Total variation is additive over adjacent subintervals and decreases under restriction

Given partitions $P_1$ of $[a,c]$ and $P_2$ of $[c,b]$, splice their point lists at $c$. The resulting partition $P$ of $[a,b]$ satisfies $V(f,P)=V(f,P_1)+V(f,P_2)$. Thus, whenever $f$ is BV on $[a,b]$, each restriction is BV and $\operatorname{Var}_{[a,c]}(f)+\operatorname{Var}_{[c,b]}(f)\le\operatorname{Var}_{[a,b]}(f)$. The assertion is unchanged when $c=a$ or $c=b$, because the singleton term is $0$.

Conversely, insert $c$ into an arbitrary partition $P$ of $[a,b]$. The refined sum splits into a sum on $[a,c]$ and one on $[c,b]$, while inserting a point does not decrease the sum, because the corresponding increment is replaced by two increments whose absolute values dominate it by the triangle inequality. Hence $V(f,P)\le\operatorname{Var}_{[a,c]}(f)+\operatorname{Var}_{[c,b]}(f)$.

Taking the supremum over $P$ in step 1.2 and combining it with step 1.1 proves the equality and the equivalence of boundedness. Applying the equality twice to $[u,v]\subseteq[a,b]$ shows $\operatorname{Var}_{[u,v]}(f)\le\operatorname{Var}_{[a,b]}(f)$.