Jordan Measure and Lebesgue Measure

Question

The Jordan outer measure $J^*(E)$ of a set $E\subseteq \mathbb{R}$ is defined as infimum of $\sum_{i=1}^n (b_i-a_i)$ where $(a_i,b_i)$ are open intervals whose union contains $E$. The Jordan inner measure $J_*(E)$ of a set $E\subseteq \mathbb{R}$ is defined as supremum of $\sum_{i=1}^n (b_i-a_i)$ where $(a_i,b_i)$ are open intervals, whose union is contained in $E$. A set is $E$ Jordan measurable if $J^*(E)=J_*(E)$.

Lebesgue measure of a set $E\subseteq \mathbb{R}$ is defined in a similar way by defining Lebesgue outer measure and inner measure, where the sums/unions in above definition are allowed to be countable.

Question: What properties of functions can be characterized by the Lebesgue measure but not the Jordan measure?

(I want a motivation of Lebesgue measure with some drawback/disadvantages of Jordan measure. I didn't find theory of Jordan measure in many books of Measure theory, although it was a motivational point towards development of Lebesgue measure and Integration.)

Answer 1

Some drawbacks (assumed for convenience to be sited in $\mathbb{R}^n$): Not all compact sets are Jordan measurable, e.g. certain generalised Cantor sets in $\mathbb{R}$ with positive Lebesgue measure; no characterisation of Riemann integrable functions; assigns non-measurability to too many sets ("$A$ is Jordan measurable iff $\partial A$ has Jordan measure $0$") thereby making it a less discriminating estimate of the "size" of sets than Lebesgue measure; lacks the powerful integral convergence theorems of Lebesgue theory.

Nevertheless, see the very interesting Multidimensional Analysis vol.2 by Duistermaat & Kolk, which pushes the Jordan theory quite far and includes the Arzela integral theorem as a practical alternative for the dominated convergence theorem.

Answer 2

Question: What properties of functions can be characterized by the Lebesgue measure but not the Jordan measure?

Answer: Jordan measure(which actually is NOT measure, though the name says “measure”. So confusing term) is very weak concept coampred to Lebesgue measure. Firstly, Jordan measure of a set E in R^d always is to be +∞ if E is unbounded. This is very unsatisfactory because it implicitly says integral with Jordan measure always has to deal with unbounded set as +∞. So whatever underlying sets and functions you try to integrate, the value can highly be +∞ as long as underlying set or range of function contains unbounded set. You can see this is why we were very nervous about doing "irregular integrals". Secondly, even if E is bounded closed set or compact, there’s an counterexample of non-Jordan measurable set. The famous one is (Q∩[0,1]). Judging from (at least) the two drawbacks of Jordan measure, one can intuitively say “Jordan measure” is very fragile for underlying sets with 1. some kinds of unboundeness 2. very intricated like fractal to define an integral theory which is stronger enough to integrate functions equipped with the above properties.

The theory of Lebesgue integral is sometimes called the completion of “Remmal integral”. I highly recommend you to read “An introduction to Measure theory” written by Terrence Tao.

Answer 3

My two cents:

Question: What properties of functions can be characterized by the Lebesgue measure but not the Jordan measure?

I'll assume the "functions" you refer to are the "Riemann integrable functions", for the Jordan measure is defined on terms of the Riemann integral: if the indicator function $\chi_{X}:A\rightarrow\{0,1\}$ of a set $X\subseteq\mathbb{R}^{n}$ is integrable in the $n$-box $A\supseteq X$ (a cartesian product of $n$ intervals) then we say $X$ is Jordan measurable and we define its volume to be $$\mathrm{vol.}X=\int_{A}\chi_{X}(x)dx.$$ The finite union an intersection od Jordan measurable sets are Jordan measurable, as well as the difference of two given Jordan measurable sets and the complement $X^{c}$ with respect to a $n$-box $A$ which contains $X.$ Note that I said finite: passage to countable unions/intersections does not hold, the previous answer cites $\mathbb{Q}\cap[0,1]$ as the classical example. In general, no dense set of $\mathbb{R}^{n}$ whose complement is also dense is Jordan measurable. This is an issue, but is a $\textit{measure}$ issue. You asked about the issues with Riemann integrable functions which are fixed by Lebesgue integration.

A function is Riemann integrable on a set if it doesn't change values too abruptly in it, which means the set of points in which the function does oscilate too much has Lebesgue measure zero (can be covered by a countable union of $n$-boxes which total volume is less than a given $\varepsilon>0$). Problem: this is not at all preserved under limits. Only way to guarantee that the limit of a sequence of integrable functions is integrable is if this convergence ignores completely what happens around the values $f_{n}(x)$ and $f(x)$ - the sequence converges uniformly. So we naturally go looking for a better tool for dealing with limits.

If you're not convinced, also realll the continuity matter. Equivalently to what I said above, a function is Riemann integrable iff its set of discontinuities has Lebesgue measure zero, which is not the case for the indicator function of dense sets with dense complement, as I mentioned. So if we wanna improve things, we must get rid of continuity, even up to a set of measure zero, as an integrability criterion. Then it comes down to integration being essentialy about measuring regions through splitting it in bits and then joining them again. You do this the Lebesgue way (partitioning the range, thus inducing a partition in the domain itself, while ignoring whatever continuity issues in there), you don't need to care at all about the continuity of the function to be integrated.

There's also the need for Jordan sets and the Riemann integrable functions to be bounded, which means you can't ("properly") integrate over sets as $(a,+\infty),$ or functions like the derivative of $$f(x)=\begin{cases}x^{2}\sin\left({\frac{1}{x^{2}}}\right),\ \text{if}\ x\neq0\\ 0,\ \text{if}\ x=0\end{cases}$$ which is unbounded in $[-1,1].$ Also, That means we can produce derivatives that are not integrable (let it be because of unboundedness of because of a set of discontinuities with positive measure), even though we have their primitives on our hands. There goes the fundamental theorem of calculus (ftc). Lebesgue Differentiation Theorem on the other hand, a generalized version of ftc for the Lebesgue integral, asks only for the integral of the function the be finite (not the function itself, only its integral).

From the top of my head, this is it. Most of it comes down to the ease of integral approximation and getting rid of continuity all together, really. If we wanna sum it up: Riemann integrability really needs well behaved functions and sets, whereas Lebesgue integrability does not require neither.