## Function of Bounded Variation that is not continuous

This is a basic example of a function of bounded variation on [0,1] but not continuous on [0,1].

The key Theorem regarding functions of bounded variation is Jordan’s Theorem: A function is of bounded variation on the closed bounded interval [a,b] iff it is the difference of two increasing functions on [a,b].

Consider $g(x)=\begin{cases}0&\text{if}\ 0\leq x<1\\ 1&\text{if}\ x=1 \end{cases}$

$h(x)\equiv 0$

Both $g$ and $h$ are increasing functions on [0,1]. Thus by Jordan’s Theorem, $f(x)=g(x)-h(x)=g(x)$ is a function of bounded variation, but it is certainly not continuous on [0,1]!