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Suppose $f$ is increasing and there exist numbers $A$ and $B$ such that for all $x$, $A \leq f(x) \leq B$. Show that for each $\epsilon >0$ the number of jumps of size exceeding $\epsilon$ is at most $(B-A)/\epsilon$.

Let $\epsilon >0$. So the jump size is $f(x+)-f(x-)$. So it seems that we divide the maximum possible height $B-A$ by $\epsilon$ to get the maximum number of jumps exceeding $\epsilon$ (e.g. dividing a large number by small number). Is this correct?

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Since $f$ is increasing, you can't "undo" a jump upwards. Therefore $n$ jumps of size at least $\epsilon$ contribute an increase of at least $n\epsilon$, and this has to be less than the total increase $B-A$.

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This question is closely related to this one; in particular, if $x_k \in (a,b)$, $k=1,\ldots,n$, are such that $f(x_k+)-f(x_k-) > \varepsilon$, we have from the latter that $ n \varepsilon < \sum\limits_{k = 1}^n {[f(x_k + ) - f(x_k - )]} \le f(b - ) - f(a + ) \le B - A, $ hence $n < (B-A) / \varepsilon$.