Lebesgue measurable sets which have the same measure when intersected with intervals of the same length have measure $0$ or cozero 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 $m(I\cap E)\geq \alpha m(I)$.
 A: 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:

Given a measurable set $E$ such that $m(E)>0$ and $\eta \in (0, 1)$, there exists a nonempty open interval $I$ such that $m(E\cap I)\geq m(I)$.

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.
