Existence of a point of continuity to the right of any point on a right-continuous function Let $f: \mathbb{R}\to\mathbb{R}$ be a right-continuous function. How may I show that for any $\epsilon>0$, any $x\in\mathbb{R}$, there exists $y$, $x<y<x+\epsilon$, such that $f$ is continuous at $y$?
This claim is used in the proof of Exercise 6.3 in this document (http://www.ma.utexas.edu/users/sirbu/probability-I/solutions-hw6.pdf).
Thank you very much!
 A: Without much extra effort, you can get someting stronger:
if the points where $f: \mathbb{R}\to\mathbb{R}$ has a one-sided limit
are dense, then the points where $f$ is continuous are too.
To prove this, assume that in every open interval there is an $x$ where $f$
has a one-sided limit. Take any $n \in \mathbb{Z}^+$ and any interval $(a, b)$.
There is an $x \in (a,b)$ where $f$ has a one-sided limit $y$, so there is an
interval 
 $I = (x -\delta, x)$ or $I = (x, x + \delta)$ such that
$f(I) \subset (y - \frac{1}{3n}, y + \frac{1}{3n})$. Then $I \cap (a, b)$ is a subinterval
of $(a, b)$ where the oscillation of $f$ is less than $\frac{1}{n}$.
Since this is true for arbitrary $(a, b)$, it shows that
$F_n = \{ x \mid \omega_f(x) \ge \frac{1}{n} \}$ is nowhere dense.
We can conclude that the discontinuity set of $f$, which is 
$\bigcup_{n=1}^{\infty} F_n$, is meagre, and by the Baire category theorem
its complement is dense.

Edit: Here $\omega_f(x)$ is the oscillation of $f$ at $x$, i. e. $$\omega_f(x) = \inf_{\delta>0} \sup_{x - \delta < y < z < x+ \delta} |f(y) - f(z)|$$
