I read the definition of total variation of a measure in the following link
https://www.encyclopediaofmath.org/index.php/Signed_measure
And then I have a question:
If $\mu$ is a vector measure in $C^*([0,T];\mathbb{R}^{m})$ which is a topological dual of the space of continuous functions. Let $\mu(t)=\mu^i$ for $t\in[t^i,t^{i+1})$, $i=0,\ldots,n-1$, where $t^0=0$, $t^n=T$. Each partition $[t^i,t^{i+1}]$ has a length of $h_n$ which tends to zero as $n\to\infty$.
Then we have the total variation of $\mu$ $$|\mu|([0,T])=\sup\left\{ \sum_{i=0}^{\infty}\|\mu([t^i,t^{i+1}])\|: \{[t^i,t^{i+1}]\} \textrm{ is a countable partition of } [0,T] \right\}.$$
Can we obtain the following?
\begin{equation}\label{1} |\mu|([0,T])=\sup\left\{ \sum_{i=0}^{\infty}\|\mu^i\|\right\}. \end{equation} So the uniform boundedness of $\mu$ implies $\sum_{i=0}^{\infty}\|\mu^i\|$ is uniformly bounded with respect to (w.r.t.) $n$? Is that correct?
Thank you so much for your time and help.