# Real Analysis, Folland Problem 1.5.29 Lebesgue measurable set

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.

• What is the definition of $N_r$? It seems to be a nonmeasurable set in part (a) but a measurable set in part (b). Jun 17, 2016 at 19:43
• $N_r$ is defined in the same way as $E_r$. Jun 17, 2016 at 19:59
• isnt b just the contrapositive of a? Apr 9, 2019 at 12:54
• @yoshi I believe so Apr 9, 2019 at 15:42

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.