This time I’m reading it out of Tao’s book Introduction to Measure Theory.

I’m thinking of a set that is Lebesgue measurable but not Jordan measurable.

Okay, well what are the sets that aren’t Jordan measurable? What was so limiting about the Jordan measure? Well first of all the Jordan measure requires cubes to approximate a volume. So you would need to construct a set that can’t be covered precisely by cubes. It would have to be something that has some holes in it, something the cubes will either over-approximate or can’t cover.

The first set I see is the set [0,1] intersect the rationals. The rationals are dense in R, but the cubes will over-approximate them. Remember, Jordan measure requires a finite number of cubes, so even if you make the cubes really small, you still can’t approximate them because it will require more and more cubes.

This set is Lebesgue measurable, however. Because now we can take a countable amount of them and since the rationals can form a countable basis of R, the Lebesgue outer measure should at least be 1 for this set right? What about the inner measure? I’m having a hard time with the inner measure because it doesn’t have that regularity thing. How would I take a cover of this set with Lebesgue inner measure that would equal 1? But is the Lebesgue measure of this set even equal to one? I’ll update when I think about it more.

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