A Mind for Madness

Musings on art, philosophy, mathematics, and physics

Lebesgue Points


Just a quick detour. I’ve found a new reason to dislike analysis. I’m trying to learn Radon-Nikodym derivatives (i.e. an attempt to take derivatives in a general measure theory sense and maintain the Fundamental Theorem of Calculus for the Lebesgue integral), and Rudin uses the approach of Lebesgue Points. Since I’ve never learned this before, I’m not sure if the other methods are easier, but this is certainly proving to be rough. Apparently we are supposed to be familiar with random facts about LPs, even though this is the very first time the definition is given. So here are the random unproven statements about Lebesgue points that I’ve encountered and my proofs to go along with them. I don’t think all of these are what Rudin had in mind, since my proofs are far more complicated than one could probably just think through.

Definition: Let f\in L^1(\mathbb{R}^k), then x is a Lebesgue point of f if \displaystyle \lim_{r\to 0}\frac{1}{m(B_r)}\int_{B_r}|f(y)-f(x)|dm(y)=0. Where m is Lebesgue measure, and that B notation is the open ball centered at x of radius r. Yeah. Not the simplest definition to be assuming knowledge of.

Claim 1: If f is continuous at x, then x is a Lebesgue point of f (under the suitable conditions on f that will always be assumed in this post). Let f be continuous at x. Then let \varepsilon>0 and choose \delta>0 such that if |x-y|<\delta, then |f(x)-f(y)|<\varepsilon. Now whenever |x-0|<\delta, we have \displaystyle \big| \frac{1}{m(B_\delta)}\int_{B_\delta}|f(y)-f(x)|dm -0 \big| \leq \frac{1}{m(B_\delta)}\int_{B_\delta}\varepsilon dm =\frac{\varepsilon m(B_\delta)}{m(B_\delta)}=\varepsilon. i.e. The limit behaves as we would like and x is a Lebesgue point.

Claim 2: If x is a Lebesgue point of f, then \displaystyle f(x)=\lim_{r\to 0}\frac{1}{m(B_r)}\int_{m(B_r)}fdm. Now I’m not sure if it is just me, but things were just moved around, so the fishy business I’m going to pull doesn’t seem necessary. Let x be a Lebesgue point of f. Then

\displaystyle 0=\lim_{r\to 0}\frac{1}{m(B_r)}\int_{B_r}|f(y)-f(x)|dx
\displaystyle\geq \lim_{r\to 0}\big|\frac{1}{m(B_r)} \int_{B_r} f(y)dm -\frac{1}{m(B_r)}\int_{B_r} f(x)dm\big|
\displaystyle =\lim_{r\to 0}\big|\frac{1}{m(B_r)}\int_{B_r}fdm-f(x)\big|. Thus since the right side is positive and less than or equal to 0 get rid of the absolute value since it must be equal to 0 and we have \displaystyle 0=\lim_{r\to 0}\frac{1}{m(B_r)}\int_{B_r}fdm - f(x) and rearrange.

I think there was a third claim, but I can’t find it now. Also, these proofs may look rather trivial now, but when you are completely unfamiliar with the definition and properties, this is rather confusing to try to work out quickly to continue reading the proof. Hopefully this post will help future readers of Rudin when they come to this.

I guess since I’ve come this far I should probably post some bonus material just to see the point.

Interesting result 1: Almost every point of f is a Lebesgue point (still assuming appropriate conditions on f).

The point is to get to the definition of the derivative, so if for all measurable sets E, we have \mu(E)=\int_E fdm for some f, then f is called the Radon-Nikodym derivative and notationally it is usually written that d\mu=f dm (for the obvious reason that if you integrate both sides you get the first form). But that notation leads us nicely to a more familiar Leibniz-type notation: f=\frac{d\mu}{dm}. Now skipping some other interesting results, some of the meat of the theory comes out in a FTC type result

Interesting result 2: If f\in L^1(\mathbb{R}^k) and F(x)=\int_{-\infty}^x fdm, then F'(x)=f(x) at every Lebesgue point of f (and by IR 1 almost everywhere).

Author: hilbertthm90

I write about math, philosophy, literature, music, science, computer science, gaming or whatever strikes my fancy that day.

2 thoughts on “Lebesgue Points

  1. Cool! These are exactly my questions in mind :D Thanks buddy.

  2. The point about Lebesgue point is that for L^1 function one does NOT know what f(x) is! So the proper definition must be:

    There is a real a such that the corresponding limit with a in place of f(x) is zero.

    Then, but only then, one could associate a with the value of f at x and – since almost any point is Lebesgue – one can think of the function as defined almost everywhere.

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