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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 $N$. Any BV function has only jump discontinuities. That is, at any point $a$, the limit $\lim_{x\rightarrow a^-}f(x)$ and $\lim_{x\rightarrow a^+}f(x)$ exist, but they may or may not be equal.

I want to use this to show that a BV function $f$ is continuous except at countably many points.

Suppose the function is discontinuous at uncountably many points. How can I choose $x_0,x_1,\ldots,x_N$ to contradict the definition of BV?

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JJ Beck
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Any BV function can be written as a sum of two monotone functinos, so there is no loss in assuming that $f$ is increasing. Suppose there were more than countably many discontinuities, at points $x_{\alpha}$; each of these are jump discontinuities. Choose a rational $r_{\alpha}$ such that

$$f(x_{\alpha}-) < r_{\alpha} < f(x_{\alpha}+)$$

Do you see why this leads to a contradiction?

  • "Any BV function can be written as a sum of two monotone functinos" I've never seen that fact before, and can't find it in Wikipedia either. How can we prove it? – JJ Beck Nov 28 '13 at 22:00
  • @JJBeck See, e.g. here. –  Nov 28 '13 at 22:02