_0$ elements of $\sigma\_{m\_1}$, call them $\widetilde{\sigma\_{m\_1}}:=\{u\_{l\_s}^{m\_1}\}\_{s=1}^{m\_0}$ are closer as we want to the elements of $\pi\_{m\_0}$, say $|r\_s^{m\_0}-u\_{j\_s}^{m\_1}|<\varepsilon\_1$.


We have to work on $\sigma\_{m\_1}$: there exists $m\_2>m\_1$ such that considering suitable elements of $\pi\_{m\_2}$, say $\widetilde{\pi\_{m\_2}}:=\{r\_{j\_s}^{m\_2}\}\_{s=1}^{m\_1}$, we can approximate the elements of $\sigma\_{m\_1}$, say $|u\_s^{m\_i}-r\_{j\_s}^{m\_2}|<\varepsilon\_2$, and so on.


Then I did the obvious:
$$
M\_{ts}^{\pi\_{m\_{2\beta}}}-M\_{ts}^{\sigma\_{m\_{2\beta+1}}}
=\left(M\_{ts}^{\pi\_{m\_{2\beta}}}-M\_{ts}^{\widetilde{\sigma\_{m\_{2\beta+1}}}}\right)
+\left(M\_{ts}^{\widetilde{\sigma\_{m\_{2\beta+1}}}}-M\_{ts}^{\widetilde{\pi\_{m\_{2(\beta+1)}}}}\right)
+\left(M\_{ts}^{\widetilde{\pi\_{m\_{2(\beta+1)}}}}-M\_{ts}^{\sigma\_{m\_{2(\beta+1)-1}}}\right)\;\;;
$$ 
now the first and the last summand go to zero as $\beta\to+\infty$, but the second term doesn't a priori, since it is an approximation of the LHS.


I thought that maybe there is a way to modify the $\widetilde{\pi}$ and $\widetilde{\sigma}$ and/or exploiting the absolute continuity of $M$ in oder to compare the LHS and the second term of RHS, and maybe writing something like 
$$
\left|M\_{ts}^{\widetilde{\sigma\_{m\_{2\beta+1}}}}-M\_{ts}^{\widetilde{\pi\_{m\_{2(\beta+1)}}}}\right|
=a\_{\beta}|M\_{ts}^{\pi\_{m\_{2\beta}}}-M\_{ts}^{\sigma\_{m\_{2\beta+1}}}|
$$
for some $0<a\_{\beta}<1$ such that $\limsup\_{\beta\to+\infty}a\_{\beta}<1$, thing that allow me to conclude, but for the moment I'm stuck. 


**ATTEMPT 2:** 


Step 1: Let us fix $m\_0\ge1$ and consider $\Pi\_{m\_0}$ (which has $k\_{m\_0}+2$ elements); then, since the meshes of both sequences of partitions go to $0$, there exists $m\_1>m\_0$, such that, looking at $\mathfrak S\_{m\_1}$ we can find in it $k\_{m\_0}+3$ elements we label as $\{u\_{j\_s}^{m\_1}\}\_{s=0}^{k\_{m\_0}+2}=:\widetilde{\mathfrak S}\_{m\_1}$ such that
\begin{align\*}
\left|B\_{r\_{l+1}^{m\_0}r\_l^{m\_0}}-B\_{u\_{j\_{l+1}}^{m\_1}u\_{j\_{l}}^{m\_1}}\right|&\le\frac1{2(k\_{m\_0}+1)^3}
\;\;,\;\;\;\;l=0,\dots,k\_{m\_0}\\
\left|B\_{u\_{j\_{k\_{m\_0}+2}}^{m\_1}u\_{j\_{k\_{m\_0}+1}}^{m\_1}}\right|&\le\frac1{2(k\_{m\_0}+1)^3}\;.
\end{align\*}
Thus we associate step $1$ to the pair $\left(\Pi\_{m\_0},\widetilde{\mathfrak S}\_{m\_1}\right)$.


