It is asked to verify that for any $a \in \mathbb{R},$ $\lambda(\mathbb{\{a\}} \times \mathbb{R}^{n-1}) = 0.$

Preliminary : 1) Define $$\lambda(I) = (b_1 - a_1)...(b_n - a_n)$$ where $I = [a_1, b_1] \times ... \times [a_n, b_n].$ Call $I$ special rectangle.

2) Outer measure $$\lambda^*(A) = \inf\{\lambda(G): G \ \mbox{is open and} \ A \subseteq G\}$$ where $\lambda(G) := \sup\{\sum_{i=1}^N \lambda(I_i): \mbox{where} \ N \in \mathbb{N}, \cup_{i=1}^N I_i \subseteq G \ \mbox{and} \ I_i \ \mbox{special rectangle with disjoint interior} (I^o_i \cap I^o_j = \phi \ \mbox{for any} \ i \neq j)\}.$

3) It can be shown that $$\lambda^*(A) = \inf\{\sum_{k=1}^\infty \lambda(I_k): A \subseteq \cup_{k=1}^\infty I_k\},$$ where each $I_k$ is a special rectangle.

Sol 1) Using the equivalent definition in 3), let $A = \{a\} \times \mathbb{R}^{n-1}$ and $I_k = [a,a] \times [-k, k]^{n-1}$, $$A \subseteq \cup_{k=1}^\infty I_k, \sum_{k=1}^\infty \lambda(I_k) = 0.$$ So $\lambda(A) = 0.$

2) Using the initial definition of $\lambda^*$, let $G$ be an open set such that $A \subseteq G$. Since $G \subseteq \mathbb{R}^n$, $G$ is a union of open balls. I am not sure how to proceed form this point. (Basically, if $G = (a-\epsilon, a+\epsilon) \times \mathbb{R}^{n-1}$, then $\lambda(G) = \infty$. So to show that $\lambda(A) = 0$, must be other open set $G$ with measure $0$. However, I cannot come up with such $G$. Any help please ?)

  • $\begingroup$ Just use the rectangles $R_k=(a-\epsilon/(2^{n-1}k^{n-1}2^k),a+\epsilon/(2^{n-1}k^{n-1}2^k))\times (-k,k)^{n-1}$. Their union is open and covers $\{a\}\times\mathbb{R}^{n-1}$ and the sum of their measures is $\leq \epsilon$. Then let $\epsilon \to 0$. $\endgroup$ – EEE Oct 22 '17 at 13:54
  • $\begingroup$ But $\lambda(G)$ is the supremum of the finite union of special rectangles that contained in $G$. I am not sure if that is enough to show that there exist an open set $G$ such that $G$ contains $A$ and $\lambda(G) = 0.$ $\endgroup$ – Both Htob Oct 22 '17 at 13:59
  • $\begingroup$ You can't show that. The measure of an open set is not zero. What you need to show are open sets with arbitrarily small measure, like $\epsilon$ for any $\epsilon>0$. $\endgroup$ – EEE Oct 22 '17 at 14:01
  • $\begingroup$ Oh, ok. You mean $A \subset \cup R_k$. Then show that $$\sum \lambda^*(R_k)\leq \epsilon$$ right ? $\endgroup$ – Both Htob Oct 22 '17 at 14:04

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