Let $f\in\operatorname{BV}(\mathbb R)$ and $$V_f(x):= \displaystyle\sup_{x_1 < x_2 < x_3 < ... < x_n=x} \sum_{j=1}^n |f(x_{j+1})-f(x_j)|$$
be the total variation of $f$ on $]-\infty, x[$.
As a step in a proof left as an exercise in class, I would need that $V_f(-\infty)=0$, whenever $f(-\infty)=0$ and the value in $-\infty$ is naturally extended by limits. By this, we mean that
$$f(-\infty):=\displaystyle\lim_{x\to-\infty} f(x)\\ V_f(-\infty):=\displaystyle\lim_{x\to-\infty} V_f(x)$$
$V_f(x)$ is defined for every $x$ as above.
We can begin a proof attempt with taking an $y$ so that $|f(x)|<\varepsilon$ for all $x<y$, $$V_f(y) = \displaystyle\sup_{x_1 < x_2 < x_3 < ... < x_n=y} \sum_{j=1}^n |f(x_{j+1})-f(x_j)| \leq \sup 2n\varepsilon = \infty,$$ so this doesn't help us.
What can I do instead?
Given $\epsilon>0$, consider the finite $V_{f}(0)$, then there exist some points $x_{1}<x_{2}<\cdots<x_{n}=0$ such that \begin{align*} V_{f}(0)-\epsilon<\sum_{j=1}^{n-1}|f(x_{j+1})-f(x_{j})|. \end{align*} Consider any points $t_{1}<t_{2}<\cdots<t_{N}=x_{1}$, then \begin{align*} \sum_{j=1}^{N-1}|f(t_{j+1})-f(t_{j})|+\sum_{j=1}^{n-1}|f(x_{j+1})-f(x_{j})|\leq V_{f}(0), \end{align*} and hence \begin{align*} \sum_{j=1}^{N-1}|f(t_{j+1})-f(t_{j})|\leq V_{f}(0)-\sum_{j=1}^{n-1}|f(x_{j+1})-f(x_{j})|<\epsilon. \end{align*} The arbitrariness of such $t_{1}<t_{2}<\cdots<t_{N}=x_{1}$ gives $V_{f}(x_{1})\leq\epsilon$. Since $x\leq x_{1}$ implies that $V_{f}(x)\leq V_{f}(x_{1})$, this shows that $\lim_{x\rightarrow-\infty}V_{f}(x)=0$.
It seems that the condition $\lim_{x\rightarrow-\infty}f(x)=0$ is no needed.