Existence of a certain increasing function

[gamma5772]

I'm wondering if there is a monotonically increasing function with a jump discontinuity at every rational (or any other dense, countable subset of the reals). Here's a specific candidate that I've come up with:

Let $$g:\mathbb{Q} \cap [0,1] \rightarrow \mathbb{R}$$ take the rational p/q (p and q coprime) to exp(-q) (or 0 if p = 0). Let $$f:[0,1] \rightarrow \mathbb{R}$$. $$f(x) = \sum_{q \in \mathbb{Q} \cap [0,x]} g(q)$$

It is monotonically increasing and bounded, and I'm pretty sure it's well defined, but I'd just like to be sure. I also believe it is continuous at every irrational and discontinuous at every rational (which can be shown using a simple delta-epsilon proof).

 

[mathman]

I am having difficulty following your derivation.

Here is a simpler (I think) example.

For each rational (p) in the interval [0,1] let f_p(x) = 0 for x < p and f_p(x) = 1 for x ≥ p. Next arrange the rationals in some countable order p1, p2, ...
Then G(x) = ∑2^-kf_pk(x) has the desired property.

 

[AlephZero]

I didn't undestand why you chose exp(-q) in your OP, when you don't seem to be using any properties of e. I think 2^-q would be just as good.

It is monotonically increasing

obviously

and bounded

because if you sum the discontiuities including the duplicates when p and q are not coprime, the sum is bounded

I also believe it is discontinuous at every rational

obviously

and continuous at every irrational (which can be shown using a simple delta-epsilon proof).

Well, I wouldn't call it a "simple" proof, but I think it this is valid:
Let $f(1) = S$. Choose $\epsilon$.
Then there are only a finite number of rationals $q_i$ whose discontinuities sum to $S-\epsilon/2$.
Therefore there is a finite length interval around any irrational point $x$ which does not contain any of the $q_i$.
So the function can not vary by more than $\epsilon/2$ in this interval
So you can find a $\delta$ to prove continuity at $x$.