I'm looking for reasonably simple examples of applications of Lebesgue measure theory outside the measure theory setting. I give an example.

Theorem: Let $X$ be a differentiable submanifold of $\mathbb{R}^n$ with codimension $\geq 3$. Then $\mathbb{R}^n\setminus X$ is simply conected.

Proof. Let $\alpha:S^1\to \mathbb{R}^n\setminus X$ be a closed $C^1$ curve. We want to show that there exists a point $p$ outside $X$ s.t. a linear homotopy between $\alpha$ and $p$ can be contructed. Well, define $F:\mathbb{R}\times S^1\times X\to \mathbb{R}^n $ by $$ F(t,s,x)= (1-t)\alpha(s)+tx. $$ Note that, 1) $F$ collects all the bad lines, i.e., the lines connecting $\alpha(s)$ and $x\in X$. 2) $F$ is $C^1$. 3) $\dim(\mathbb{R}\times S^1\times X)\leq n-1$. So, by Sard's theorem, the set $F(\mathbb{R} \times S^1\times X)$ has zero Lebesgue measure, and therefore its complement is non-empty. This easily implies the result.

Also as an example we have the the weak form of the Whitney immersion theorem, for which one can use the same kind of argument in the proof.

I want to know more "simple" applications of the type the above mentioned, in areas other than measure theory. But not too complicated ones!

Sorry if this question is too basic.

  • Presumably you want to exclude probability and analysis as well? – Robert Israel Aug 10 at 0:11
  • Not necessarily, but I'm looking for unexpected examples, as I think the above examples are to me... However the expression "unexpected" is a kind of personal... – Eduardo Aug 10 at 0:15
  • 1
    The argument you have given looks more like an application of Sard's lemma than that of "Lebesgue measure theory", but I cannot follow it anyway. What is $M$? What is $f$? – Kostya_I Aug 10 at 8:08
  • 2
    my apologies, $M=X$ and $f=F$. – Eduardo Aug 10 at 13:17
  • @Kostya_I: Sard's theorem has Lebesgue measure in its very statement! – Nate Eldredge Aug 14 at 0:27

This was American Mathematical Monthly Problem #11526, circa 2010:

Proposition. There is no function $f$ from $\mathbb R^3$ to $\mathbb R^2$ with the property that $|f(x)-f(y)| \ge |x-y|$ for all $x,y \in \mathbb R^3$.

Proof. (Mouse over below...)

Such $f$ would be injective, and its inverse $f^{-1}$ would be a surjective Lipschitz map from a subset of $\mathbb{R}^2$ onto $\mathbb{R}^3$. But Lipschitz maps don't increase Hausdorff dimension, so the image of $f^{-1}$ must have Lebesgue measure zero, contradicting surjectivity.

The existence of normal numbers.

  • Much of the field of probabilistic combinatorics has a similar flavor to this: you want to prove the existence of an object with some property X, so you construct an appropriate measure on the space of all objects, and prove that the set of those with property X has positive measure. – Nate Eldredge Aug 11 at 14:07

This isn't a direct answer, but it may be on topic, depending on the motivation for the question.

When I teach measure theory, I feel I owe the students an explanation of why they should have to learn about the Lebesgue integral when they already know the Riemann integral. What is the new application that you need Lebesgue for? Perhaps you ask this question because you face the same difficulty.

I guess you can cook up examples of functions that are Lebesgue but not Riemann integrable. But that is going to appear contrived. What I tell the class is that it's similar to asking "Why should I learn about real numbers, when I already know the rationals?" One answer is that you can find examples of infinite series that don't have a sum in $\mathbb{Q}$ but do in $\mathbb{R}$. A more sophisticated way to say this is that $\mathbb{R}$ is complete as a metric space, and this carries so many benefits that it is just obviously worthwhile.

The punchline is then that the passage from Riemann to Lebesgue can be seen as a "completion" in much the same way that $\mathbb{R}$ is a completion of $\mathbb{Q}$. If you define the distance between two functions on $[0,1]$ to be $d(f,g) = \int_0^1 |f - g|\, dx$ then the set of Riemann integrable functions isn't complete. Completing it yields the space of Lebesgue integrable functions (modulo functions which vanish off a null set, but that is a topic to be discussed later).

