lebesgue measurable subset of $\mathbb{R}$ such that every subset of $S$ is measurable, Then measure of $S$ equal to [closed]

$S$ 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 $S$

closed as off-topic by Andrés E. Caicedo, azimut, Paul, TZakrevskiy, Willie WongAug 27 '13 at 17:27

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework questions must seek to understand the concepts being taught, not just demand a solution. For help writing a good homework question, see: How to ask a homework question?." – Andrés E. Caicedo, azimut, Paul, TZakrevskiy, Willie Wong
If this question can be reworded to fit the rules in the help center, please edit the question.

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

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