# Bounded variation function is countinuous except at countably many points

A function $f:\mathbb{R}\rightarrow\mathbb{R}$ is a BV function if there exists $M<\infty$ for which $$\sum_{k=1}^N|f(x_k)-f(x_{k-1})|\leq M$$ for every sequence $x_0<x_1<\ldots<x_N$ and every $N$.

I want to show that a BV function $f$ is continuous except at countably many points, and to identify a left-continuous function $g$ that agrees with $f$ at all but countably many points.

(Note: Left-continuous means that $\lim_{x\rightarrow a^-}f(x)=f(a)$.)

Suppose $f$ is discontinuous at uncountably many points. How can I choose $x_0<x_1<\ldots<x_N$ such that $\sum_{k=1}^N|f(x_k)-f(x_{k-1})|$ is larger with each choice of $x_0,\ldots,x_N$.

• There are two main ideas. 1) a monotone function has a countable number of discontinuities in any finite interval. 2) A BV function can be expressed as sum of two monotone functions. Both these ideas and various properties of BV functions are presented in many real-analysis texts and I have an exposition of the same in my blog post paramanands.blogspot.com/2012/07/… – Paramanand Singh Nov 28 '13 at 7:52

For every function $f: [a,b] \longrightarrow \mathbb{R}$ of bounded variation there exists a monotone increasing functions $g,h :[a,b] \longrightarrow \mathbb{R}$ such that $f=g-h.$
From the above theorem it follows that the restriction $f|_{[a,b]}$ of function $f$ to any interval $[a,b]$ has at most a countably set of discontinuities. And since $\mathbb{R} =\bigcup_{n=1}^{\infty } [-n,n] ,$ hence $f$ has at most a countably set of discontinuities.
Note that it is a property of functions of bounded variation that these functions have only jump discontinuities, i.e., both right- and left- limits exist at each point, but (when there is a discontinuity) , the limits are different. If you have uncountably-many points of discontinuity, these will be jump discontinuities, in which the right- and left- limits exist. Then the sum $f(x_i^{+})-f(x_i^{-})$; $i$ in an uncountable index set $I$ will necessarily diverge, .i.e., the variation will not be bounded.
• But you can only take a countable set $I$ of indices – JJ Beck Nov 28 '13 at 15:53