My question is how to prove that a function f:[a,b]→R with bounded variation and the intermediate value property (IVP) is continuous.
I have seen Bounded Variation $+$ Intermediate Value Theorem implies Continuous but it is not clear to me.
So I think there must be an easy way using the fact that $f = g- h$ where $g$ and $h$ are monotonic increasing. A monotonic function only has jump discontinuities, so $f$, $g$ and $h$ all have this property. Also a monotonic function with IVP must be continuous. But I can’t be sure $g$ and $h$ have IVP even if $f$ does. Furthermore $f$ has IVP and only jump discontinuities but because it may not be monotonic I can’t immediately say it is continuous.
Answer
Suppose $f$ is discontinuous at $c$.
Since a BV function is the difference of increasing functions, the one-sided limits $f(c+)$ and $f(c-)$ exist. WLOG we have $\Delta = f(c+) - f(c-) > 0$ and
$$f(c+) - \frac{\Delta}{3} = f(c-) + \frac{2\Delta}{3} > f(c-) + \frac{\Delta}{3}.$$
But there exists $\delta > 0$ such that $f(x) < f(c-)+ \frac{\Delta}{3}$ when $c-\delta < x < c$ and $f(x) > f(c+) - \frac{\Delta}{3}$ when $c < x < c+\delta$.
Consequently, if $f(c-) + \frac{\Delta}{3} < K < f(c+)- \frac{\Delta}{3}$ and $K \neq f(c)$, then there exists no $x$ such that $f(x) = K$, contradicting the intermediate value property.
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