lebesgue-measuremeasure-theory
I was wondering if the following is true:
Let $m$ be the Lebesgue measure and let $E$ be a Lebesgue-measurable subset of the reals $\mathbb R$.
Suppose that for every two intervals
$I, J$ such that $m(I)=m(J)$ we have $m(E\cap I)=m(E\cap J)$. Then
either $m(E)=0$ or $m(\mathbb R\setminus E)=0$.
My intuition tells me that this is true, but I was not able to prove it.
I do know that for every $\alpha \in (0, 1)$ there exists an interval $I$ such that $m(I\cap E)\geq \alpha m(I)$.
For 1: There is an open set $U$ containing all rationals such that $m(U)< (b-a).$ Thus $m([a,b]\setminus U) > 0.$ The set $[a,b]\setminus U$ is compact, is a subset of $[a,b],$ and since it contains no rational, has no interior.
Firstly, I think that the easiest order of implications is $1) \implies 3)$, then $3) \implies 2)$, then $2)\implies 1)$. Before drafting a full answer, I will give you the sketch of each argument.
For $1) \implies 3)$, you essentially want to appeal to the definition of the outer measure. You can find a collection of intervals $(I_i)_i$ which covers $A$ and for which $m^*(A) \leq m^*\big(\bigcup_i I_i\big) < m^*(A) + \varepsilon$ by definition of the infimum. Then the set $G_\varepsilon = \bigcup_i I_i$ is the one you want, and you use the measurability of $A$ to argue that the desired inequality holds.
Side remark: almost all of these arguments are most direct to show in the case where $m^*(A) < \infty$; to handle the case where $m^*(A) = \infty$ what I believe you want to do is apply the result to each of the sets $A_n = [-n,n] \cap A$ and then pass to the limit.
For $3) \implies 2)$, you can find an open set $G_\varepsilon$ for each $\varepsilon > 0$ satisfying the hypothesis; choose $B = \bigcap_\varepsilon G_\varepsilon$ to be your borel set, and argue the result holds by monotonicity.
For $2) \implies 1)$, first prove the lemma that all borel sets are measurable (if the open intervals are measurable then the sigma algebra generated by the open intervals, i.e. the borel algebra, is contained in the sigma algebra of measurable sets). Then, the lemma that sets of outer measure zero are measurable. Finally, stitch these together to get the desired equation for $A$ to hold.
$1) \implies 3)$
Suppose $A$ is a set such that $\forall E \ \big( m^*(E) = m^*(E \cap A) + m^*(E \cap A^c)\big)$. In the first case, suppose $m^*(A) < \infty$ and let $\varepsilon > 0$.
By definition of the infimum, one can find a collection of intervals $(I_i)_i$ such that $A \subset \bigcup_i I_i$ and
$$m^*(A) \leq \sum_i \ell(I_i) < m^*(A) + \varepsilon.$$
Let $G = \bigcup_i I_i$; we will show that $m^*(G \setminus A) < \varepsilon$. First, apply the hypothesis $1)$ with the choice $E = G$. Then we conclude
$$m^*(G) = m^*(G \cap A) + m^*(G \cap A^c) = m^*(A) + m^*(G \setminus A)$$
by the fact that $A \subset G$. Next, we have the following estimation by the countable subadditivity of the outer measure.
\begin{align*} m^*(G) &= m^*\big(\bigcup_i I_i\big)\\ &\leq \sum_i m^*(I_i)\\ &< m^*(A) + \varepsilon \end{align*}
The above also assumes that $m^*(I) = \ell(I)$ for an interval $I$.
We conclude that $m^*(G \setminus A) = m^*(G) - m^*(A) < \varepsilon$. Observe the necessity of the hypothesis $m^*(A) < \infty$ in this argument, else the difference is undefined.
