# Monotone function with predescribed discontinuities

1. Feb 18, 2008

### e_to_the_i_pi

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..

2. Feb 18, 2008

### EnumaElish

Posting your proof for x in (a,b) may help the helpers.

3. Feb 18, 2008

### e_to_the_i_pi

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.