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I have proved the following statement and I would like to know if my proof is correct and/or/if it can be improved, thanks:

"Prove that there exists a bounded set $A\subset\mathbb{R}$ such that $|F|≤|A|−1$ for every closed set $F\subset A$"

My proof:

We know that a set $A\subset\mathbb{R}$ is Lebesgue measurable iff for every $\varepsilon>0$ there exists $F\subset A$ closed such that $|A\setminus F|<\varepsilon$ and we want to prove that there exists some $A\subset\mathbb{R}$ such that $|F|\leq |A|-1$ i.e. such that $|A|-|F|=|A\setminus F|\geq 1$ so such an $A$ cannot be Lebesgue measurable. Let now $V\subset [-1,1]$ denote the Vitali set: this set must have positive Lebesgue measure, $|V|>0$ (for if it were $0$ it would be measurable, a contradiction) and if we consider the set $A:=\frac{2}{|V|}V=\{\frac{2}{|V|} x:x\in V\}\subset [-\frac{2}{|V|},\frac{2}{|V|}]$ we have that $|A|=|\frac{2}{|V|}|V||=\frac{2}{|V|}|V|=2$.

Now, let $E$ be a Borel subset of $A$. Suppose that $|E|>0$: then $\bigcup_{r\in\mathbb{Q}, r\in [-1,1]} (E+r)\subset [-\frac{2}{|V|}-1,\frac{2}{|V|}+1]$ so $|\bigcup_{r\in\mathbb{Q}, r\in [-1,1]} (E+r)|\leq 2(\frac{2}{|V|}+1)$ but being a union of disjoint sets with $|E+r|=|E|>0$ it is also $|\bigcup_{r\in\mathbb{Q}, r\in [-1,1]} (E+r)|=\infty$, contradiction. Thus if $E\subset A$ is a Borel set it must be $|E|=0$ which implies that for any closed set $F\subset A$ it must be $|A\setminus F|=|A|-|F|=|A|=2\geq 1$, as desired. $\square$

Credits: Measurable subset of Vitaly set has measure zero. Proof. for the proof that every measurable subset of the Vitali set has Lebesgue measure $0$.

lorenzo
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    Are you using $|A|$ to denote the outer measure then? – Brian Moehring Aug 28 '21 at 19:20
  • @BrianMoehring Thank you for your interest in my question; yes, I am using $|A|$ to denote the outer measure of set $A$ – lorenzo Aug 28 '21 at 19:37
  • What is your notation for inner measure, then? Usually I see $\lambda^$ and $\lambda_$ for the outer and inner Lebesgue measures, respectively, leaving $|\cdot |$ for a measure on a measure space. – Brian Moehring Aug 28 '21 at 19:53
  • @BrianMoehring I am using the convention from Axler's MIRA book, where $\mu$ denotes a generic measure on a measure space and $|\cdot|$ is used for outer measure. I haven't encountered the concept of inner measure yet (nor is it explained in the chapter where this exercise is from) – lorenzo Aug 28 '21 at 19:55
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    Fair enough. To elaborate on why I'm asking, the question directly relates to the theorem "$A$ is not measurable if and only if the inner measure of $A$ is strictly less than the outer measure of $A$". Using that theorem along with a common characterization of inner measure, the proof is a short three or four lines long, but without the inner measure, that proof would be inappropriate. – Brian Moehring Aug 28 '21 at 20:02
  • @BrianMoehring Ah I see, interesting. Thanks for the information, I will look up the concept of inner measure and the proof! – lorenzo Aug 28 '21 at 20:04

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Let $V \subseteq [0, 1]$ be such that $V$ is not Lebesgue measurable. Therefore, there exists $\epsilon_0 > 0$ such that for every closed $F \subseteq V$, $|V \setminus F| = |V| - |F| \geq \epsilon_0$.

Next, let $t > 0$ be such that $t\epsilon_0 > 1$. Then, it follows that $tV$ is not Lebesgue measurable. For if it were Lebesgue measurable, then $V = \frac{1}{t}(tV)$ would be Lebesgue measurable (dilation invariance of Lebesgue measurable sets). Moreover, $tV \subseteq [0, t]$, hence it is bounded.

Now, if $F' \subseteq tV$ is closed, then $F = \frac{1}{t}F'$ is closed (dilation invariance of closed sets) and $F \subseteq V$. Therefore, \begin{align*} |tV| - |F'| = |tV| - |tF| = |t|(|V| - |F|) \geq t\epsilon_0 > 1 \end{align*} as desired.

Sid Malik
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