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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).

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    $\begingroup$ 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. $\endgroup$ – Daniel Fischer Jul 21 '15 at 15:08
  • $\begingroup$ Thanks for the suggestion ;) @DanielFischer $\endgroup$ – Jose Antonio Jul 21 '15 at 15:16
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Not much to say except you have done everything correctly. Since you are still learning I would say fill in a detail or two in (ii) to (iii) and (iii) to (i) but the ideas are correct.

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  • $\begingroup$ Thanks. I really appreciate it. A take note of your suggestions. $\endgroup$ – Jose Antonio Jul 21 '15 at 15:16

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