In the second case, suppose $m^*(A) = \infty$ and again let $\varepsilon > 0$. For each natural $n > 0$, let $A_n = [-n,n] \cap A$. Observe that for each $n$, we have $m^*(A_n) < \infty$ in which case (by the above argument) there is an open set $G_n$ such that $A_n \subset G_n$ and $m^*(G_n \setminus A_n) < \varepsilon / 2^n$. Observe the scaling of $\varepsilon$, which hints that in the end we will sum over all of these differences.
Indeed, let $G = \bigcup_n G_n$; we argue that this is the desired $G$. First, we have the lemma
$$\Big( \bigcup_n G_n \Big) \setminus \Big(\bigcup_n A_n\Big) \subset \bigcup_n \big(G_n \setminus A_n\big).$$
If a point $x$ belongs to the LHS, then $x \in G_m$ for some $m$ while $x\notin A_n$ for any $n$. In particular, $x \notin A_m$ and therefore $x \in G_m \setminus A_m$, hence $x$ belongs to the RHS. The inclusion follows.
In light of this, we have the following estimation, again relying on the (monotonicity and) countable subadditivity of the outer measure.
\begin{align*} m^*(G \setminus A) &= m^* \Big(\Big( \bigcup_n G_n \Big) \setminus \Big(\bigcup_n A_n\Big)\Big) \\ &\leq m^*\Big(\bigcup_n \big(G_n \setminus A_n\big)\Big)\\ &\leq \sum_n m^*(G_n \setminus A_n)\\ &< \sum_n \varepsilon /2^n\\ &= \varepsilon \end{align*}
This concludes the first implication.
$2) \implies 1)$
I shall preface this section with the observation that this proof is a bit of a headache, not to understand but to come up with, once you've forgotten the order in which things must be done. Drawing pictures of $A$, $B$ and $E$ venn-diagram style helps immensely to determine which set must be split by which other set.
I shall leave it to you to cover the lemmas that each Borel set is measurable (because each open interval is measurable, plus sigma-algebra-generating tricks), as is each set with outer measure zero.
So let $A \subset \mathbb R$ and suppose $B \subset A$ is a Borel set such that $m^*(A \setminus B) = 0$. Observe that both $B$ and $A \setminus B$ are therefore measurable. Let $E \subset \mathbb R$ be arbitrary and apply the measurability of $B$; we have that
$$m^*(E) = m^*(E \cap B) + m^*(E \cap B^c).$$
We then claim that $m^*(E \cap A) = m^*(E \cap B)$ and $m^*(E \cap A^c) = m^*(E \cap B^c)$. For the first, apply the measurability of the set $B$ to the set $E \cap A$; we have
$$m^*(E \cap A) = m^*(E \cap A \cap B) + m^*(E \cap A \cap B^c) = m^*(E \cap B) + 0.$$
The latter equality follows from the fact that $B \subset A$ together with $(E \cap A \cap B^c) \subset (A \setminus B)$ which hence has outer measure zero by hypothesis + monotonicity.
For the last step, we apply the measurability of $A \setminus B$ to the set $E \cap B^c$. We obtain
$$m^*(E \cap B^c) = m^*(E \cap B^c \cap (A\setminus B)) + m^*(E \cap B^c \cap (A\setminus B)^c) = 0 + m^*(E \cap A^c).$$
The latter equality follows again from hypothesis + monotonicity, together with set equality which can be argued by chasing the logic out of a venn diagram. Specifically,
$$E \cap B^c \cap (A\setminus B)^c = E \cap A^c.$$
We finally conclude $1)$. This proof may also be (carefully) adapted to the dual case, where $B$ is a borel set such that $A \subset B$ and $m^*(B \setminus A) = 0$.
This concludes the second implication.
$3) \implies 2)$
As I mentioned in the comments, I think it would be generally easier to replace condition $2)$ with the dual statement (that there exists a Borel set $B \supset A$ such that $m^*(B \setminus A) = 0$), prove that these three are equivalent, then recover the inner Borel approximation in a separate argument. Nevertheless, as these are all ultimately equivalent, it is theoretically possible to go directly from $3)$ to $2)$ in its current form, and so here is an argument of that.
