Prove that for any sequence $\{x_n\}$ of bounded variation there exist increasing bounded sequences $\{a_n\}$, $\{b_n\}$ such that $x_n = a_n - b_n$ for $n\in \Bbb N$

This problem appears in the context of theese two problems. Those two posts show some properties related to the bounded variation of the sequences. Now I want to show the last part, namely what's in the problem statement.

I was thinking of the following sketch. Let's suppose $x_n$ may be presented as a difference of the two sequences, namely: $$ x_n = a_n - b_n $$ Then what is left to show is that both $a_n$ and $b_n$ are increasing. Since the variation is bounded then it must follow: $$ \exists \lim_{n\to\infty}\sigma_n = L\\ \exists \lim_{n\to\infty}x_n = x $$

Then: $$ \lim_{n\to\infty}(\sigma_n - \sigma_{n-1}) = 0 \\ \begin{align} \sigma_n - \sigma_{n-1} &= |x_{n+1} - x_n| \\ &= |(a_{n+1} - b_{n+1}) - (a_n - b_n)| \\ &= |(a_{n+1} - a_n) - (b_{n+1} - b_n)| \end{align} $$

It follows that: $$ \lim_{n\to\infty} |(a_{n+1} - a_n) - (b_{n+1} - b_n)| = 0 \implies \\ \lim_{n\to\infty} (a_{n+1} - a_n) = \lim_{n\to\infty}(b_{n+1} - b_n) = C \tag1 $$

But this seems to be a road to nowhere. What would be the way to show $a_n$ and $b_n$ are increasing and bounded? Because according to $(1)$ it looks like $a_n$ and $b_n$ are not even bounded.


You can just define $a_n=x_1^{+}+(x_2-x_1)^{+}+\cdots+(x_n-x_{n-1})^{+}$ and $b_n=x_1^{-}+(x_2-x_1)^{-}+\cdots+(x_n-x_{n-1})^{-}$ where $x^{+}=\max\{x,0\}$ and $x^{-}=-\min \{x,0\}$.


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