Help with the proof that $E\subset \mathbb{R}$ with finite perimeter and area has to be equal to the finite union of bounded intervals I'm preparing for an exams and I've problems solving this exercise from an old exam, any help will be welcomed
Let $E\subset \mathbb{R}$ be a measurable set with $\mathfrak{L}^1(E)<\infty$ and $$P(E) = \sup\biggl\{\int_E\varphi'(x)\,dx:\varphi \in C_c^1(\mathbb{R})\wedge |\varphi|_{\infty}\le 1\biggr\}<\infty.$$

Prove that $E$ is (equivalent to, up to a set of measure $0$) the union of a finite number of bounded
  intervals.

In here $P(E)$ is just a formal way to define the perimeter, so the problem is proving that, if a subset of the real line $\mathbb{R}$ has finite length and finite perimeter then it is the union of a finite number on bounded intervals.
My problem is how to put together the two datas we have on $E$ (the finite perimeter and finite length).
I actually found a proof of this fact but it seems quite convoluted so I was hoping to find some help in understanding it or even better state it in a simpler way
This come from the book "Sets of finite perimeter and geometric variational problems" from Francesco Maggi

EDIT: My idea is to pass to the closure and then argue with connected components of the former, also using the fact that for balls $P(E)=H^{n-1}(E)$ the $(n-1)$-dimensional Hausdorff measure, this ball will need to be bounded and each of them giving a finite contribution to the perimeter therefore in a finite number.
However to do so I need $\overline{E}\setminus E$ to have measure 0 and even if I'm quite sure about this in the comments of this own post it doesn't seem like so.
 A: In case you fancy Lebesgue density point there is a fairly simple argument that goes as follows: Suppose $a<b$ are Lebesgue density points of $E$ and $E^c$, respectively. Then by Lebesgue density we may to any $\epsilon>0$ find $\delta>0$ so that $\lambda([a,a+\delta]\cap E)\geq (1-\epsilon)\delta$ and $\lambda([b-\delta,b]\cap E)\leq \epsilon \delta$. 
Now construct a bump-function $\phi\in C_c^1[a,b]$ so that $\phi(x)=1$ on $[a+\delta,b-\delta]$ which is almost linearly ("almost" in order to make the function $C^1$, costing some extra $\epsilon$ below)  increasing, respectively decreasing on the small intervals to the left and to the right.
Then
$$ \int_a^b \phi' 1_E =\int_a^{a+\delta} \phi' 1_E + \int_{b-\delta}^b \phi' 1_E \geq (1-2\epsilon) \int_a^{a+\delta} \phi' - 2\epsilon \int_{b-\delta}^b \phi'\geq 1-4\epsilon.$$
Now, if $a_1<b_1<a_2<b_2<...<a_n<b_n$ is an intertwined sequence of density points in $E$ and $E^c$, respectively, you get by simply adding the corresponding bump-functions:
  $$P(E)=\sup\bigg \{ \int_E \phi' : \phi\in C_c^1({\Bbb R}), |\phi|_\infty\leq 1\bigg\} \geq n .$$
Thus, $n$ must be finite. In general, if $I$ is any nontrivial interval and $$0<\lambda(I\cap E)<\lambda(I)$$ then $I$ contains density points of both of the above types. So there can only be finitely many disjoint intervals that verifies this inequality. From this, the result follows easily.
Remark: Using a more clever construction of functions $\phi$ being zero at $\pm \infty$ and varying between $-1$ and $+1$ at a sequence of intertwined density points you realize that $P(E)$ simply counts the number of essential boundary points of $E$. 
