 Rota---Baxter operator of nonzero weight $\lambda $ then
$\tau' = \frac{1}{\lambda }\tau $ is a Rota---Baxter operator of weight 1. Hence,
there are two essentially different cases: $\lambda = 0$ (zero weight) and $\lambda =1$ (unit weight).

The following statement was proved in \cite{GubKol2013} in the case of binary operations ($\nu(f)=2$).
By means of the approach presented in this paper, the proof becomes clear in the general case.

Given a  $\Sigma $-algebra $A$ equipped with
 a Rota---Baxter operator $\tau $, denote by $A^{(\tau )}$
the $\Sigma^{(3)}$-algebra defined on the space $A$ by
\[
 f^H(a_1,\dots, a_n) = f(a_1^H,\dots, a_n^H),
\]
where
$f\in \Sigma$, $i=1,\dots, n$, $a_1,\dots, a_n \in A$,
$a_i^H$ are given by \eqref{eq:a^H_iDefn}.

The same relations restricted to $|H|=1$ define a $\Sigma^{(2)}$-algebra structure on $A$
also denoted by $A^{(\tau)}$.

\begin{theorem}\label{thm:RB-operators}
 \begin{enumerate}
\item If $A\in \mathfrak M$ and $\tau $ is a Rota---Baxter operator of zero weight on $A$
then $A^{(\tau )}$ is a $\text{\rm pre-}\mathfrak M$-algebra.

\item If $A\in \mathfrak M$ and $\tau $ is a Rota---Baxter operator of unit weight on $A$
then $A^{(\tau )}$ is a $\text{\rm post-}\mathfrak M$-algebra.

\item Every $D\in \text{\rm pre-}\mathfrak M$ may be embedded into $A^{(\tau)}$
for an appropriate $A\in \mathfrak M$ equipped with Rota---Baxter operator $\tau $ of zero weight.

\item Every $T\in \text{\rm post-}\mathfrak M$ may be embedded into $A^{(\tau)}$
for an appropriate $A\in \mathfrak M$ equipped with Rota---Baxter operator $\tau $ of unit weight.
\end{enumerate}
\end{theorem}

\begin{proof}
 As in Theorem \ref{thm:AveragingOper}, let us prove (2) and (4).

For (2), it is enough to consider $C\boxtimes A^{(\tau )}$ for any $C\in \mathrm{ComTrias}$, and note
(by induction on $m\ge 1$) that
\[
 \Phi(y_1\otimes a_1,\dots, y_m\otimes a_m)
 = \sum\limits_{H\in \mathcal P(m)} e^{(m)}(y_1,\dots, y_m)\otimes \Phi (a_1^H,\dots, a_m^H)
\]
for every $\Phi \in \mathfrak F(m)$.

Hence, $C\boxtimes A^{(\tau )}\in \mathfrak M$.

To prove (4), consider $A=C_2\boxtimes T \in \mathfrak M$, where $C_2$ is the algebra from Example \ref{exmp:C_2},
 and define
\begin{equation}\label{eq:RB_Hat}
\tau (e_1\otimes a) = -e_1\otimes a, \quad \tau(e_2\otimes a)= e_1\otimes a, \quad a\in T.
\end{equation}
Let us show that \eqref{eq:RB_Hat} is a Rota---Baxter operator of unit weight
on $C_2\boxtimes T$. Indeed, suppose $f\in \Sigma $, $\nu(f)=n$,
$u_i=e_{k_i}\otimes a_i$, $k_i\in \{1,2\}$, $a_i\in T$, $i=1,\dots, n$.
Evaluate the left-hand side of \eqref{eq:Rota-BaxterGeneral}:
\[
 f(\tau(u_1), \dots, \tau(u_n)) = (-1)^{|K|}f(e_1\otimes a_1,\dots, e_1\otimes a_n),
\]
where $K=\{i\mid k_i=1\}$.
On the other hand,
\begin{multline}\nonumber
 f(u_1^H, \dots, u_n^H) =
 (-1)^{ |K\setminus H| } f(e_{k_1'}\otimes a_1,\dots, e_{k_n'} \otimes a_n) \\
= (-1)^{ |K\setminus H| } \sum\limits_{M\in \mathcal P(n)}
 e_M^{(n)} (e_{k_1'},\dots, e_{k_n'})\otimes f^M(a_1,\dots, a_n),
\end{multline}
where $k'_i=\begin{cases}
             k_i, & i\in H, \\
             1, & i\notin H
            \end{cases}$.
Nonzero summands appear in two cases: (1)  $k'_i=1$ for all $i=1,\dots, n$; (2) $k'_i=2$ if and only if $i\in M$.
The first case occurs if and only if $H\subseteq K$, the second one corresponds to $M=H\setminus K$.
Hence,
\[
 f(u_1^H, \dots, u_n^H)= \begin{cases}
 (-1)^{|K|-|H|} e_1 \otimes \sum\limits_{M\in \mathcal P(n)} f^M(a_1,\dots, a_n), & H\subseteq K, \\
 (-1)^{|K\setminus H|} e_2\otimes f^{H\setminus K}(a_1,\dots, a_n), & H\not\subseteq K.
\end{cases}
\]
Let us evaluate the right-hand side of \eqref{eq:Rota-BaxterGeneral}:
\begin{multline}\label{eq:RB_relation_check}
 \sum\limits_{H\in \mathcal P(n)}\tau (f(u_1^H, \dots, u_n^H)) 
=
\sum\limits_{\emptyset\ne H\subseteq K}(-1)^{|K|-|H|+1} f(e_1\otimes a_1,\dots, e_1\otimes a_n) \\
+
\sum\limits_{H\not\subseteq K} (-1)^{|K\setminus H|} e_1\otimes f^{H\setminus K}(a_1,\dots, a_n).
\end{multline}
The first summand in the right-hand side of \eqref{eq:RB_relation_check}
is equal to $(-1)^{|K|}f(e_1\otimes a_1,\dots, e_1\otimes a_n)$
since
\[
 \sum\limits_{H\subseteq K}(-1)^{|H|} = 1 + \sum\limits_{\emptyset\ne H\subseteq K}(-1)^{|H|} = 0.
\]
In the second summand, present $H\not\subseteq K$ as
$H=U\cup M$, $U\subseteq K$, $M\ne \emptyset$, $M\cap K=\emptyset$. Then
\[
\sum\limits_{U\subseteq K}\sum\limits_{\begin{subarray}{c} M\ne \emptyset \\ M\cap K=\emptyset\end{subarray}}
 (-1)^{|K|-|U|} e_1\otimes f^{M}(a_1,\dots, a_n) = 0
\]
by the same reasons.

We have proved that \eqref{eq:Rota-BaxterGeneral} holds for $\tau $ ($\lambda =1$), and thus it is a Rota---Baxter operator of unit weight.
\end{proof}

\begin{remark}
Theorem \ref{thm:RB-operators} implies that Definition \ref{defn:postAlg_via_ComTrias}
provides an equivalent description of the same class of systems as
the splitting procedure described in
in \cite{BaiGuoNi2012}: $\text{\rm pre-}\mathfrak M = \EuScript{A}\mathrm{Sp}(\mathfrak M)$,
$\text{\rm post-}\mathfrak M = \EuScript{B}\mathrm{Sp}(\mathfrak M)$.

In the binary case, $\text{\rm pre-}\mathfrak M$ and $\text{\rm post-}\mathfrak M$ coincide