Step 2: Let us consider $\mathfrak S\_{m\_1}$ (which has $h\_{m\_1}+2$ elements); then there exists $m\_2>m\_1$ such that looking at $\Pi\_{m\_2}$, we can find in it $h\_{m\_1}+3$ elements we label as $\{r\_{j\_s}^{m\_2}\}\_{j=0}^{h\_{m\_1}+3}=:\widetilde{\Pi}\_{m\_2}$ such that
\begin{align\*}
\left|B\_{u\_{l+1}^{m\_1}u\_l^{m\_1}}-B\_{r\_{j\_{l+1}}^{m\_2}r\_{j\_{l}}^{m\_2}}\right|&\le\frac1{2(h\_{m\_1}+1)^3}
\;\;,\;\;\;\;l=0,\dots,h\_{m\_1}\\
\left|B\_{r\_{j\_{h\_{m\_1}+2}}^{m\_2}r\_{j\_{h\_{m\_1}+1}}^{m\_2}}\right|&\le\frac1{2(h\_{m\_1}+1)^3}\;.
\end{align\*}
Thus we associate step $2$ to the pair $\left(\mathfrak S\_{m\_1},\widetilde{\Pi}\_{m\_2}\right)$.


Going on with this construction, we have 
\begin{align\*}
&{Step 2N}\leadsto \left(\mathfrak S\_{m\_{2N-1}},\widetilde{\Pi}\_{m\_{2N}}\right)\\
&{Step 2N+1}\leadsto \left(\Pi\_{m\_{2N}},\widetilde{\mathfrak S}\_{m\_{2N+1}}\right).
\end{align\*}
Now it is well know that given a converging sequence (and it is known that $\{M\_{ts}^{\Pi\_n}\}\_{n\ge1}$, up to passing to a subsequence, is such), every its subsequence converges to the same limit. Since we want to prove that
$$
\left|M\_{ts}^{\Pi\_n}-M\_{ts}^{\mathfrak S\_n}\right|\stackrel{n\to+\infty}{\longrightarrow}0
$$
we can work with any subsequence of $\{M\_{ts}^{\Pi\_n}\}\_{n\ge1}$ and $\{M\_{ts}^{\mathfrak S\_n}\}\_{n\ge1}$.
So, let us prove that 
$$
\left|M\_{ts}^{\Pi\_{m\_{2N}}}-M\_{ts}^{\mathfrak S\_{m\_{2N+1}}}\right|\stackrel{N\to+\infty}{\longrightarrow}0.
$$


Now, observing that $k\_m,h\_m\ge m$ and $m\_N\ge N$, we have
\begin{align\*}
\left|M\_{ts}^{\Pi\_{m\_{2N}}}-M\_{ts}^{\mathfrak S\_{m\_{2N+1}}}\right|
&\le\left|M\_{ts}^{\Pi\_{m\_{2N}}}-M\_{ts}^{\widetilde{\mathfrak S}\_{m\_{2N+1}}}\right|
+\left|M\_{ts}^{\widetilde{\mathfrak S}\_{m\_{2N+1}}}-M\_{ts}^{\widetilde{\Pi}\_{m\_{2(N+1)}}}\right|
+\left|M\_{ts}^{\widetilde{\Pi}\_{m\_{2(N+1)}}}-M\_{ts}^{\mathfrak S\_{m\_{2N+1}}}\right|\\
&\le\frac1{(k\_{m\_{2N}}+1)^2}
+\left|M\_{ts}^{\widetilde{\mathfrak S}\_{m\_{2N+1}}}-M\_{ts}^{\widetilde{\Pi}\_{m\_{2(N+1)}}}\right|
+\frac1{(h\_{m\_{2N+1}}+1)^2}\\
&\le\frac1{(2N+1)^2}
+\frac1{(2N+2)^2}
+\left|M\_{ts}^{\widetilde{\mathfrak S}\_{m\_{2N+1}}}-M\_{ts}^{\widetilde{\Pi}\_{m\_{2(N+1)}}}\right|.
\end{align\*}


but here I am stuck, because in order to repeat the argument for the last summand I should pass to a suitable subsequence wrt to $N$ in the whole inequality
$$
\left|M\_{ts}^{\Pi\_{m\_{2N}}}-M\_{ts}^{\mathfrak S\_{m\_{2N+1}}}\right|
\le\frac1{(2N+1)^2}
+\frac1{(2N+2)^2}
+\left|M\_{ts}^{\widetilde{\mathfrak S}\_{m\_{2N+1}}}-M\_{ts}^{\widetilde{\Pi}\_{m\_{2(N+1)}}}\right|
$$
but I don't know how to do it in a rigorous way.


