Lebesgue measure vs. simple length on the real line

Question

I have been studying measure on my own using Folland. I can generally follow the math, but I have lost sight of the big picture, at least on the real line. Basically, I am asking myself how Lebesgue measure on the real line is different from our everyday sense of length. What are we getting with Lebesgue measure that we cannot get from the everyday idea of length?

Provided of course that simple length has the property that the length of a countable union of disjoint sets is the sum of the individual lengths, what are the differences between simple length and Lebesgue measure?

Specifically, what are some sets we can measure with Lebesgue measure that we cannot with simple length? What are some sets that Lebesgue measures correctly, that everyday length does not?

Informally speaking, is it best to think of Lebesgue measure as a formalization of everyday length, a generalization of everyday length, or as a “correction” of everyday length?

I would certainly appreciate any comments and insights.

Note: We don’t need Lebesgue measure to calculate the measure of the Cantor set. We just add up the 1/3 parts we take out, which sum to 1, so the parts left (which are the Cantor set) have to measure zero. We can do all that with just simple length, and the every day idea that the length of a set must equal the sum of its disjoint parts.

Similarly, we don’t need Lebesgue measure to calculate the measure of the rationals since that measure is just the countable sum of points of zero measure.

Answer 1

You wrote: "We don't need Lebesgue measure to calculate the measure of the Cantor set". But, in fact, your calculation of the measure of the Cantor set used some incredibly powerful properties of measurable sets and their Lebesgue measures:

1. If $\{A_1,A_2,A_3,\ldots\}$ is a countable collection of measurable sets then its union $\cup_{i=1}^\infty A_i$ is a measurable set.
2. If furthermore the collection $\{A_1,A_2,A_3,\ldots\}$ is pairwise disjoint then $\text{Length}\cup_{i=1}^\infty A_i = \sum_{i=1}^\infty \text{Length}(A_i)$.

You used these properties correctly in your calculation of the measure of the Cantor set, which is evidence that you have correctly learned the properties of Lebesgue measure and their application. If all you want to do is to apply the tools of Lebesgue measure and to ignore the theory, then you've done a good job.

And yes, you should think of Lebesgue measure in all three ways that you suggest: as a formalization, a generalization, and a correction of everyday length.

But having learned how to apply some tools, the mathematician in you ought to be very, very suspicious, and that mathematician should immediately start asking hard questions about how those tools and their properties:

• How is the concept of a measurable set defined? What are all of the measurable sets? What is the collection of all measurable sets?
• How do I prove property 1?
• How is the measure of a measurable set defined?
• How do I prove property 2?

Until you grapple with these issues, your notion of "everyday length" or "simple measure" is specious. The difference between that and Lebesgue measure is that the latter is a rigorously developed mathematical theory, and the former is a collection of loose intuitions. The mathematician in you should not be satisfied with loose intuitions, and should struggle to see the logical unity which makes those loose intuitions into applicable properties of a rigorous theory.

Also, if you do not grapple with these issues, then you are left with hard questions. For instance, you decomposed the Cantor set in a nice way which facilitated your calculation. But if I gave you an entirely different decomposition of the Cantor set, are you sure that the result would be the same? Would the sum of measures still be zero?

This question arises even about old safe familiar sets like $[0,1]$. For example:

• Suppose that I gave you two countable, pairwise disjoint collections of measurable sets $\{A_1,A_2,\ldots\}$ and $\{B_1,B_2,\ldots\}$. Now, the only extra tiny bit of information that I give you is that $$\cup_{n=1}^\infty A_n = \cup_{n-1}^\infty B_n $$ For example, maybe that union is $[0,1]$. Are you really confident that $\sum_{n=1}^\infty \text{Length}(A_n) = \sum_{n=1}^\infty \text{Length}(B_n)$?