Are the following implications true:
(I) if $g$ is continuous on $[a,b]$, and if $g$ satisfies the Banach (S) property on $[a,b]$, then $g$ has bounded variation on $[a,b]$? Here we assume that $a,b\in {\bf R}$ and $a<b$ (i.e., $[a,b]$ is a compact interval such that $0<\lambda([a,b])<+\infty$).
(II) if $g\in {\rm C}({\bf R})\cap {\rm L}^1({\bf R})\cap {\rm L}^{\infty}({\bf R})$, and if $g$ satisfies the Banach (S) property on ${\bf R}$, then $g$ has bounded variation on ${\bf R}$?
(III) if $g\in {\rm C}({\bf R})\cap {\rm L}^1({\bf R})\cap {\rm L}^{\infty}({\bf R})$, if $g$ is uniformly continuous on ${\bf R}$, and if $g$ satisfies the Banach (S) property on ${\bf R}$, then $g$ has bounded variation on ${\bf R}$?
(IV) if $g\in {\rm C}({\bf R})\cap {\rm L}^1({\bf R})$, and if $g$ satisfies the Banach (S) property on ${\bf R}$, and if $\lim_{|s|\rightarrow+\infty}|g(s)|=0$, then $g$ has bounded variation on ${\bf R}$?
(V) if $g\in {\rm C}{(\bf R})\cap {\rm L}^1({\bf R})$, if $g$ is uniformly continuous on ${\bf R}$, and if $g$ satisfies the Banach (S) property on ${\bf R}$, and if $\lim_{|s|\rightarrow+\infty}|g(s)|=0$, then $g$ has bounded variation on ${\bf R}$?
Reminder. We say that a measurable function $g:\Omega\rightarrow {\bf R}$, where $\Omega\subseteq {\bf R}$ is a measurable set, satisfies the Banach (S) property on $\Omega$ if the following holds: for every $\varepsilon>0$ there exists $\delta>0$ such that for every measurable set $E\subseteq\Omega$ it holds that $\lambda(E)\leq\delta$ implies $\lambda(g(E))\leq\varepsilon$.
${\bf DISCUSSION.}$
Remark(0). We note that, if $g\in {\rm C}([a,b])$, then $g\in {\rm L}^{1}(a,b)$ and $g\in {\rm L}^{\infty}(a,b)$ (more precisely: ${\rm sup}_{s\in [a,b]}|g(s)|<+\infty$, so $g$ is not merely essentially bounded, it is actually bounded). This line of reasoning fails if we consider the domain ${\bf R}$. Hence, I added assumptions $g\in {\rm L}^{1}({\bf R})$ and $g\in {\rm L}^{\infty}({\bf R})$. Note that (II) implies (III), (II) implies (IV), (III) implies (V), and (IV) implies (V). So, in (V), we have the strongest assumptions on $g$.
Remark(1). Regarding (I): It is known that, if $g$ is continuous on $[a,b]$ and if $g$ satisfies the Luzin (N) property on $[a,b]$, then, in general, $g$ does not satisfy the Banach (S) property on $[a,b]$, see Continuous function which satisfies the Luzin N property, but which does not satisfy the Banach S property
Remark(2). Regarding (I): By the Banach-Zaretsky theorem (see Theorem 3.41, page 87 or Corollary 3.49, page 89, in textbook G. Leoni, A first course in Sobolev Spaces. 2nd ed., Graduate Studies in Mathematics 105, American Mathematical Society, Providence RI, 2017), we known that, if $g\in {\rm C}([a,b])$ and if $g$ satisfies the Luzin (N) property, then it does not follow that $g\in {\rm BV}([a,b])$. This is because, if the opposite was true, this would mean that every $g\in {\rm C}([a,b])$ which satisfies the Luzin (N) property, also satisfies $g\in {\rm AC}([a,b])$ (i.e., $g$ is absolutely continuous on $[a,b]$). So, in particular, it would follows that $g$ satisfies the Banach (S) property (see the proof of assertion (v) in question Extension of Luzin N-property of absolutely continuous functions and measure preserving mappings ), which, by Remark1, is not true.
But, in our context, we are assuming a little bit more than the Luzin (N) property (since the Banach (S) property implies the Luzin (N) property).
Remark(3). Regarding (I): By the Banach-Zaretsky theorem, if the implication (I) is true, then it follows that, in fact, we have $g\in {\rm AC}([a,b])$.
