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

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.

Answer 1

I am going to prove, by reductio ad absurdum (contradiction), that the sought for result is basically a consequence of the Archimedean property of real numbers: despite being relatively short, it is a "pseudo-elementary" proof since it involves the use of Vitali's Covering Lemma in the form given by Gordon ([2], chapter 4, pp. 52-54).

New edit: after the comment by @Del I realized that the first step in equation \eqref{2}, should be fully justified: that step is implied by the following relation $$ E\cap\bigcup^\infty_{k=1} I_k\underset{{\mu_\mathfrak{L}}}{\simeq}\,\bigcup^\infty_{k=1} I_k\iff \mu_\mathfrak{L}\left(E\cap\bigcup^\infty_{k=1} I_k\right) = \mu_\mathfrak{L}\left(\bigcup^\infty_{k=1} I_k\right) \label{3}\tag{N} $$ and a full justification of \eqref{3}, using the definition of Lebesgue's outer measure and Caratheodory's criterion for measurability of a set is given in the appendix section.

Notation.

• In the development below we consider only finite i.e. bounded or proper intervals.
• In order to simplify the (and by abuse of) notation, we use the symbol $\Bbb N$ for the set of positive integers, excluding $0$.
• $\mu_\mathfrak{L}$ stands for the Lebesgue measure and $\mu_\mathfrak{L}^\ast$ for the corresponding outer measure: obviously, $E$ is (Lebesgue) measurable if and only if $\mu_\mathfrak{L}^\ast=\mu_\mathfrak{L}$.

The perimeter of an interval. Let $I=[a,b]\subset \mathbb{R}$ with $-\infty<a\leq b<+\infty$ be a finite interval: then $$ \begin{split} P(I) & = \sup\Biggl\{\int\limits_I\varphi'(x)\,\mathrm{d}x:\varphi \in C_c^1(\mathbb{R})\wedge |\varphi|_{\infty}\le 1\Biggr\}\\ &=\sup\biggl\{\varphi(b)-\varphi(a):\varphi \in C_c^1(\mathbb{R})\wedge |\varphi|_{\infty}\le 1\biggr\}= \begin{cases} 0 &\text{if }a=b\\ 2 &\text{if }a\neq b \end{cases}. \end{split}\label{1}\tag{1} $$ Thus the perimeter of an interval is always $2$ unless its Lebesgue measure is zero.

The structure of sets of finite perimeter on the real line. Let's introduce the deeper tool used in this answer: a strong form of Vitali's Covering lemma which holds true for any subset of the real line and involves only the analysis of its outer measure.

Definition (Gordon [2], p. 5). Let $E\subseteq\mathbb{R}$. A collection (family) $\mathscr{I}$ of intervals is a Vitali covering of $E$ if for each $x\in E$ and $\epsilon>0$ there exists an interval $I\in\mathscr{I}$ such that $x\in I$ and $\mu_\mathfrak{L}(I)<\epsilon$.

