TITLE: Bounded variation functions have jump-type discontinuities
QUESTION [5 upvotes]: I read on the Wikipedia page that bounded variation (BV) functions have only jump-type discontinuities. Why is that? Suppose at some $a\in\mathbb{R}$, the limit $\lim_{x\rightarrow a^+}f(x)$ doesn't exist (or is infinite). Why would such a function $f$ not be BV?

REPLY [8 votes]: If $f$ is unbounded near $a$, it can clearly not have bounded variation, since
$$V(f,[a,b]) \geqslant \lvert f(x)-f(a)\rvert$$
for all $x \in (a,b]$.
If $-\infty < m = \liminf\limits_{x \searrow a} f(x) < \limsup\limits_{x\searrow a} f(x) = M < +\infty$, let $\varepsilon = (M-m)/3$. Then we can find sequences $x_n \searrow a$ and $y_n \searrow a$ with $x_n > y_n > x_{n+1}$ and $f(y_n) < m+\varepsilon$, $f(x_n) > M-\varepsilon$ for all $n$, so
$$V(f,[a,b]) \geqslant \sum_{n=1}^N \lvert f(x_n) - f(y_n)\rvert \geqslant N\cdot \varepsilon.$$
Letting $N\to \infty$, we see that then $V(f,[a,b]) = +\infty$.