Non-decreasing functions and continuity I have the following situation:


*

*$f\colon\mathbb{R}\to\mathbb{R}$ is a non-decreasing

*$g\colon\mathbb{R}\to\mathbb{R}$ is defined as $\ g(x):=\lim_{t\to x^+}f(t)$


I have proved that also $g$ is non-decreasing and right continuous, but I am not able to show rigorously that $f$ and $g$ have exactly the same points of continuity.
Do you have some ideas?
 A: Just look at one-sided limits of $f$ and $g$ at every point. You gave already proved that for every $x$:
$$
\lim_{t \to x^+} f(t) = g(x) = \lim_{t \to x^+} g(t).
$$
Now one possible way to continue is like this.


*

*Prove the silimar fact about left-sided limits:
$$
  \lim_{t \to x^-} f(t) = \lim_{t \to x^-} g(t).
  $$

*Note that a non-decreasing function $h(x)$ is continuous at $x$ if and only if its one-sided limits at $x$ are equal to each other.

*$f$ and $g$ have the same right-sided limits at each point, and the same left-sided limits. It follows from step 2 that they have the same points of continuity.

A: $f$ is continuous at $x$ exactly if $$
  \lim_{t \to x^-} f(t) = \lim_{t\to x^+} f(t) \text{,}
$$
$g$ is continuous at $x$ exactly if $$
  \lim_{t \to x^-} g(t) = \lim_{t \to x^+} g(t) \text{.}
$$
Since you already know that $g$ is right-continuous, you thus need to show that $$
  \lim_{t\to x^-} \underbrace{\lim_{u\to t^+} f(u)}_{=g(t)} = \underbrace{\lim_{t\to x^+} f(t)}_{=g(x)}
  \Leftrightarrow
  \lim_{t \to x^-} f(t) = \lim_{t\to x^+} f(t) \text{.}
$$
A: I don't understand your answers.
Since $f$ and $g$ are nondecreasing, a result in real analysis tells us that the sets of discontinuity points of $f$ and $g$ are respectively
$D_{f}=\{x\in\mathbb{R}:\lim_{t\to x^-}f(t)<\lim_{t\to x^+}f(t)\}$
$D_{g}=\{x\in\mathbb{R}:\lim_{t\to x^-}f(g)<\lim_{t\to x^+}g(t)\}$
and that these sets are countable.
We have to prove that $D_{f}=D_{g}$ or, equivalently, that $ \mathbb{R}\setminus D_{f} = \mathbb{R} \setminus D_{g}$.
There are two implications to prove. One is by definition, that is $x\in D_{f}\Rightarrow x\in D_{g}$.
I am not able to show the other one: $x\in D_{g}\Rightarrow x\in D_{f}$.