Firstly, for each $n$ there is an open set $G_n \supset A$ such that $m^*(G_n \setminus A) < 1/n$. Let $B = \bigcap_n G_n$, which is Borel. We have $A \subset B$ since $A \subset G_n$ for every $n$, and therefore by monotonicity we have
$$0 \leq m^*(B \setminus A) \leq m^*(G_n \setminus A) < 1/n$$
for every $n$, hence $m^*(B \setminus A) = 0$ by the squeeze theorem. This is the inverse of what we aim to prove, but here is how the gap may be bridged: we shall first use the set $B$ to argue that $A$ satisfies $1)$ by identical argument as $2) \implies 1)$ above, then hence $A^c$ satisfies $3)$ and the above construction can be repeated to finally produce the desired Borel set $B'' \subset A$.
Indeed, we repeat that any Borel set and any set of outer measure zero is measurable, so $B$ and $B \setminus A$ are measurable. Let $E \subset \mathbb R$ be arbitrary; by the measurability of $B$ we have
$$m^*(E) = m^*(E \cap B) + m^*(E \cap B^c).$$
Then we apply the measurability of $B \setminus A$ to the set $E \cap B$ and deduce
$$m^*(E \cap B) = m^*(E \cap B \cap (B \setminus A)) + m^*(E \cap B \cap (B \setminus A)^c) = 0 + m^*(E \cap A).$$
Here we have used that $E \cap B \cap (B \setminus A) \subset B \setminus A$ and therefore has outer measure zero by monotonicity. We have also used the fact that $E \cap B \cap (B \setminus A)^c = E \cap A$ which can be easily proven under the assumption that $A \subset B$.
Indeed, if $x$ is a point belonging to $E \cap B \cap (B\setminus A)^c$ then we have $x \in E$ and $x \in B$ and $X \notin B \setminus A$, the third statement being equivalent to $x \notin B$ or $x \in A$. Since $x \in B$ is already known, then we must have $x \in A$ and hence $x \in E \cap A$. This demonstrates that $E \cap B \cap (B\setminus A)^c \subset E \cap A$; the reverse inclusion is checked similarly.
We then apply the measurability of $B$ to the set $E \cap A^c$ and deduce
$$m^*(E \cap A^c) = m^*(E \cap A^c \cap B) + m^*(E \cap A^c \cap B^c) = 0 + m^*(E \cap B^c).$$
Here again we use $E \cap A^c \cap B \subset A^c \cap B = B \setminus A$ which has outer measure zero, as well as $E \cap A^c \cap B^c = E \cap B^c$. This follows because $A \subset B$ implies $B^c \subset A^c$, in which case $A^c \cap B^c = B^c$.
We have therefore shown that $m^*(E \cap B) = m^*(E \cap A)$ and $m^*(E \cap B^c) = m^*(E \cap A^c)$, hence
$$m^*(E) = m^*(E \cap B) + m^*(E \cap B^c) = m^*(E \cap A) + m^*(E \cap A^c).$$
It therefore follows that $A$ satisfies $1)$.
By symmetry in $1)$ we have that $A^c$ satisfies $1)$ as well. Then by earlier argument, $A^c$ satisfies $3)$. Finally, repeat the construction at the beginning to find a Borel set $B' \supset A^c$ such that $m^*(B' \setminus A^c) = 0$, and let $B'' = (B')^c$. This is the desired Borel set for the condition $2)$.
Firstly, $B'' \subset A$ is clear by definition. Then, one may unpack complements to conclude the set equality $A\setminus B'' = B' \setminus A^c$, in which case
$$m^*(A \setminus B'') = m^*(B' \setminus A) = 0.$$
This completes the theorem.
Best Answer
Edit: I have done it without employing Lebesgue's density theorem.
