Motivation: I am currently in a rather uncomfortable spot in my Analysis studies. In class we introduced the Jordan measure in a very vague way, meaning no proofs, no examples. (Because next Semester we study the better Lebesgue-Measure theory)

I do understand the concepts of it as a generalization of the 1 Dimensional case.

That is, approximating a given bounded set $\Omega\subset \mathbb{R}^n$ from above and below by rectangles (boxes, simple sets) and if it happens that the smallest of all the approximations from above is equal to the largest of all the approximations from below, we say by definition that $\Omega$ is Jordan measurable.

However when it comes to applying said intuitive idea I run into troubles

Problem: Show that the set $\Omega := \lbrace (u,v) \in \mathbb{R}^2 \mid |u| + |v| \leq 1 \rbrace \subset \mathbb{R}^2$ is Jordan measurable.

My approach: Clearly the set is bounded, from below by $0$ and from above $1$. I don't know how to work with the definition as explained above in the motivation to show that said set is Jordan measurable.

The best thing I could do was to cheat my way around by computing the following picture:

visualisation of the set

The red line shows the boundary $\partial \Omega$ of the set. Since it is only a line segment in $\mathbb{R}^2$ it has Jordan measure zero respective to the topology in $\mathbb{R}^2$.

According to How to prove $E\subset R^n$ Jordan measurable is equivalent to $\bar{E}-E$ is Jordan measured null this would 'show' that $\Omega$ is Jordan measurable.

My question is if I can formalize this idea into an actual proof rather than my hand wavy explanation above.

Or even better, is it possible to show that $\Omega$ is Jordan measurable by just relying on the definition of above/below approximations and check the equality of the infimum and supremum? Because said definition is so far what I understand best, the iff statement linked is only covered in the $\implies$ direction in my course.


2 Answers 2


The segment $\sigma$ connecting $(0,1)$ with $(1,0)$ can obviously be covered by $N$ little squares of area ${1\over N^2}$ each. It follows that the boundary of your set $\Omega$ can be covered with $4N$ little squares of total area ${4\over N}$. The latter number can be made arbitrarily small.

  • $\begingroup$ If I understand this correctly then the line segment can be covered by $N$ little squares of area $1/N^2$ and it follows that the boundary $\partial \Omega$ can be covered by $4N$ such little squares. I hope I am not mistaking this. $\endgroup$
    – Spaced
    May 10, 2015 at 19:23

Let $j$ denotes the Jordan measure. Then consider the partitions of the plane $Q_n = \{ [i{1 \over 2^n},(i+1){1 \over 2^n}]\times [j{1 \over 2^n},(j+1){1 \over 2^n}))\}_{i,j \in \mathbb{Z}}$.

Let $A_n = \{ R \in Q_n| R \subset \Omega^\circ \}$, $B_n = \{ R \in Q_n| R \cap \Omega \neq \emptyset\}$. We have $A_n \subset B_n$ and $\alpha_n=\cup_{R \in A_n} R \subset Q_n \subset \beta_n = \cup_{R \in B_n} R$. Furthermore $\alpha_n$ and $\beta_n$ are increasing.

Some tedious work shows that $|B_n \setminus A_n| \le 8 \cdot 2^n$ and so $j(\beta_n \setminus \alpha_n) \le 8 \cdot 2^n {1 \over 2^{2n}}$.

Since $j(\beta_n) = J(\alpha_n)+j(\beta_n \setminus \alpha_n)$, a little more work shows that $\Omega$ is Jordan measurable.


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