Let $f:[a, b] \to \mathbb R$ and $V:[a, b] \to \mathbb R \cup \{+\infty\}$ be such that $V(x) := \operatorname{Var}_{[a, x]} f$ for all $x \in [a, b]$. I'm trying to prove a continuity relation between $f$ and $V$, i.e.,
Theorem Let $f$ be of bounded variation and $c \in [a, b]$. Then $f$ is continuous at $c$ IFF $V$ is continuous at $c$.
Could you have a check on my below attempt?
Proof Notice that $V$ is real-valued.
- Let $f$ be continuous at $c$. By symmetry, it suffices to prove $V$ is right continuous at $c$.
Fix $\varepsilon>0$. There is $\delta>0$ such that $0 < x-c<\delta$ implies $|f(x)-f(c)| <\varepsilon$. There is a partition $c=x_1 < x_2 < \cdots < x_n = b$ of $[c, b]$ such that $$ \operatorname{Var}_{[c, b]} f < \sum_{i=1}^{n-1} |f(x_{i+1})-f(x_i)| + \varepsilon. $$
Let $c<x< \min \{x_2, c+\delta\}$. By this lemma, we have the decomposition $$ V(x) -V(c) =\operatorname{Var}_{[c, x]} f = \operatorname{Var}_{[c, b]} f - \operatorname{Var}_{[x, b]} f. $$
Notice that $x < x_2 <x_3 < \cdots < x_n$ is a partition of $[x, b]$. So $$ \operatorname{Var}_{[x, b]} f \ge |f(x_2)-f(x)| + \sum_{i=2}^{n-1} |f(x_{i+1})-f(x_i)| . $$
As such, $$ V(x) -V(c) \le |f(x)-f(x_1)| + \varepsilon < 2 \varepsilon. $$
- Let $V$ is continuous at $c$. By symmetry, it suffices to prove $f$ is right continuous at $c$.
Let $x>c$. We have $|f(x)-f(c)| \le \operatorname{Var}_{[c, x]} f = V(x)-V(c)$. The claim then follows easily. $\tag*{$\blacksquare$}$
Corollary If $f$ is continuous and of bounded variation, then $f$ can be written as the difference of two strictly increasing continuous functions.
Proof We have $V, V+f, V-f$ are increasing. Also, $$ f (x) = (V(x) +x) - [(V(x)-f(x)) +x ]. $$
The claim then follows by taking $f=f_1-f_2$ with $f_1 (x) := V(x)+x$ and $f_2(x) = (V(x)-f(x)) +x$. $\tag*{$\blacksquare$}$
