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Let $(a_n)_{n\geq 1}$ be a sequence with $a_n \in [0,1]$ for all $n \geq 1$.

Show that $$f\colon[0,1]\to [0,1], \quad f(x) = \sum_{\{k : a_k\leq x \}}2^{-k-1}$$

is a monotone function and that the set of its discontinuities is equal to $(a_n)_{n\geq 1}$.


I tried around with some sequences and think that the function starts at 0, is monotonically increasing and is constant until x is equal to the next highest element of the sequence and then it jumps to a higher value. But I don't know where to start to prove it, could someone provide me a hint?

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Suppose $x,y \in [0,1]$ with $x \leq y$. Then we clearly have $\{k : a_k\leq x\} \subseteq \{k : a_k \leq y\}$. This means $\sum_{\{k : a_k\leq x \}}2^{-k-1}\leq \sum_{\{k : a_k\leq y \}}2^{-k-1}$, and so $f(x) \leq f(y)$.


As for the set of discontinuities your intuition seems on the right track (except for the "starts at $0$" bit) but unless we change how either the sequence or function are defined, we can merely say that the set of discontinuities of $f$ is contained in the range of the sequence $\{a_n\}_{n=1}^\infty$. Why? Consider the problem as if $\{a_n\}_{n=1}^\infty$ has no nonzero term or even just a finite number of nonzero terms.

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    $\begingroup$ Thank you! So it is not equal to the sequence in those cases, because it has no jump discontinuity at 0? $\endgroup$ – strelsol Jan 22 at 12:23
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    $\begingroup$ for instance if the sequence $\{a_n\}_{n=1}^\infty$ has no nonzero term, then $f \equiv \frac12$. $\endgroup$ – Matt A Pelto Jan 22 at 19:12

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