  • I like this approach that to get elements of the completion as functions on [0,1], you introduce Lebesgue measurable functions etc. There is also an exercise in Rudin's "Real and Complex Analysis" which says: if $0\leq f_n(x)\leq 1$ is a sequence of continuous functions on [0,1], tending pointwise to $0$, then the sequence of integrals $\int _0 ^1 f_n(x)dx$ tends to zero. This is immediate from the dominated convergence theorem , and Rudin says this illustrates the power of the Lebesgue integral (it is hard to prove this directly). – Venkataramana Aug 10 at 17:54
  • @Nik Weaver I'm not facing this kind of difficulty, yet. However I find your comment of great value, and I intend to use it when the time comes. – Eduardo Aug 12 at 17:23

I liked the following example.

Theorem: Let R be a rectangle in the plane with sides parallel to the axes. Suppose R is cut up into countably many smaller rectangles whose sides are also parallel to the axes, such that the length of at least one of the sides of the smaller rectangles is an integer. Then the big rectangle R also has the same property.

Proof: Consider the complex measure $d\mu =dxdye^{2\pi i (x+y)}$ on the plane. The hypotheses imply that each of the smaller rectangles has $\mu$ measure zero. By summing up, the big rectangle R also has $\mu$ measure zero.

  • Wouldn't it work just equally well with any pre-Lebesgue theory of integration (Jordan measure, Riemann integral)? – Kostya_I Aug 10 at 14:37
  • @Kostya_l: "countable additivity" may be a problem – Venkataramana Aug 10 at 14:38
  • ah, I mistook "countably many" for "finitely many". Nice indeed! – Kostya_I Aug 10 at 14:41
  • @Kostya_l: Thank you! – Venkataramana Aug 10 at 14:55

1) Averaging tricks of all kinds, e. g., the entire area of integral geometry. For a specific simple example, see Crofton formula and its consequences listed in the linked article. A version of the formula can be used to prove that if a curve on the sphere is not contained in any hemisphere, then its length is at least $2\pi r$, and there are many more serious application too. To answer a possible objection, even for smooth curves the integrand is neither continuous nor bounded; good luck working out the proof with a weaker version of the integral.

Of a similar spirit is e. g. Weyl's unitarian trick.

2) Lebesgue measure gives a lazy way to construct an infinite sequence of independent scalar random variables with prescribed distributions, which is (kind of) important for Probability. Indeed, binary digits on $[0,1]$ give you an infinite sequence of i. i. d. Bernoullis variables. Rearranging them, you get an infinite sequence of independent infinite sequences of i. i. d. Bernoullis, which is the same as and infinite sequence of uniform random variables on $[0,1]$. Post-composing with functions gives arbitrary distributions.

3) A proper integration theory is essential for completeness and duality in $L^p$ spaces, which of course have tons of applications: to prove that some function exists, it is enough to either construct a corresponding functional, or a Cauchy sequence. See this answer for an elementary example, or the $L^2$ projection proof of existence of conditional expectation in Williams' book.

I do not know what the protocol is when you give another answer; this has nothing to do with my previous example. The most spectacular applications of measure theory that I know come from Margulis' work. For example, suppose $\Gamma \subset SL_3(\mathbb R)$ is a discrete subgroup with compact quotient. Then Margulis shows that every non-trivial normal subgroup of $\Gamma $ has finite index in $\Gamma$. The proof uses measure theory ( and a lot else besides) in a serious way. His proof that such a $\Gamma$ is arithmetic also uses ergodic theory (and measure theory). These purely "algebraic" statements were proved by use of measure theory.

  • I think multiple answers are fine in cases like this. I liked your other answer too. – Nik Weaver Aug 11 at 4:18
  • @Nik Weaver: thank you! – Venkataramana Aug 11 at 5:23

Your Answer

 
discard

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.