Remark(4). Here we consider the definition: we say that a function $g$ is of bounded variation on $[a,b]$ (on ${\bf R}$, resp.) (which we write as $g\in {\rm BV}([a,b])$ ($g\in {\rm BV}({\rm R})$, resp.) if its total pointwise variation is finite $$ \mathrm{Var}(g) := \sup\left\{\sum_{i=1}^{n}|g(x_i)-g(x_{i-1})|\right\}, $$ where the supremum is taken over all finite partitions of $[a,b]$ ($\mathbf{R}$, resp.) (I included the definition just to be precise; although I think that, under the assumption that $g$ is continuous, it does not matter much, we could probably use essential variation as well, but this is beyond the point of the question). Note that, if we use the definition above, there exist functions $g\in {\rm BV}({\bf R})$ which do not belong to ${\rm L}^{1}({\bf R})$ (for example, constant functions). For this reason, I added assumption $g\in {\rm L}^{1}({\bf R})$.
Remark(5). Here are somewhat related questions:
Example of a function that has the Luzin $n$-property and is not absolutely continuous.
What are examples of uniformly continuous functions with unbounded variation?
Bounded variation on the entire real line
Remark(6). Open for re-checking: Regarding (II), I observe that, if $g\in {\rm BV}({\bf R})\cap {\rm L}^{1}({\bf R})$, then it follows that $g\in {\rm L}^{\infty}({\bf R})$ (since we know that $g$ is bounded on $[-R,R]$ for arbitrarily large $R>0$, see
Does bounded variation imply boundedness
$f$ is Of Bounded Variation $\Rightarrow$ $f$ is Bounded
, and, under assumption $g\in {\rm L}^{1}({\bf R})$, we also know that $\lim_{R\rightarrow+\infty}||g||_{{\rm L}^{\infty}(R,\infty)}=0$, see
Do BV functions vanish at infinity?
, see also Theorem 3.27, (c), page 103, in textbook G. B. Folland, Real Analysis, John Wiley and Sons, Inc., 1984, which establishes the existence of limits of any given $BV({\bf R})$-function at $-\infty$ and $+\infty$). I actually think that (and this is also a conclusion open for re-checking) that we have a bit more then $\lim_{R\rightarrow+\infty}||g||_{{\rm L}^{\infty}(R,\infty)}=0$, we actually have $\lim_{R\rightarrow+\infty}||g||_{{\rm L}^{\infty}((-\infty,-R)\cup(R,\infty))}=0$, which, in particular, by continuity of $g$, implies $\lim_{|s|\rightarrow+\infty}g(s)=0$.
Remark(7). As a consequence, a necessary condition for the assertion (II) to be true, is that the following implication is true:
(VI) if $g\in {\rm C}({\bf R})\cap {\rm L}^{1}({\bf R})\cap {\rm L}^{\infty}({\bf R})$ and if $g$ satisfies the Banach (S) property on ${\bf R}$, then it follows that $\lim_{R\rightarrow+\infty}||g||_{{\rm L}^{\infty}((-\infty,-R)\cup(R,\infty))}=0$.
But I do not know if (VI) is true, or if it is useful. Hence, I formulated versions (IV) and (V) of the question, whereby condition $g\in {\rm L}^{\infty}({\bf R})$ is replaced by the condition $\lim_{|s|\rightarrow+\infty}|g(s)|=0$ (which, since $g\in {\rm C}({\bf R})$, implies $g\in {\rm L}^{\infty}({\bf R})$).
Remark(8). Regarding Remark(7). In a quite similar way we can formulate a further necessary condition which links assertions (III), (IV) and (V), with the condition $\lim_{R\rightarrow+\infty}||g||_{{\rm L}^{\infty}((-\infty,-R)\cup(R,\infty))}=0$.
Remark(9). Regarding Remark(7) and Remark(8). One peripheral question here is: can we replace the condition $\lim_{R\rightarrow+\infty}||g||_{{\rm L}^{\infty}((-\infty,-R)\cup(R,\infty))}=0$ with similar condition of the pointwise type, say $\lim_{|s|\rightarrow+\infty}|g(s)|=0$. Again, for $g\in {\rm C}({\bf R})\cap {\rm L}^{1}({\bf R})\cap{\rm BV}({\bf R})$, these two conditions are probably equivalent.
Remark(10). I noticed that, by the Barbalat lemma (see Theorem 1. in https://arxiv.org/pdf/1411.1611 ), $g\in {\rm L}^{1}({\bf R})$ and uniform continuity of $g$ on ${\bf R}$ already imply that we have $\lim_{|s|\rightarrow+\infty}g(s)=0$. So, I guess, assumptions in (III) are equivalent to assumptions in (V).
