$S$ be a non empty lebesgue measurable subset of $\mathbb{R}$ such that every subset of $S$ is measurable, Then measure of $S$ equal to

$1.$ measure of any bounded subset of $S$.

$2.$ measure of any closed subset of $S$

$3.$ measure of any subset of $S$

$4.$ measure of any countable subset of $S$


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Do you know the construction of the vitali set?

Just assume that $\mu(S)\neq 0$ and use the vitali set construction on $S$.

So you see the only subsets of $\mathbb{R}$ with this proberty are null sets.

A rough sketch for the vitali set:

Define on $[0,1]$ an equivalence relation via $$x\sim y \iff x-y\in \mathbb{Q}$$ Let $(x_\alpha)_{\alpha \in I}$ be a complete representative system of $[0,1]$, meaning you have exactly one representative of every equivalence class.

Now we define the set $$N:=\bigcup_{\alpha \in I} \{x_\alpha\}$$

Furthermore we know that $[0,1]\cap \mathbb{Q}$ is countable so we take an enumeration $(v_n)_{n\in \mathbb{N}}$. In addition we define $$N_n:= N+v_n =\{ x\in\mathbb{R}: x=y+v_n \text{ with } y\in N\}$$ and see that $$\mu(N_n)=\mu(N)$$ as the Lebesgue measure is translation invariant.

From the construction we see that $$ [0,1]\subseteq \bigcup_{n\in \mathbb{N}} N_n \subseteq [-1,2],$$ hence $$1 \leq \mu\Big( \bigcup_{n=1}^\infty N_n\Big)\leq 3.$$ On the other hand the $N_n$ are pairwise disjoint meaning $N_n \cap N_m=\varnothing$ when $n\neq m$.

From the $\sigma$-additivity we know that $$\mu\Big( \bigcup_{n=1}^\infty N_n\Big)=\sum_{n=1}^\infty \mu(N_n)$$ and as $\mu(N_n)=\mu(N)$ we sum over a constant.

Now there are 2 cases: $\mu(N)=0$, then the sum will be $0$ and not greater than $1$.

When $\mu(N)>0$, then the sum will be $\infty$ and not lower than $3$, hence $N$ can't be measurable

  • $\begingroup$ I don't know the construction of vitali set $\endgroup$ – Marso Aug 27 '13 at 6:28
  • $\begingroup$ Do you know any non measurable set? $\endgroup$ – Dominic Michaelis Aug 27 '13 at 6:28
  • $\begingroup$ Do you know the theorem of Caratheodory that every subset of $\mathbb{R}$ with positive measure has a non-measurable subset? $\endgroup$ – André Nicolas Aug 27 '13 at 6:30
  • $\begingroup$ No sir, I don't know that too $\endgroup$ – Marso Aug 27 '13 at 6:31
  • $\begingroup$ @AndréNicolas are there proofs of the Caratheodory theorem not using the Vitali set construction $\endgroup$ – Dominic Michaelis Aug 27 '13 at 6:31

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