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]!

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Author: mathtuition88

Math and Education Blog

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