# Bounded Variation $+$ Intermediate Value Theorem implies Continuous

Let $I=[a,b]$ with $a<b$ and let $u:I\rightarrow\mathbb{R}$ be a function with bounded pointwise variation, i.e. $$Var_I u=\sup\{\sum_{i=1}^n|u(x_i)-u(x_{i-1})|\}<\infty$$

where the supremum is taken over all partition $P=\{a=x_0<x_1<...<x_{n-1}<b=x_n\}$. How can I prove that if $u$ satisfies the intermediate value theorem (IVT), then $u$ is continuous?

My try: $u$ can be written as a difference of two increasing functions $f_1,f_2$. I know that a increasing function that satisfies the (ITV) is continuous, hence, if I prove that $f_1,f_2$ satisfies the (ITV) the assertion follows. But, is this true? I mean, $f_1,f_2$ satisfies (ITV)?

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Begin by proving that every function of bounded variation has finite one-sided limits at every point. (Decomposition into monotone functions does this in one line.) Then observe that the intermediate value property fails unless $f(a+)=f(a-)=f(a)$.
Thank you @5pm. Now I am just curious, do you know if $f_1$ and $f_2$ satisfies (ITV)? – Tomás Feb 22 '13 at 20:35
@Tomás Not if you allow arbitrary increasing functions: for example $0=f_1-f_1$ for any increasing function $f_1$. But for the canonical ("least increasing") $f_1$ and $f_2$ this is true. Recall that these $f_1$ and $f_2$ are indefinite integrals of the positive and negative parts of $u'$, understood as a signed measure. So the argument is: since $u$ is continuous, $u'$ has no atoms, hence its positive and negative parts have no atoms either. An antiderivative of an atomless measure is continuous. – user53153 Feb 22 '13 at 20:45