Vitali's Covering Lemma (Gordon ([2], chapter 4, lemma 4.5, pp. 52-54). Let $E\subseteq\mathbb{R}$ with $\mu_\mathfrak{L}^\ast(E)<\infty$. If $\mathscr{I}$ is a Vitali covering of $E$, then for each $\epsilon >0$ there exists a finite collection $\{I_k\}_{k\in (n)}=\{I_k: 1\le k\le n\}$ of disjoint intervals in $\mathscr{I}$ such that $$ \mu_\mathfrak{L}^\ast\left(E\setminus\bigcup^n_{k=1} I_k\right)<\epsilon. $$ In addition, there exists a sequence $\{I_k\}_{k\in \Bbb N}$ of disjoint intervals in $\mathscr{I}$ such that $$ \mu_\mathfrak{L}^\ast\left(E\setminus\bigcup^\infty_{k=1} I_k\right)=0. $$ Now, considering the additivity of the integral as a set function, \eqref{1} and a sequence $\{I_k\}_{k\in \Bbb N}$ which satisfies the Covering lemma, we have: $$ \begin{split} P(E) &= \sup\Biggl\{\,\int\limits_E\varphi'(x)\,\mathrm{d}x:\varphi \in C_c^1(\mathbb{R})\wedge |\varphi|_{\infty}\le 1\Biggr\}\\ & = \sup\Biggl\{\:\int\limits_{\bigcup^\infty_{k=1} I_k}\!\!\!\varphi'(x)\,\mathrm{d}x \;\;\,+\!\!\!\! \int\limits_{E\setminus\bigcup^\infty_{k=1} I_k}\!\!\!\!\!\!\!\varphi'(x)\,\mathrm{d}x:\varphi \in C_c^1(\mathbb{R})\wedge |\varphi|_{\infty}\le 1\Biggr\}\\ & = \sup\Biggl\{\:\int\limits_{\bigcup^\infty_{k=1} I_k}\!\!\!\varphi'(x)\,\mathrm{d}x :\varphi \in C_c^1(\mathbb{R})\wedge |\varphi|_{\infty}\le 1\Biggr\}\\ &= \sum_{k=1}^\infty P(I_k)= 2 \sum_{k=1}^\infty \delta({I_k}) \end{split}\label{2}\tag{2} $$ where the set function $\delta(A)$ is defined as follows $$ \delta_{A}= \begin{cases} 0 &\text{if }A=\emptyset\\ 1 &\text{if }A\neq\emptyset \end{cases} $$ Now, from \eqref{2} we immediately see that $I_k$ must be empty for $k$ larger than some $k>n\in\Bbb N$ for if otherwise the perimeter of $E$ cannot be finite: thus $E$ is (equivalent to, up to a set of measure $0$) to the union of a finite number of bounded intervals.

Appendix: proof of the relation \eqref{3}.

Let's recall the definition of outer measure of a set: $$ \mu^\ast(E)=\inf_{\mathscr{C}=\mathcal{C}(E)}\mu^\ast\left(\bigcup_{I_k\in \mathscr{C}}I_k\right) $$ where $\mathcal{C}$ is the set of all covering of $E$ made of intervals. Thus, by elementary considerations from real analysis, for all $\varepsilon>0$ there exist a covering $\mathscr{I}$ of $E$ such that $$ \mu^\ast\left(\bigcup_{I_k\in \mathscr{I}}I_k\right) < \mu^\ast(E)+\varepsilon\label{4}\tag{A1} $$ Now, $\mathscr{I}$ can be extended to a Vitali covering by simply adding to it the family of intervals $\{[-2^{-n}+x_o,2^{-n}+x_o]\cap I_k\}_{n\in\Bbb N}$ for all $x_o\in E$, where $I_k$ can be any interval containing $x_o$ chosen from the covering $\mathscr{I}$. Therefore we can apply Vitali's covering lemma to the family $\mathscr{I}$ in \eqref{4} and obtain a sequence of disjoint intervals intervals $\{I_k\}_{k\in\Bbb N}\subseteq\mathscr{I}$ such that $$ \mu^\ast\left(\bigcup_{k=1}^\infty I_k\right)\le \mu^\ast\left(\bigcup_{I_k\in \mathscr{I}}I_k\right) \le \mu^\ast(E)+\varepsilon\iff \mu^\ast\left(\bigcup_{k=1}^\infty I_k\right)\le \mu^\ast(E) \label{5}\tag{A2} $$ Now, since $$ E\setminus\bigcup^\infty_{k=1} I_k = E\cup\left(\bigcup^\infty_{k=1} I_k\!\right)^{\!\!c}, $$ we can apply \eqref{5} and Caratheodory's condition (since $E$ is measurable) and get $$ \begin{split} \mu^\ast\left(\bigcup_{k=1}^\infty I_k\right)&\le \mu^\ast(E) \\ & = \mu^\ast\left(E\cap\bigcup_{k=1}^\infty I_k\right)+\mu^\ast\left(E\cup\left(\bigcup^\infty_{k=1} I_k\!\right)^{\!\!c}\right)\\ & = \mu^\ast\left(E\cap\bigcup_{k=1}^\infty I_k\right)+\mu^\ast\left(E\setminus\bigcup^\infty_{k=1} I_k\right)\\ &= \mu^\ast\left(E\cap\bigcup_{k=1}^\infty I_k\right). \end{split} $$ But since $E\cap\bigcup_{k=1}^\infty I_k\subseteq \bigcup_{k=1}^\infty I_k$, then $$ \mu^\ast\left(E\cap\bigcup_{k=1}^\infty I_k\right) \le \mu^\ast\left(\bigcup_{k=1}^\infty I_k\right) \iff \mu^\ast\left(E\cap\bigcup_{k=1}^\infty I_k\right) =\mu^\ast\left(\bigcup_{k=1}^\infty I_k\right)\blacksquare $$ Final notes