Original post:
As @Salcio suggested, the result is true and may be obtained by using Lebesgue's density theorem. If both $m(E)$ and $m(\mathbb R\setminus E)$ have positive measure, then there exist Lebesgue density points $x$, $y$ for $E$, $\mathbb R\setminus E$ (respectively). So there exists $h>0$ such that both $\frac{m((x-h, x+h)\cap E)}{2h}$ and $\frac{m((y-h, y+h)\setminus E)}{2h}$ are strictly bigger than $\frac{1}{2}$. But then:
$$1<\frac{m((y-h, y+h)\setminus E)}{2h}+\frac{m((x-h, x+h)\cap E)}{2h}$$ $$=\frac{m((y-h, y+h)\setminus E)+m((y-h, y+h)\cap E)}{2h}=\frac{m((y-h, y+h))}{2h}=1,$$
a contradiction.
Now I wonder if there is a more direct proof that does not rely on this theorem.
Edit:
I am going to use the following fact that I mentioned in the question:
Proof: by passing to a bounded subset of $E$ of positive measure we may suppose that $E$ is bounded. In particular, $m(E)<\infty$. There exists a open set $U$ containing $E$ such that $m(U)<\frac{1}{\eta}m(E)$, so $\eta m(U)<m(E)$. Since $E$ is bounded, we may suppose that $U$ is bounded. It is well known that $U$ can be written as a countable union of pairwise disjoint open intervals, $U=\bigcup_{n \in \mathbb N}I_n$ (some but not all of $I_n$'s may be empty). Thus, $\sum_{n \in \mathbb N}\eta m(I_n)=\eta m(U)\leq m(E)=m(E\cap U)\sum_{n \in \mathbb N}m(E\cap I_n)$. This implies that for at least one $n$, $\eta m(I_n)\leq m(E\cap I_n)$ and $m(I_n)>0$ (or we would have the reversed strict inequality between the terms in the edge of the previous expression) as intended. $\square$
Now we proceed to prove the main result. Suppose that $m(E)>0$. It suffices to prove that for every integer $n$, $m(E\cap (n, n+1))=1$. By the hypothesis, all we need to do is to prove that $m(E\cap (0, 1))=1$. We will show that given $\eta\in (0, 1)$, $m(E\cap (0, 1))\geq \eta$, which completes the proof. Fix $\eta$. We will show that given $\epsilon\in (0, 1)$, $m(E\cap (0, 1)\geq \eta-\epsilon$, which completes the proof.
By the claim, there exists a nonempty interval $I$ such that $m(I\cap E)\geq \eta. m(I)$.
Fix $\epsilon$. Fix a rational number $q=\frac{n}{m}$ where $n, m$ are positive integers such that $\frac{1-\epsilon}{m(I)}<q<\frac{1}{m(I)}$.
Write $I=(a, b)$ and, for $i\in \mathbb N$, let $x_i=a+ i\frac{b-a}{m}$ and $I_i=(x_i, x_{i+1})$. All intervals $I_i$ have length $\xi=\frac{b-a}{m}=\frac{m(I)}{m}$ and $I=\dot \bigcup_{i=0}^{m-1}I_i\cup\{x_1, \dots, x_{m-1}\}$. Let $l$ be the common measure of the $m(E\cap I_i)$'s. It follows that, $\mu.m.\xi=\mu. m(I)\leq \mu.m(E\cap I)=\sum_{i=0}^{m-1}m(E\cap I_i)=m.l$, so $\mu \xi\leq l$.
Now for each $i \in \mathbb N$, let $J_i=(i.\xi, i+1.\xi)$. Notice that $1-\xi<q m(I)=n\xi<1$, so $\bigcup_{i=0}^{n-1}J_n\subseteq (0, 1)$ and:
$m(E\cap (0, 1))\geq m(E\cap \bigcup_{i=0}^{n-1}J_i)=n.l\geq \mu.n.\xi>\mu(1-\epsilon)=\mu-\mu\epsilon>\mu-\epsilon$. This completes the proof.