Let be $V_1$ the length of real valued intervals of type $(a,b]$.
Show that if $\bigcup\limits_{i=1}^m(c_i,d_i]$ is a disjoint union of $\bigcup\limits_{j=1}^n(a_j,b_j]$, then $\sum\limits_{i=1}^mV_1((c_i,d_i])\leq \sum\limits_{j=1}^nV_1((a_j,b_j])$.
(Note that $\bigcup\limits_{j=1}^n(a_j,b_j]$ might not be necessarily disjoint).
My approach by induction:
If $n=1$ then the statement is trivially true. So we assume the statement for a $n\in\mathbb{N}$.
Induction step $n\to n+1$:
We consider the not necessarily disjoint union $\bigcup\limits_{j=1}^{n+1}(a_j,b_j]$.
If $(a_{n+1},b_{n+1}]\cap \bigcup\limits_{j=1}^{n}(a_j,b_j] =\emptyset$ or $(a_{n+1},b_{n+1}]\cap \bigcup\limits_{j=1}^{n}(a_j,b_j] =(a_{n+1},b_{n+1}]$, then we can immediately appy the induction hypothesis and we are done.
If $(a_{n+1},b_{n+1}]\cap \bigcup\limits_{j=1}^{n}(a_j,b_j] \neq \emptyset$ and $(a_{n+1},b_{n+1}]\cap \bigcup\limits_{j=1}^{n}(a_j,b_j]\neq (a_{n+1},b_{n+1}]$, then we can represent $(a_{n+1},b_{n+1}]\setminus \bigcup\limits_{j=1}^{n}(a_j,b_j]$ by a disjoint union $(a_{n+1},b_{n+1}]\setminus \bigcup\limits_{j=1}^{n}(a_j,b_j]= \bigcup\limits_{k=1}^{r}(c_k,d_k]$ (We know this statement from lecture).
As $\bigcup\limits_{k=1}^{r}(c_k,d_k]\subseteq (a_{n+1},b_{n+1}]$, we know from lecture that $\sum\limits_{k=1}^rV_1((c_k,d_k])\leq V_1((a_{n+1},b_{n+1}])$.
By induction hypothesis there exists a disjoint union $\bigcup\limits_{i=1}^m(c_i,d_i]$ of $\bigcup\limits_{j=1}^{n}(a_j,b_j]$ such that $\sum\limits_{i=1}^mV_1((c_i,d_i])\leq \sum\limits_{j=1}^nV_1((a_j,b_j])$.
Finally, $\bigcup\limits_{i=1}^m(c_i,d_i]\cup \bigcup\limits_{k=1}^{r}(c_k,d_k]$ is a disjoint union of $\bigcup\limits_{j=1}^{n+1}(a_j,b_j]$ and if we use the conclusions from above we get
$\sum\limits_{i=1}^{m+r}V_1((c_i,d_i])\leq \sum\limits_{j=1}^{n+1}V_1((a_j,b_j])$.
This proves the induction step and we are done.
Is this correct? Any feedback is welcome!