Part a.) First notice that for any set $E \subset \mathbb{R}$, any finite closed covering, $E \subset \bigcup_{k=1}^{N}I_k$, must be a closed set. For any $E \subset \mathbb{R}$, choose an arbitrary convergent sequence, $\lbrace x_n \rbrace \subset E$. By definition, $x_n \to x \in \overline{E}$. Notice also, that $\lbrace x_n \rbrace \subset E \Rightarrow \lbrace x_n \rbrace \subset \bigcup_{k=1}^{N}I_k$. Since $\bigcup_{k=1}^{N}I_k$ is closed, $x_n \to x \in \bigcup_{k=1}^{N}I_k$.
Thus, any finite closed covering of an arbitrary set $E$ must also contain $\overline{E}$. Therefore, $J_*{(E)} = J_*{(\overline{E})}$.
Part b.) Consider $\mathbb{Q} \cap [0,1]$. Certainly, $m(\mathbb{Q} \cap [0,1]) = 0$. However, $\overline{\mathbb{Q} \cap [0,1]} = [0,1]$. The result from part (a) gives $J_*{(\mathbb{Q} \cap [0,1])} = J_*{([0,1])} = 1$.
I am wondering why you just take the closed covering? The problem reads "finite covering" rather than "finite closed covering".
ReplyDeleteThis is a good point. You need closedness to enforce Jordan measure to work. the closed finite covering for the set of rational numbers in [0,1] is [0,1]; in which, by Jordan outter measure, the infimum of the sum from i=1 to 1 (because we only has 1 closed set) of the set [0,1] is equal to 1 (obviously). The Lebeque's measure of rational numbers in [0,1], you can use the open ball argument of epsilon/2^j and Theorem 3.2 in the book (states that if we have a countable union of measurable sets then the measure of the union of those sets is exactly the sum of the individual set measure) which allows you to drive the measure of set E to zero as desired.
DeleteBur how
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