Right continuity and bounded variation. Given $f \in BV[a, b],$ define $g(x) = f(x+)$ for $a \leq x < b$ and $g(b) = f(b).$
Prove that $g$ is right continuous and of bounded variation on $[ a, b ] $.
I tried to use Jordan's Theorem stating that a function is of bounded variation iff it can be expressed as the difference between two monotonically increasing functions. Then it suffices to show that the "right limit" function of any monotone functions on [a,b] is monotone and right continuous. The former is not hard since we can write the right limit for a monotonically increasing function $f$ as $f(x+) = \inf \{f(t) \mid t > x\}$ and then it follows $y>x \Rightarrow f(y+)\geq f(x+).$ However, I'm stuck on showing the right continuity.
 A: Firstly, we consider the case that $f$ is increasing. For each $x\in[a,b)$,
the right-hand limit $\lim_{t\rightarrow x+}f(t)=\inf\{f(t)\mid t>x\}$
exists, so $g(x)$ is well-defined. Moreover, for each $t>x$, we
have $f(t)\geq f(x)$, so $\lim_{t\rightarrow x+}f(t)\geq f(x)$,
i.e., $g(x)\geq f(x)$.
Claim 1: $g$ is increasing on $[a,b]$. Let $x_{1},x_{2}\in[a,b]$
with $x_{1}<x_{2}$. For any $t\in(x_{1},x_{2})$, we have that $f(t)\leq f(x_{2})$.
Letting $t\rightarrow x_{1}+$ yields $g(x_{1})\leq f(x_{2})\leq g(x_{2})$.
Claim 2: $g$ is right-continuous. Let $x_{0}\in[a,b]$ be arbitrary.
If $x_{0}=b$, $g$ is automatically right-continuous at $x_{0}$.
Suppose that $x_{0}<b$. Let $\varepsilon>0$ be arbitrary, then there
exists $\delta>0$ such that $\left|f(t)-g(x_{0})\right|<\varepsilon$
whenever $t\in(x_{0},x_{0}+\delta)$. Let $x_{1}\in(x_{0},x_{0}+\delta)$
be arbitrary. For any $t\in(x_{1},x_{0}+\delta)$, we still have $\left|f(t)-g(x_{0})\right|<\varepsilon$.
Now, leting $t\rightarrow x_{1}+$, then we obtain $\left|g(x_{1})-g(x_{0})\right|\leq\varepsilon$.
This shows that $g$ is right-continuous at $x_{0}$.

For the general case that $f$ is of bounded variation, we write $f=f_{1}-f_{2}$
for some increasing functions $f_{1},f_{2}$ defined on $[a,b]$.
Define $g_{1}(x)=f_{1}(x+)$ and $g_{2}(x)=f_{2}(x+)$. By the above
discussion, $g_{1},g_{2}$ are increasing and right-continuous. Moreover,
for $x\in[a,b)$, $g(x)=\lim_{t\rightarrow x+}\left[f_{1}(t)-f_{2}(t)\right]=\lim_{t\rightarrow x+}f_{1}(t)-\lim_{t\rightarrow x+}f_{2}(t)=g_{1}(x)-g_{2}(x)$.
That is, $g=g_{1}-g_{2}$, the difference of two right-continuous
functions, so $g$ is also right-continuous.
