If $\phi$ is finite and convex in $(a, b)$, then $\phi$ is continuous in $(a, b)$. Moreover, $\phi'$ exists except at most in a countable set and is monotone increasing.

Theorem 2.8 Every function of bounded variation has at most a countable number of discontinuities, and they are all of the first kind (jump or removable discontinuities).

  • How does $(7.41)$ show in particular that $\phi$ is continuous in $(a,b)$? Don't we need to show $D^-\phi(x) = D^+\phi(x)$ as $h\to 0+$? The book shows $D^-\phi(x) = D^+\phi(x)$ at the end of the proof.

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  • $\begingroup$ @mfl This is Theorem 7.40 on page 157 of Measure and integral by Wheeden and Zygmund. The theorem on page 157 doesn't state that $\phi$ is finite, but since they say $D^+\phi$ is of bounded variation I looked up the beginning of this section on page 154, and they suppose $\phi$ is finite on an open interval $(a,b)$. Also see the comment by Michael Grant on this post math.stackexchange.com/q/258511/426645. $\endgroup$
    – user398843
    Aug 10, 2018 at 22:32
  • $\begingroup$ @mfl The interval here is open, the one in the counterexample is compact, and the compactness plays a role there. $\endgroup$
    – amsmath
    Aug 10, 2018 at 22:37

1 Answer 1


The existence of the limit $D^+\phi(x)$ tells us in particular that $\phi(x+h)\to\phi(x)$ as $h\to 0+$. Similarly, the existence of $D^-\phi(x)$ implies $\phi(x-h)\to\phi(x)$ as $h\to 0+$. These together yield that $\phi$ is continuous at $x$.


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