Let $Z:\mathbb{R}_+\rightarrow\mathbb{R}$ be cadlag and of local bounded variation with $Z(0)=0$ and $V_Z(t)$ denotes the value of the total variation of $Z$ on $[0,t]$ for all $t\in\mathbb{R}_+$. I want to show that there is a unique decomposition $Z=Z_1-Z_2$ of monotonically increasing and right-continuous functions $Z_1,Z_2:\mathbb{R}_+\rightarrow \mathbb{R}_+$, such that $Z_1(0)=0=Z_2(0)$ and $V_Z=Z_1+Z_2$, more precisely
$$Z_1=\frac{V_Z+Z}{2},\quad Z_2=\frac{V_Z+Z}{2}-Z.$$
I know that $V_Z-Z$ is monotonically increasing and right-continuous with $V_Z(0)-Z(0)=0$. Therefore it holds also for
$$Z_2=\frac{V_Z-Z}{2}$$
But I do not know how to show monotonicity and right-continuity for $Z_1$. Also I do not know how to show uniqueness. I would appreciate any help. Thanks in advance!
According to MisterRiemann's comment one has
$$\forall\ 0\le s\le t: V_Z[0,t]\ge V_Z[0,s]+Z(s)-Z(t)$$
which is equivalent to
$$\forall\ 0\le s\le t: V_Z[0,t]+Z(t)\ge V_Z[0,s]+Z(s)$$
$Z_1(0)=0$ and right-continuity of $Z_1$ are obvious from definition of $V_Z$ and $Z$.
Suppose there exists another decomposition $X_1,X_2$ with the properties stated above, i.e. $$\exists\ t\in\mathbb{R}_+:X_1(t)>Z_1(t)$$
From $X_1(t)-X_2(t)=Z_1(t)-Z_2(t)$ follows $X_2(t)>Z_2(t)$ and we get $$V_Z(t)=Z_1(t)+Z_2(t)<X_1(t)+X_2(t)=V_Z(t),$$ which is a contradicition. The other case $X_1(t)<Z_1(t)$ can be proven in an analogue way.