Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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)?

share|cite|improve this question
up vote 3 down vote accepted

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 $f(a+)=f(a-)=f(a)$.

share|cite|improve this answer
@JonasMeyer Thanks, fixed. – user53153 Feb 22 '13 at 20:32
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.