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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,

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

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

  1. 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,

  1. $X\subset\mathbb{R}^n$ 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 $\Rightarrow \lambda(X)\in(0,\infty)$.

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    $\begingroup$ For (3) consider $\mathbb{Q}\cap[0,1]$ $\endgroup$
    – morrowmh
    Mar 13, 2021 at 18:50
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    $\begingroup$ $1$ depends on your notion of measurable, unfortunately it is not a canonical notion (sometimes it depends on the extension to a complete space). $2$ and $3$ are obviously true. $\endgroup$
    – John B
    Mar 13, 2021 at 18:51
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    $\begingroup$ The Lebesgue measure of a countable subset of $\mathbb{R}$ is zero. Thus $\mathbb{Q}\cap[0,1]$ is a countably infinite and bounded subset of $\mathbb{R}$ with Lebesgue measure zero. This contradicts (3). $\endgroup$
    – morrowmh
    Mar 13, 2021 at 18:58
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    $\begingroup$ In that case it can have positive measure. Take $[0,1]$ for example. It has Lebesgue measure $1$, and it is an uncountable bounded subset of $\mathbb{R}$. $\endgroup$
    – morrowmh
    Mar 13, 2021 at 19:01
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    $\begingroup$ It can also have zero measure. Google "cantor set". $\endgroup$
    – morrowmh
    Mar 13, 2021 at 19:03

1 Answer 1

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  1. 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.

  2. 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 $$\mu(X)=\begin{cases}\operatorname{card} X&\text{if }X\text{ is finite}\\ \infty&\text{if }X\text{ is an infinite set}\end{cases}$$

  3. 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.

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