Three simple questions on measure theory A couple of days ago, I posted this question, which remains unanswered. The comments that I have received make me think that this is so because it is ill-defined. Hence, I post this other question with the hope of understanding the basics of measure theory so I can go back to the other question and edit it accordingly.
Let $(\mathbb{R}^n,\mathcal{A})$ be any measurable space. Then,

*

*$X\in\mathcal{A}\Leftrightarrow X$ measurable.

Let $(\mathbb{R}^n,\mathcal{L})$ be a measurable space, where $\mathcal{L}$ is the Lebesge $\sigma$-algebra. Then,


*There exist two measures $\mu:\mathcal{L}\to\mathbb{R}$ and $\mu':\mathcal{L}\to\mathbb{R}$ satisfying $\mu\neq\mu'$.

Let $(\mathbb{R}^n,\mathcal{L},\lambda)$ be a measure space where $\lambda$ is the Lebesgue measure. Then,


*$X\subset\mathbb{R}^n$ infinite and bounded $\Rightarrow \lambda(X)\in(0,\infty)$.

Are (1), (2) and (3) true statements? I think that (1) is true by definition of $\sigma$-algebra. I think that (2) is also true because as shown in the accepted answer to this question, one can define a measure that is not the Lebesgue measure on $(\mathbb{R}^n,\mathcal{L})$. For (3), I do not have an intuition and I'm not sure whether it's true.
Thank you all!
EDIT! As noted in the comments, (3) is false. Hence, let me replace (3) for (3'). Let $(\mathbb{R}^n,\mathcal{L},\lambda)$ be a measure space where $\lambda$ is the Lebesgue measure. Then,
3'. $X\subset\mathbb{R}^n$ infinite, uncountable and bounded $\Rightarrow \lambda(X)\in(0,\infty)$.
 A: *

*In a measurable space $(Y,\mathcal A)$, the term $\mathcal A$-measurable sets indicates the elements of $\mathcal A$, and obviously the term is shortened to measurable sets when there isn't any ambiguity.


*There are always at least two measures on a non-empty measurable space $(Y,\mathcal A)$: the zero measure and the restriction to $\mathcal A$ of the counting measure, namely $$\mu(X)=\begin{cases}\operatorname{card} X&\text{if }X\text{ is finite}\\ \infty&\text{if }X\text{ is an infinite set}\end{cases}$$


*Some bounded sets aren't Lebesgue-measurable, therefore discussing their Lebesgue measure is somewhat idle talk (it's still possible to talk about their Lebesgue outer measure, but I digress). For Lebesgue-measurable subsets, $A\subseteq B$ implies $\lambda(A)\le\lambda(B)$ and therefore, since $\lambda(a,b)=b-a$, bounded Lebesgue-measurable subsets have finite Lebesgue measure. Obviously, countable additivity makes it so that countably infinite subsets are Lebesgue-null sets but, more to the point, Cantor's set is a compact and uncountable Lebesgue-null set.
