Multiple statements on the Lebesgue measure I try to prove or disprove some statements on the Lebesgue measure $\lambda$ on the Borel $\sigma$-algebra $\mathfrak{B}(\mathbb{R})$. And I want to know if my thoughts are right:
Let $A \in \mathfrak{B}(\mathbb{R})$


*

*If $B\subseteq A$ then $B \in \mathfrak{B}(\mathbb{R})$


False: From the fact that $\mathfrak{B}(\mathbb{R})\neq P(\mathbb{R})$ we know that there exist sets $A\subseteq \mathbb{R}$ which are not measurable.


*If $\lambda(A) = \infty$ then A is an unbounded set.


True: For every measurable set $A$ we can find a compact interval $I$ with $I \subset A$ with $sup\{\lambda(I)\}=\lambda(A)$. $I$ is of the form $[a,b]$ but by the definition of the Lebesgue measure we have $\lambda(I)=|b-a|=\infty$ which can only be true if $a=\infty$ or $b=\infty$


*If $\lambda(A)<\infty$, then $A$ is a bounded set.


True: for every bounded set we find a finite Interval $I$ with $A\subset I$ with $\lambda(I)<\infty$


*If $\lambda(A)=0$ then $A$ is bounded


False: $\lambda(\mathbb{Q})=0$ but $\mathbb{Q}$ is infinte and not bounded.


*If $A$ is an open set then $\lambda(A)>0$


False: $\lambda(\emptyset)=0$ and $\emptyset$ is open.


*If $\lambda(A) > 0$ then $A$ has a non-empty interior.


True: For any $\lambda(A)>0$ we find a compact interval $I$ with $I\subseteq A$ with $\lambda(I)>0$


*If $\lambda(A \cap ]0,1[)=1$ then $A \cap ]0,1[$ is dense in $]0,1[$.


I don't know.
Thank you.
 A: 1) Correct
2) It's false, but the justification is not really clear. You can very easily prove that if $A$ is bounded, then $\lambda(A)<\infty $. Therefore, the contrapositive is also true.
3) No, it's false. For example $\mathbb Q$.
4) Strange that you answered correct here but not at 3)
5) correct.
6) No, it's false. You can construct a set $C$ as cantor set but such that $\lambda(C)>0$. And this set has empty interior. But if there is a $a>0$ s.t. $\lambda(A)>a$, then the interior is non empty. The set $\mathbb R\backslash \mathbb Q$ work as well (as suggested by daw)
7) Yes, it's true. Let $B=A\cap ]0,1[$. Suppose by contradiction that $B$ is note dense in $]0,1[$. Then, there is an $x\in (0,1)$ such that $$\underbrace{]x,\varepsilon,x+\varepsilon[}_{=:I_{x,\varepsilon}}\cap B=\emptyset.$$
Since $(0,1)$ is open, you can take $\varepsilon$ small enough to have $I_{x,\varepsilon}\subset (0,1)$. In particular, $$(0,1)=I_{x,\varepsilon}\cup(0,x-\varepsilon]\cup [x+\varepsilon,1)$$ and since $B\subset (0,1)$, you get that $$B\subset (0,x-\varepsilon]\cup [x+\varepsilon,1)$$
in other word, $$\lambda(B)\leq x-\varepsilon+1-x-\varepsilon=1-2\varepsilon<1$$ 
which is a contradiction with $\lambda(B)=1$. Therefore $B$ is dense in $(0,1)$.
