Let $$I=I_1\times\cdots\times I_n=[a_1,b_1]\times\cdots\times[a_n,b_n]$$ be a $n-$rectangle. Let $f:\ I\longrightarrow\mathbb R$ be bounded and $|f|\leq M$ on $I$.
Let $P=P_1\times\cdots\times P_n$ be a partition of $I$ where each $P_i$ is a partition of the interval $I_i$, which can be written as $$P_i=\big\{a_i=p_{i0},\,p_{i1},\cdots,p_{i{m_i}}=b_i \big\}.$$ Suppose $P''$ be a partition obtained by adjoining a single point $q $ to one of $P_i$, which means $$P''= P_1\times\cdots\times\big(P_i\cup\{q\}\big)\times\cdots\times P_n. $$ I want to show that $$0\leq L(f,P'')-L(f,P)\leq 2M\|P\|\big(\text{width }I\big)^{n-1}. $$ where $\|P\|$ is the mesh of partition $P$, which is the maximum length of edge of subrectangles correspond to $P$.
My attempt :
First, we observe that since $P''$ is a refinement of $P$, so each subrectangle $S$ corresponds to $P''$ must be contained in some subrectangle $R$ corresponds to $P$. Thus, we have \begin{align} L(f,P'')=\sum_{S\in H(P'')}\inf_Sf|S|=\sum_{R\in H(P)}\sum_{\substack{S\in H(P'')\\ S\subseteq R}}\inf_Sf|S| \end{align} This implies \begin{align} L(f,P'')-L(f,P)&=\sum_{R\in H(P)}\sum_{\substack{S\in H(P'')\\ S\subseteq R}}\inf_Sf|S|-\sum_{R\in H(P)}\inf_Rf|R| \\ &=\sum_{R\in H(P)}\left(\sum_{\substack{S\in H(P'')\\ S\subseteq R}}\inf_Sf|S|-\inf_Rf|R| \right) \\ &=\sum_{R\in H(P)}\left(\sum_{\substack{S\in H(P'')\\ S=R}}\inf_Sf|S|-\inf_Rf|R| \right)+\sum_{R\in H(P)}\left(\sum_{\substack{S\in H(P'')\\ S\subset R,\ S\ne R}}\inf_Sf|S|-\inf_Rf|R| \right) \\ &=\sum_{R\in H(P)}\left(\sum_{\substack{S\in H(P'')\\ S\subset R,\ S\ne R}}\inf_Sf|S|-\inf_Rf|R| \right) \end{align} since those terms in the sum involving $S=R$ will cancel.
Now, since $P''$ is obtained by adjoining a point $q$ in some $P_i$, so without loss of generality, we can assume that $q$ is added to $P_1$. Thus, if $q$ is added to some subinterval $R_1$ of $I_1$ and the subrectangle $R$ is of the form \begin{align} R&=R_1\times R_2\times\cdots\times R_n \\ &=\big[p_{1j_1},p_{1(j_1+1)}\big]\times\big[p_{2j_2},p_{2(j_2+1)}\big]\times\cdots\times\big[p_{nj_n},p_{n(j_n+1)}\big] \end{align} and notice that $R_1$ is devided into at most 2 new intervals, so if $S\subsetneq R$, it will be of the form $$S=S_1\times R_2\times\cdots\times R_n\ \text{ where }\ S_1=\big[p_{1j_1},q\big]\ \text{ or }\ \big[q,p_{1(j_1+1)}\big] $$ Therefore, \begin{align} L(f,P'')-L(f,P)&=\sum_{R\in H(P)}\left[\left(\sum_{\substack{S\in H(P'')\\ S\subsetneq R}}\inf_Sf|S_1|-\inf_Rf|R_1|\right)|R_2|\cdots|R_n|\right] \end{align} Now, we consider this term \begin{align} \sum_{\substack{S\in H(P'')\\ S\subsetneq R}}\inf_Sf|S_1|-\inf_Rf|R_1| &=\inf_{\big[p_{1j_1},q\big]}f\,\big(q-p_{1j_1} \big)+ \inf_{\big[q,p_{1(j_1+1)} \big]}f\,\big(p_{1(j_1+1)}-q \big)-\inf_Rf\,|R_1| \\ &\leq M\big(q-p_{1j_1} \big)+M\big(p_{1(j_1+1)}-q \big)+M|R_1| \\ &=M|R_1|+M|R_1|=2M|R_1| \end{align} Therefore, \begin{align} L(f,P'')-L(f,P)&\leq 2M|R_1||R_2|\cdots|R_n|\leq 2M\|P\|\big(\text{width }I \big)^{n-1}. \end{align}
Is my proof correct ? And if so, how can I show that if $P''$ is obtained by adjoining at most $N$ points to $P''$, then $$L(f,P'')-L(f,P)\leq 2MN\|P\|\big(\text{width }I \big)^{n-1}\ ? $$ Could anyone give me some hints ? Thank you so much.