Relative measure of a set in all intervals

Suppose $A \subset \mathbb {R}$ is Lebesgue measurable and assume that

for any $a, b \in \mathbb {R}$, $a < b$. Prove that $m(A) = 0$.

Prove that there does not exist a Lebesgue-measurable subset $A$ of the real line $\mathbb {R}$ such that for any interval $\Delta$

Let $0<\theta <1, E \subset \mathbb {R}^{m}, 0<\lambda_{m}(E)<\infty$. Prove that there exists a cube $\Delta$ such that $\theta \lambda_{m}(\Delta )<\lambda_{m}(E \cap \Delta )$.

Prove the existence of a set $A \subset [0,1]$ such that $\lambda (A \cap \Delta )>0$ and $\lambda (\Delta \backslash A)>0$ for every (nonempty) interval $\Delta \subset (0,1)$.

Prove that there is no Lebesgue measurable set $A \subset \mathbb {R}$ with the property that for any interval $I$,

Let $E \in \mathcal{M}$ and $0<m\left(E\right)<\infty$. Prove that for every $\alpha \in (0,1)$ there exists an interval $I$ for which $m\left(E \cap I\right)>\alpha m\left(I\right)$.

Let $A$ be a measurable subset of $\mathbb {R}$ and suppose that there is $\alpha >0$ such that $m(A \cap [a, b]) \geq \alpha (b-a)$, whenever $a<b$. Prove that $m\left(A^{C}\right)=0$

Construct a Borel set $E \subset \mathbb {R}^{1}$ such that

for every nonempty segment $I$. Is it possible to have $m(E) < \infty$ for such a set?

Assume the result of Exercise 8 [problem 7.4.8 above]. Show that there cannot exist a set $E \in \mathcal{B}$ such that $\lambda \left(E \cap [a, b]\right)=\frac{1}{2}(b-a)$ for all real $a<b$. [Write the left side as $\int_{[a,b]} \; \mathbb {1}_{E} \, d\lambda$, and write the right side as $\int_{[a,b]} \frac{1}{2} \, d\lambda$. Use the previous exercise, working by contradiction.]

$\uparrow$ There is no Borel set $A$ in $(0,1)$ such that $a \lambda (I) \leq \lambda (A \cap I) \leq b \lambda (I)$ for every open interval $l$ in $(0,1)$, where $0<a \leq b<1$. In fact prove:

Suppose that $A \in \mathscr {R}^{1}, \lambda (A)>0$, and $0<\theta <1$. Show that there is a bounded open interval $I$ such that $\lambda (A \cap I) \geq \theta \lambda (I)$. Hint: Show that $\lambda (A)$ may be assumed finite, and choose an open $G$ such that $A \subset G$ and $\lambda (A) \geq \theta \lambda (G)$. Now $G=\bigcup_{n} I_{n}$ for disjoint open intervals $I_{n}$ [A12], and $\Sigma_{n} \lambda \left(A \cap I_{n}\right) \geq \theta \sum_{n} \lambda \left(I_{n}\right)$; use an $I_{n}$.