# Equivalent conditions of Lebesgue measurable sets

Hi I'd appreciate if someone can check the following exercise any suggestions are welcome. Thanks ;)

Let $$A$$ a subset of $${\bf{R}}^d$$ show that the following conditions are equivalent:

(i) $$A$$ is Lebesgue measurable

(ii) $$A$$ is union of a F$$_{\sigma}$$ and a set of Lebesgue measure zero

(iii) There is a set $$B$$ that is a F$$_{\sigma}$$ and satisfies $$\lambda^*(A\triangle B)=0$$

Proof: (i) $$\Rightarrow$$ (ii) Suppose $$\lambda (A)<+\infty$$. For each natural number $$n$$ we can choose a compact set $$K_n$$ such that $$K_n\subset A$$ and $$\lambda(A)-2^{-n}<\lambda(K_n)$$. Let $$K=\bigcup_n K_n$$. Then $$K$$ is a F$$_{\sigma}$$, $$K\subset A$$ and the relation

$$\lambda(K)\ge\lambda(K_n)>\lambda(A)-2^{-n}$$

holds for all $$n$$, so $$\lambda(K)=\lambda(A)$$. Then $$\lambda(A-K)=0$$; furthermore $$A=K\cup A-K$$. Thus the assertion is proved in the case where $$\lambda (A)<+\infty$$.

If $$A$$ is an arbitrary Lebesgue measurable set, then $$A$$ is the union of a sequence $$\{A_n\}$$ of Lebesgue measurable sets of finite Lebesgue measure (since $${\bf{R}}^d$$ is sigma finite, $$A=\bigcup_n A\cap B_n$$, where $${B_n}$$ is a sequence of measurable sets whose union is $${\bf{R}}^d$$ and $$\lambda (B_n)<+\infty$$). For any positive natural number, we have that $$A_n=F_n\cup Z_n$$ where $$F_n$$ is a F$$_{\sigma}$$ and $$Z_n$$ is a set of Lebesgue measure zero. The set $$F$$ and $$Z$$ defined by $$F=\bigcup_n F_n$$ and $$Z=\bigcup_n Z_n$$, satisfies that $$F$$ is F$$_{\sigma}$$, $$Z$$ is a zero set and their union is $$A$$.

(ii) $$\Rightarrow$$ (iii) Follows immediately

(iii) $$\Rightarrow$$ (i) Every F$$_{\sigma}$$ is Lebesgue measurable. From the condition $$\lambda^*(A\Delta B)=0$$, we can derive that $$\lambda^*(A-B)=0=\lambda^*(B-A)$$. So $$A=(B\cup A-B)-(B-A)$$ is Lebesgue measurable (since any null set is Lebesgue measurable).

• Suggestion: Instead of writing $X \cup Y - Z$, write $X \cup (Y - Z)$ [or $X\cup (Y\setminus Z)$]. That doesn't add much clutter, and makes parsing easier since one doesn't need to recall the operator precedence. Jul 21, 2015 at 15:08
• Thanks for the suggestion ;) @DanielFischer Jul 21, 2015 at 15:16