Property of the variation of a function I need help with the following:
given $ f:[a,b]\rightarrow \mathbb{R}$, show that $$V_f (a;b)=V_f(a;c)+V_f (c;b)$$ with $a< c <b$. 
We know that $$V_f(a;b) \geq \sum_i |f(x_i)-f(x_{i-1})|$$ for any finite partition $P_c$ of $[a,b]$ such that there exist $m$ with $x_m=c$. Then $$V_f(a;b) \geq \sup_{P_c} \{\sum_i |f(x_i)-f(x_{i-1})|\}.$$ But now I don't know what to do and also I can't prove $V_f(a;b) \leq V_f(a;c)+V_f(c;b)$.
 A: For a given partition $\Pi =\{ x_0<\ldots<x_n\}$, we denote by
$$V_f(\Pi) = \sum_{i=1}^n |f(x_i)-f(x_{i-1})|$$
the total variation sum.
Here is (a sketch of) the proof:


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*$V_f(a,b) \leq V_f(a,c)+V_f(c,b)$: Let $\Pi$ be a partition of $[a,b]$. If $c \notin \Pi$, then we set $\Pi' := \Pi \cup \{c\}$. Obviously, we can write $\Pi' = \Pi_1 \cup \Pi_2$ where $\Pi_1$ is a partion of $[a,c]$ and $\Pi_2$ a partiton of $[c,b]$. It follows from the triangle inequality that $$V_f(\Pi) \leq V_f(\Pi') = V_f(\Pi_1)+V_f(\Pi_2)$$ Taking the supremum over all partitions yields the claim.

*$V_f(a,b) \geq V_f(a,c) + V_f(c,b)$: Fix $\varepsilon>0$ and let $\Pi_1$ ($\Pi_2$) be a partition of $[a,c]$ ($[c,b]$) such that $$V_f([a,c]) \leq V_f(\Pi_1)+\varepsilon \qquad \quad V_f([c,b]) \leq V_f(\Pi_2)+\varepsilon.$$ Obviously, $\Pi := \Pi_1 \cup \Pi_2$ defines a partition of $[a,b]$ and $$V_f(\Pi) = V_f(\Pi_1)+V_f(\Pi_2) \geq V_f([a,c])+V_f([c,b])-2\varepsilon.$$ Since $\varepsilon>0$ is arbitrary, this finishes the proof.

A: For any $\epsilon>0$, there is a partition such that
$$V_f(a,b)-\epsilon < \sum_{k=1}^{n}|f(x_k)-f(x_{k-1})|<V_f(a,b).$$
If $c$ is not one of the partition points, then add it and the inequalities still hold. So assume $c = x_j$, then 
$$V_f(a,b)-\epsilon < \sum_{k=1}^{j}|f(x_k)-f(x_{k-1})|+ \sum_{k=j+1}^{n}|f(x_k)-f(x_{k-1})|<V_f(a,b).$$
Then,
$$V_f(a,b)-\epsilon < \sum_{k=1}^{j}|f(x_k)-f(x_{k-1})|+ \sum_{k=j+1}^{n}|f(x_k)-f(x_{k-1})|<V_f(a,c)+V_f(b,c)$$
If we can show $V_f(a,c)+V_f(b,c) \leq V_f(a,b)$, then it will follow that  $V_f(a,c)+V_f(b,c) = V_f(a,b)$ since $\epsilon $ is arbitrary.
For any partition $P = P' \cup P''$ where $P'=(x_0,...,x_j=c)$ and $P''=(c=x_{j+1},...,x_n)$
$$\sum_{k=1}^{j}|f(x_k)-f(x_{k-1})|\leq V_f(a,b) -\sum_{k=j+1}^{n}|f(x_k)-f(x_{k-1})|$$
Taking the supremum over $P'$ we get
$$V_f(a,c)\leq V_f(a,b) -\sum_{k=j+1}^{n}|f(x_k)-f(x_{k-1})|,$$
and
$$\sum_{k=j+1}^{n}|f(x_k)-f(x_{k-1})| \leq V_f(a,b)- V_f(a,c),$$
Taking the supremum over $P''$ we get
$$V_f(c,b)\leq V_f(a,b)- V_f(a,c),$$
and we have 
$$V_f(a,b)-\epsilon < V_f(a,c)+V_f(b,c) \leq V_f(a,b)$$
