Does Right continuity at each point implies continuous a.e? Suppose f is a function from non negative real numbers to extended non negative real numbers which is right continuous at each point,does it imply that f is continuous almost everywhere in its domain?
 A: The result is actually stronger.

Let $f [0, +\infty) \rightarrow [0, +\infty]$ be a right continuous function. Then the set of points where $f$ is discontinuous is a countable set.

Proof:
Let $f [0, +\infty) \rightarrow [0, +\infty]$ be a right continuous function. For all $n \in \Bbb N$, $n\geqslant 1$,  let
$$ \Gamma_n = \left \{ I \subseteq  (0, +\infty) : I \textrm{ is an open interval and, for all } x,y \in I, |f(y)-f(x)|< 1/n    \right \} $$
Let $A_n = \bigcup_{I \in \Gamma_n} I$. We have that $A_n$ is an open set.
Let $F_n = A_n^c$. $F_n$ is closed. Now, given any $x \in  F_n$, since $f$ is right continuous, it is easy to see that there is $\delta_x>0$, such that $(x, x+\delta_x) \subseteq A_n$. It means, $(x, x+\delta_x) \cap F_n =\emptyset$. It follows that $F_n$ is countable (see Remark).
Now, $A= \bigcap_{n\geqslant 1}A_n$ is the set of points where $f$ is continuous and we have that $A^c= \bigcup_{n\geqslant 1}F_n$ is countable.
Remark: If $F_n$ is uncountable, then there is $k \in \Bbb N$ such that $F_n \cap [0, k)$ is uncountable, but this is not possible, since for each $x \in  F_n$, there is  $\delta_x>0$ such that $(x, x+\delta_x) \cap F_n =\emptyset$.
As pointed out by @EvangelopoulosF. another way to prove that $F_n$ is countable is, for each $x \in F_n$, to pick a $q_x \in \Bbb Q \cap (x, x+\delta_x)$. It is easy to see that the map $F_n \rightarrow  \Bbb Q$ defined by $x \mapsto q_x$ is injective.
