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The following are the definitions I have been working with,

A set $E \subset \mathbb{R}^n$ has $\textbf{content zero}$ if for every $\epsilon >0$ there exist $J_1,J_2, \dots, J_N$ "blocks" such that $E \subset \bigcup J_i$ and $\sum vol(J_i) < \epsilon$.

$\textbf{Jordan measure}$ of $E$ is $c(E)=$$\int_{E} 1$. A set $E \subset \mathbb{R}^n$ is Jordan measurable if $E$ is bounded and $\partial E$ has content zero.

$\textbf{My Attempt:}$ Show that $E$ having content zero implies that the Jordan measure of $E$ is also zero.

Suppose $E$ has content zero. Then $\forall \epsilon >0$, $\exists$ boxes $\{J_1, \dots, J_N\}$ such that $E \subset \bigcup_{i=1}^{N} J_i$ and $\sum_{i=1}^{N} vol(J_i) < \epsilon$. It is clear that $\int_{E} 1 \le \int_{\bigcup_{i=1}^{N} J_i} 1$. Since $1$ is a constant function, we have \begin{equation*}\int_{\bigcup_{i=1}^{N} J_i} 1=\sum_{i=1}^{N} vol(J_i) < \epsilon \end{equation*}. Thus \begin{equation*} c(E)=\int_{E} 1 < \epsilon \end{equation*}

user7090
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1 Answers1

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This is not true. For example, consider the set of rational numbers $\mathbb{Q}$, which has measure zero but not content zero. To see this set has measure zero, consider the following $\mathbb{Q} \subset \cup^{\infty}_{i = 1} \{A_i\}$, where $A_i = (x_i - \frac{\epsilon}{2^i+1}, x_i + \frac{\epsilon}{2^i+1})$, and each $x_i$ is a rational number. As the rational numbers are countable, that means we can be sure that this infinite union covers $\mathbb{Q}$. The volume of $A_i$ is $v(A_i) = \frac{2\epsilon}{2^{i+1}} = \frac{\epsilon}{2}$. Then $v(\mathbb{Q}) \leq \sum ^{\infty}_{i=1}v(A_i) = \sum ^{\infty}_{i=1}\frac{\epsilon}{2^i} = \epsilon\frac{1}{1-{\frac{1}{2}}} = 2\epsilon, \forall\epsilon$. Thus we have found a cover of \mathbb{Q}, whose volume, $v(Q) = 2\epsilon$, can be made arbitrarily small.

However, there is no way we can find a finite number of sets whose union contains $\mathbb{Q}$, and whose volume is zero. Thus, $\mathbb{Q}$, does not have content zero.

michael_fortunato
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