Continuity and Bounded Variation for absolute function Assume that $f$ is a continuous function on $[a,b]$ and $\vert f \vert$ has a bounded variation on the given domain. We have to prove that $f$  has bounded variation on $[a, b]$.
The main motivation here is to implement the intermediate value theorem find a partition such that for all indices $i$, either $f(x_{i+1}), f(x_i)\ge 0$ or $f(x_{i+1}), f(x_i) \le 0$ holds true. Surely, such construction of partition will ensure the bounded variation of function $f$ over the sums of $\vert f(x_{i+1})-f(x_i)\vert$. However, what is not clear is how such "forced" partition could be ever constructed via IVT, and doesn't the definition of bounded variation touch all the possible partitions on $[a, b]$, not only specific ones we make up to?!
 A: Bounded variation is defined in terms of a supremum over all partitions. But a partition that is a refinement of another partition can never have smaller variation than the unrefined partition. Hence, switching to a refined partition poses no problem.

That is, given a function $f$ defined on an interval $I$ and a non-empty finite subset $P\subset I$, we define
$$ V_P(f):=\sum_{i=1}^{n-1}|f(x_{i++1})-f(x_i)|$$
where $P=\{x_1,\ldots, x_n\}$ and $x_1<\ldots< x_n$. Next we define
$$ V_I(f):=\sup_{P\in \mathcal P}V_P(f)$$
where $\mathcal P$ is the set of all non-empty finite subsets of $I$,
and say tha t$f$ is of bounded variation on $I$ if $V_I(f)<\infty$.
Now assume $f$ is continuous and $|f|$ of bounded variation on $I$. I claim that
$$ V_I(f)\le V_I(|f|).$$
(In fact, "$\ge$" is clear so that we can conclude "$=$").
So let $P=\{x_1,\ldots,x_n\}\in\mathcal P$ with $x_1<\ldots< x_n$. For every $1\le i<n$ with $f(x_i)f(x_{i+1})<0$, use the IVT to pick $y_i\in(x_i,x_{i+1})$ with $f(y_i)=0$. Let $P'$ be the union of $P$ and all those $y_i$. Note that $P'$ i finte (has $n'<2n$ elements).
Then
$$\tag1 V_{P}(f)\le V_{P'}(|f|).$$
Indeed, if there is such an $y_i$ between $x_i$ and $x_{i+1}$, then we have
$$\tag2 |f(x_i)-f(x_{i+1})|=|f(x_i)|+|f(x_{i+1}|=\bigl||f(x_i)|-|f(y_i)|\bigr|+\bigl||f(y_i)|-|f(x_{i+1})|\bigr|$$
and if there is no such $y_i$, i.e., $f(x_i)$ and $f(x_{i+1})$ have same sign, then
$$\tag3 |f(x_i)-f(x_{i+1})|=\bigl||f(x_i)|-|f(x_{i+1}|\bigr|.$$
Summing all instances of $(2)$ and $(3)$, we immediately arrive at $(1)$.
Hence
$$ V_P(f)\le V_I(|f|)$$
holds for all $P$, hence the claim.
