Monotone function with predescribed discontinuities

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The discussion focuses on the properties of the function f defined using the unit jump function I(x) and a countable subset of real numbers. It establishes that as x approaches negative infinity, f(x) converges to 0, and as x approaches positive infinity, f(x) converges to 1. The proof is built on the monotonicity of f and the boundedness of the series defined by the positive weights c_n. It highlights that within a bounded interval (a,b), the function is monotone increasing and converges to its limits at the endpoints. The discussion emphasizes the need for a complete proof when extending the results beyond the bounded interval.
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Let I(x) (the unit jump function) be defined piecewise by I(x)=0 iff x<0 and I(x)=1 iff x>=0.

Let {x_n} be a countable subset of R and {c_n} a sequence of positive real numbers satisfying \sum{c_n}=1. Let f:R \to R be defined by f(x)=\sum{c_n * I(x - x_n)}.

Prove that \lim_{x\to-\infty}{f(x)}=0 and \lim_{x\to\infty}{f(x)}=1.

I can prove this when {x_n} is a countable subset of (a,b) where a,b \in R. It seems as though the proof here should be very similar, but for some reason I just can't finish it..
 
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Posting your proof for x in (a,b) may help the helpers.
 
Since 0 <= c_n * I(x-x_n) <= c_n, for all x \in [a,b], s_n(x) = \sum^n_{k=1}{c_k * I(x-x_k)} <= \sum^n_{k=1}{c_k} <= \sum^\infty_{k=1}c_k.

So for x \in [a,b], {s_n(x)} is monotone increasing and bounded above, thus convergent.

Since I(x-x_n) <= I(y-x_n), for all x<y, f is monotone increasing on [a,b].

Now, since x_n > a for all n, I(a-x_n)=0 for all n. Thus, f(a)=0. Also, I(b-x_n)=1 for all n, so f(b)=\sum{c_k}=1.
 
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