• I do not need to use geometric measure theory prove the result: however, the proof is not entirely elementary since versions of Vitali's Covering Lemma for unbounded sets seem not so easy to found in the literature. Indeed, apart from Gordon ([2], chapter 4, pp. 52-54), Kannan and Krueger ([3] §0.3, Th. 0.3.21, pp. 15-17) offer a proof (which also deals explicitly $n$-dimensional sets) but requires a more elaborate definition of Vitali covering ([2] §0.3, Def. 0.3.20, p. 15): the latter authors acknowledge Saks ([1], §IV.3, Th. 3.1, pp. 109-111) for the proof. Finally, Ambrosio, Fusco and Pallara ([1] §2.4, Th. 2.18, pp. 52-53) give a proof similar to the ones of Kannan and Kruger/Saks but for general (Radon) measures.
• Note that the perimeter of a subset of the real line, when finite, is always an even integer, and this (at least in my opinion) discloses perhaps intriguing relations between the act of counting (in dimension one) and the finiteness of the perimeter (in dimension $n>1$).
• Note on the appendix. The intuition behind the proof of the relation \eqref{3} is the following one: let $E$ be one of the following sets $$ E= \begin{cases} ]-1, 0[\cap \big\{\frac{1}{n}:n\in\Bbb N\big\}\\ \qquad\text{ or}\\ ]-1,0[\cap \mathcal{C} \end{cases} $$ where $\mathcal{C}$ is the Cantor set. Then, while it is not possible to cover $E$ by a family of disjoint of intervals which has the same Lebesgue outer measure, nevertheless by applying Vitali's covering theorem you can find a family of such kind that has exactly the same measure since it avoids covering subsets of zero measure.

References

[1] Ambrosio, Luigi and Fusco, Nicola and Pallara, Diego (2000), Functions of bounded variation and free discontinuity problems. Oxford Mathematical Monographs. New York and Oxford: The Clarendon Press/Oxford University Press, New York, pp. xviii+434, ISBN 0-19-850245-1, MR1857292, Zbl 0957.49001.

[2] Gordon, Russell A., The integrals of Lebesgue, Denjoy, Perron, and Henstock, Graduate Studies in Mathematics. 4. Providence, RI: American Mathematical Society (AMS). xi, 395 p. (1994), ISBN: 0-8218-3805-9, MR1288751, Zbl 0807.26004.

[3] Kannan, Rangachary and Krueger, Carole King, Advanced analysis on the real line, Universitext. New York, NY: Springer. ix, 259 p. (1996), ISBN: 0-387-94642-X, MR1390758, Zbl 0855.26001.

[4] Saks, Stanisław, Theory of the integral, 2nd, revised ed. Engl. translat. by L. C. Young. With two additional notes by Stefan Banach. (English), Monografie Matematyczne Tom. 7. New York: G. E. Stechert & Co. pp. vi+347 (1937), JFM 63.0183.05, MR0167578 (review of the Dover ed.), Zbl 0017.30004.

Answer 2

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 $E$.