Jordan measure and Riemann integral

Question

I am studying Riemann integral, and I haven't understood what is the relation between the concepts of Riemann integral and Jordan measure. Is Riemann integral a special case of Jordan Measure? What is the precise definition of Jordan Measure? Thank so much to who will help me!

Answer 1

Recall that for a bounded subset $A$ of $\mathbb{R}^d$, we define the Jordan upper approximation by:

$$J^{R_n}(A) = \sum_{R_n \cap A} \mathrm{Vol}(R_n)$$

Here the indexing is over all rectangles that intersect $A$, where $R_n$ is a finite set of $d$-dimensional rectangles that cover $A$. We can see why this is called the upper approximation: you cover your set with things that may have more mass than it. You get extra mass, but you don't miss any either. Finally, to find the upper approximation of this, you take a limit by taking smaller and smaller covers, i.e we define:

$$ J^* (A) = \inf\{J^{R_n}(A) : R_n \text{ is a finite rectangle cover of $A$}\} $$ Similarly, we define lower approximations by: $$ J_{R_n}(A) = \sum_{R_n \subset A} \mathrm{Vol}(R_n) $$ Here, each rectangle is a subset. So now we are measuring the size from the inside, and hence the true size should be what we get when we take successively larger volumes, i.e:

$$ J_* (A) = \sup\{J_{R_n}(A) : R_n \subset A\} $$

The Jordan measure is defined when: $$ J^* (A) = \inf\{J^{R_n}(A) : \{R_n\}_{n=1}^{N} \text{ is a finite rectangle cover of $A$}\} = J_* (A) = \sup\{J_{R_n}(A) : \{R_n\}_{n=1}^{N}\subset \mathcal{P}(A), \text{i.e. each } R_n \subset A\} $$ i.e when the lower approximations converge to the same limit as the upper ones. Let's try now and see how we could calculate the volume of a set $A$ through integration.

Define the indicator function of a set $\mathbf{1}_A$ as: $$ \mathbf{1}_A = \begin{cases}1 & x\in A \\ 0 & x \not \in A \end{cases} $$ This function "indicates" when a point is in a set or not. Hence, a natural definition for the volume of a set in terms of integration is: $$ \mathrm{vol}(A) = \int_{R} \mathbf{1}_A(x) \mathrm{d} x $$ where $R$ is a rectangle that contains $A$ (as $A$ is bounded). This is true provided the integral converges. Let's show now that if $A$ is jordan measurable, then this integral converges and its value is whatever the jordan measure of $A$ is. First, lets take the upper and lower sums of the integral. Recall that the upper sums for a partition $P$ of our domain $R$ and our function $\mathbf{1}_A$ (cutting $R$ up into smaller rectangles) is given by: $$ U(\mathbf{1}_A, P) = \sum_{n} \sup_{x \in R_n} \mathbf{1}_A (x)\mathrm{Vol}(R_n) $$ The lower sums are given by: $$ L(\mathbf{1}_A, P) = \sum_{n} \inf_{x \in R_n} \mathbf{1}_A (x)\mathrm{Vol}(R_n) $$ Note that by the definition of our function $\mathbf{1}_A$, a couple things can happen. If for a rectangle $R_n$ in our partition, we have that $R_n \subset A$, then $\inf_{x \in R_n} \mathbf{1}_A = 1$, because every point in $R_n$ is in $A$ and so the indicator function is $1$ everywhere. Similarly, if we see that one of our rectangles $R_n$ intersects $A$, then $\sup_{x \in R_n} \mathbf{1}_A(x) = 1$, as the maximum value is $1$ and it is attained. Hence, we can rewrite the Riemann sums as follows: $$U(\mathbf{1}_A, P) = \sum_{R_n \cap A}\mathrm{Vol}(R_n) = J^{R_n}(A)$$ And similarly for the lower sums:
$$ L(\mathbf{1}_A, P) = \sum_{R_n \subset A}\mathrm{Vol}(R_n) = J_{R_n}(A) $$ So the, upper and lower Riemann sums of the integral of the indicator function of $A$ are the upper and lower Jordan sums! And hence we see that the inf of the upper Riemann sums exists, because it is the inf of the upper Riemann sums, and similarly for the lower Riemann sums. Hence, the jordan content of the set is the Riemann integral of the indicator function of that set!

This was a bit of a long explanation so please let me know if anything needs clarification.