Moreover, even doing this, and getting something like 
$$
\left|M\_{ts}^{\Pi\_{m\_{2N}}}-M\_{ts}^{\mathfrak S\_{m\_{2N+1}}}\right|
\le\sum\_{k\ge N}a\_k+\left|M\_{ts}^{\Pi\_{m\_{X}}}-M\_{ts}^{\mathfrak S\_{m\_{Y}}}\right|
$$
where $\sum\_ka\_k$ is a convergent series, how can we control the last summand?


PS: this is taken from a proof contained in the 2010 paper by [M.Gubinelli and S. Tindel ROUGH EVOLUTION EQUATIONS](https://arxiv.org/pdf/0803.0552.pdf) at the end of page 9 (they say to see another paper, but I didn't found nothing on it!)



Well definition of a function
Let us define then $\mathfrak {U}\_n:=\Pi\_n\cup\mathfrak S\_n$ and $u\_n+2$ as the cardinality of $\mathfrak U\_n$; then, relabeling the elements of $\mathfrak U\_n$, we can write $\mathfrak U\_n=\{s=t\_0^n<t\_1^n<\cdots<t\_{u\_n}^n<t\_{u\_n+1}^n=t\}$ 


We recall that
\begin{align\*}
M\_{ts}^{\Pi\_n}=B\_{ts}-\sum\_{i=0}^{k\_n}B\_{r\_{i+1}^nr\_i^n}\;\;\;\;\;\;\;\mbox{and}\;\;\;\;\;\;\;\;\;\;
M\_{ts}^{\mathfrak S\_n}=B\_{ts}-\sum\_{i=0}^{h\_n}B\_{u\_{i+1}^nu\_i^n}\;.
\end{align\*}


We want to prove that
$$
\left|M\_{ts}^{\Pi\_n}-M\_{ts}^{\mathfrak S\_n}\right|\stackrel{n\to+\infty}{\longrightarrow}0.
$$
So, let us consider
$$
\left|M\_{ts}^{\Pi\_n}-M\_{ts}^{\mathfrak S\_n}\right|
\le \left|M\_{ts}^{\Pi\_n}-M\_{ts}^{\mathfrak U\_n}\right|+\left|M\_{ts}^{\mathfrak U\_n}-M\_{ts}^{\mathfrak S\_n}\right|.
$$
Now
\begin{align\*}
\left|M\_{ts}^{\Pi\_n}-M\_{ts}^{\mathfrak U\_n}\right|
&=\left|\sum\_{i=0}^{k\_n}B\_{r\_{i+1}^nr\_i^n}-\sum\_{j=0}^{u\_n}B\_{t\_{j+1}^nt\_j^n}\right|\\
&=\left|\sum\_{i=0}^{k\_n}\left(B\_{r\_{i+1}^nr\_i^n}-\sum\_{\substack{t\_j^n,t\_{j+1}^n\in\mathfrak U\_n\\r\_i^n\le t\_j^n<t\_{j+1}^n\le r\_{l+1}^n}}B\_{t\_{j+1}^nt\_j^n}\right)\right|\\
&\le\sum\_{i=0}^{k\_n}\left|B\_{r\_{i+1}^nr\_i^n}-\sum\_{\substack{t\_j^n,t\_{j+1}^n\in\mathfrak U\_n\\r\_i^n\le t\_j^n<t\_{j+1}^n\le r\_{l+1}^n}}B\_{t\_{j+1}^nt\_j^n}\right|\\
&\le\sum\_{i=0}^{k\_n}\left|M\_{r\_{i+1}^nr\_i^n}^{\mathfrak U\_n\cap[r\_i^n,r\_{i+1}^n]}\right|\\
&\le c\_{h,\lambda}\sum\_{i=0}^{k\_n}|r\_{i+1}^n-r\_i^n|^{\lambda}\\
&\le c\_{h,\lambda}|\Pi\_n|^{\lambda-1}\sum\_{i=0}^{k\_n}|r\_{i+1}^n-r\_i^n|\\
&= c\_{h,\lambda}|\Pi\_n|^{\lambda-1}|t-s|\stackrel{n\to+\infty}{\longrightarrow}0.
\end{align\*}
The proof that even $\left|M\_{ts}^{\mathfrak U\_n}-M\_{ts}^{\mathfrak S\_n}\right|$ goes to 0 is the same because of the simmetry of $\Pi\_n$ and $\mathfrak S\_n$ with respect to $\mathfrak U\_n$.


