Real Analysis, Folland Problem 1.5.29 Lebesgue measurable set

Question

1.5.29 - Let $E$ be a Lebesgue measurable set.

a.) If $E\subset N$ where $N$ is the nonmeasurable set described in section 1.1, then $m(E) = 0$.

b.) If $m(E) > 0$, then $E$ contains a nonmeasurable set. (It suffices to assume $E\subset [0,1]$. In the notation of section 1.1, $E = \bigcup_{r\in R}E\cap N_r$).

Proof a.) Let $E\subset N$ and $N\subset [0,1)$. In the notation of section 1.1, let $$E_r = \{x + r: x\in E\cap [0,1-r)\}\cup \{x+r-1:x\in E\cap [1-r,1)\}$$ Then for all $r\in R = \mathbb{Q}\cap [0,1)$, since $E_r \subset N_r$ and the collection $\{N_r\}_{r\in R}$ is a disjoint collection, it follows that $\{E_r\}_{r\in R}$ is a disjoint collection. By the translation invariance and finite additivty of Lebesgue measure, we have that each $E_r$ is measurable and $m(E_r) = m(E)$ for all $r\in\mathbb{R}$. So $$1\geq m\left(\bigcup_{r\in R}E_r\right) = \sum_{r\in R}m(E_r) = \sum_{r\in R}m(E)$$ and therefore $m(E) = 0$? Not sure if this is right.

Proof b.) Suppose that $E\subset [0,1]$ is a subset with the property that $E$ is measurable. Then for each $r\in R$, the set $E\cap N_r$ is measurable. By the translation invariance and finite additivity of Lebesgue measure, the set $E_{1-r}\cap N$ is therefore measurable, and hence must have measure $0$ by part a.). Since the collection $\{N_r\}_{r\in R}$ is disjoint we have that $$m(E) = m\left(\bigcup_{r\in R}(E\cap N_r)\right) = \sum_{r\in R}m(E\cap N_r) = \sum_{r\in R}m(E_{1-r}\cap N) = 0$$

Not sure if this is right either. Any suggestions on these is greatly appreciated.

Answer 1

Your proof of part a.) is correct. I copy it here just to add a small comment.

On the other hand, your proof of part b.) needs some corrections.

1.5.29 - Let $E$ be a Lebesgue measurable set.

a.) If $E\subset N$ where $N$ is the non-measurable set described in section 1.1, then $m(E) = 0$.

b.) If $m(E) > 0$, then $E$ contains a nonmeasurable set. (It suffices to assume $E\subset [0,1]$. In the notation of section 1.1, $E = \bigcup_{r\in R}E\cap N_r$).

Proof a.) Let $E\subset N$ and $N\subset [0,1)$. In the notation of section 1.1, let $$E_r = \{x + r: x\in E\cap [0,1-r)\}\cup \{x+r-1:x\in E\cap [1-r,1)\}$$ Then for all $r\in R = \mathbb{Q}\cap [0,1)$, since $E_r \subset N_r$ and the collection $\{N_r\}_{r\in R}$ is a disjoint collection, it follows that $\{E_r\}_{r\in R}$ is a disjoint collection. By the translation invariance and finite additivty of Lebesgue measure, we have that each $E_r$ is measurable and $m(E_r) = m(E)$ for all $r\in\mathbb{R}$. So $$1\geq m\left(\bigcup_{r\in R}E_r\right) = \sum_{r\in R}m(E_r) = \sum_{r\in R}m(E)$$

Since $R$ is countable infinite, we have that $m(E)=0$.

Proof b.) If $S$ is any set and $S \subset N_r$, for some $r\in R$, then, let $$S_{-r}=\{x-r : x \in S \cap [r,1)\}\cup \{x-r+1 : x \in S \cap [0,r)\}$$ Since $S \subset N_r$, we have $S_{-r} \subset (N_r)_{-r} =N$.

So, if $F$ is measurable and $F \subset N_r$, for some $r\in R$, we have that $F_{-r}$ is measurable, $\mu(F)=\mu(F_{-r})$ and $F_{-r} \subset N$. From part a.) we get $\mu(F_{-r})=0$ and so $\mu(F)=0$.

So we have proved that, if $F$ is measurable and $F \subset N_r$, for some $r\in R$, then $\mu(F)=0$.

Now, suppose that $E\subset [0,1]$ is a subset with the property that $E$ is measurable and $\mu(E)>0$. Suppose, that for all $r \in R$, $E \cap N_r$ are measurable. Since $E \cap N_r \subset N_r$, we have $\mu(E \cap N_r)=0$.
But, since $[0,1)$ is the disjoint union of $N_r$'s, we have $E\cap [0,1)$ is the disjoint union of $E \cap N_r$'s. So we get

$$ 0< \mu(E)=\mu(E\cap [0,1))=\sum_{r \in R} \mu(E \cap N_r)= 0$$

Contradiction. So, there is $r \in R$, such that $E \cap N_r$ is not measurable. So $E$ contains a non-measurable set.