making a function càdlàg Suppose $f:[a,b]\to\mathbb{R}$ has a right limit $f(x+)$ at all $a \le x \lt b$ and a left limit $f(x-)$ at all $a\lt x\le b$.
Is the function $$g[a,b]\to\mathbb{R}:x\mapsto\begin{cases}f(x+)&a\le x\lt b\\ f(b)&x=b\end{cases}$$ càdlàg? and is the set $\lbrace x\in[a,b]:f(x)\neq g(x)\rbrace$ at most countable (like you would expect)?
 A: $\newcommand{\ve}{\varepsilon}$ I'll try to give an $\ve$-$\delta$ proof.
You assume that $f(x+)$ exists for each $x$.
You have defined
$$g[a,b]\to\mathbb{R}:x\mapsto\begin{cases}f(x+)&a\le x\lt b\\ f(b)&x=b\end{cases}$$
and you're asking whether at each point both one-sided limits $g(x-)$ and $g(x+)$ exist and whether $g(x+)=g(x)$.
Clearly, if we show that $f(x+)=g(x+)$ and $f(x-)=g(x-)$, then this is true.
Fix a point $x_0\in[a,b)$
Let $r=f(x_0+)$, i.e. for each $\ve>0$ there is a $\delta>0$ such that $|f(x)-r|<\ve$ for each $x\in(x_0,x_0+\delta)$. This clearly implies that
$|g(x)-r|=|f(x+)-r|\le\ve$ for each $x\in(x_0,x_0+\delta/2)$. This shows that $g(x_0+)=r=f(x_0+)$.

Let $l=f(x_0-)$, i.e. for each $\ve>0$ there is a $\delta>0$ such that $|f(x)-l|<\ve$ for each $x\in(x_0-\delta,x_0)$. Now if $x\in(x_0-\delta,x_0)$ then we have an interval $(x,x_0)$ on the right from $x$ such that $|f(x')-l|<\ve$ for each $x'$ in this interval. From this we get $|g(x)-l|=|f(x+)-l|\le\ve$ for each $x\in(x_0-\delta,x_0)$. This shows that $g(x_0-)=l=f(x_0-)$

There is a result which says that for any real function there exist only countably
many points $x$ of $\mathbb R$ for which $f$ is not continuous at $x$ but $f(x+)$ exists.
See e.g. van Rooij, Schikhof: A Second Course on Real Functions,
Theorem 7.7, p.45.
Clearly $f(x)\ne g(x)=f(x+)$ implies that $f$ is not continuous at $x$. By the above result, there is only countably many such points.
A: By result that Martin said $g $ is continuous except at a contable set $C$ and in $[a,b) \backslash C, \ g(t) = f^{+}(t) = f^{-}(t)= f(t) $ . Hence in $C$, $g$ is càdlàgle se $t \in C \backslash \{b\} .$ By $C$ to be contable, $t$ is a accumulation point and
\begin{eqnarray*}
g(t^{+}) & = & \lim_{s \downarrow t} g(s)\\
& = & \lim_{s \downarrow t} f^{+}(s)\\
& = & \lim_{s\downarrow t} f(s)\\
& = & f(t^{+}) \\
& = & g(t)\\
\end{eqnarray*}
Analogously, $g(t^{-})$ exist and is equals to $f(t^{-})$. Hence $g$ is càdlàg. Excuse my bad English.
