TITLE: 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
QUESTION [6 upvotes]: 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.

REPLY [4 votes]: 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 $E$.