The jumps of a variation function equal the absolute jumps of the original function

Since $V_f$ is nondecreasing and bounded above by $V_f(b)$, its one-sided limits exist. By [L1] and [L2], $V_f(x)-V_f(c)\ge|f(x)-f(c)|$ for $x>c$; passage to the right limit gives $V_f(c+)-V_f(c)\ge|f(c+)-f(c)|$.

For the reverse inequality, fix $x_0>c$ and put $x_n=c+(x_0-c)2^{-n}$. Let $a_n=\operatorname{Var}_{[x_{n+1},x_n]}(f)$. By repeated additivity, every partial sum of the nonnegative series $\sum_na_n$ is $\operatorname{Var}_{[x_N,x_0]}(f)$ for a suitable $N$, hence is bounded by $\operatorname{Var}_{[c,x_0]}(f)$. Its tails therefore tend to zero by [L6], while $x_n\downarrow c$ by [L7].

Given $\varepsilon>0$, take $N$ so large that the series tail from $N$ is below $\varepsilon$ and $|f(y)-f(c+)|<\varepsilon$ whenever $c<y\le x_N$. For any partition $c=t_0<t_1<\cdots<t_k=x_N$, choose $m\ge N$ with $x_{m+1}<t_1\le x_m$. The part after its first increment is at most [step 1.2, L1, L2, L3, L6] $\operatorname{Var}_{[t_1,x_N]}(f)\le\operatorname{Var}_{[x_{m+1},x_N]}(f)=\sum_{n=N}^{m}a_n<\varepsilon,$ while $|f(t_1)-f(c)|\le|f(c+)-f(c)|+\varepsilon$. Taking the supremum over partitions gives $\operatorname{Var}_{[c,x_N]}(f)\le|f(c+)-f(c)|+2\varepsilon$. Restriction gives the same bound for $c<x\le x_N$, and [L2] gives the reverse bound in the limit.

Thus $\lim_{x\downarrow c}\operatorname{Var}_{[c,x]}(f)=|f(c+)-f(c)|$, and [L1] proves the right-hand formula. Applying steps 1.2–2.1 to the reversed interval proves the left-hand formula. If $f$ is continuous at $c$, both absolute jumps vanish by [L5], so $V_f$ is continuous there. Endpoint cases use only the available side.