The Lebesgue differentiation theorem says that for certain metric spaces $X$ (see below), any Borel measure $\mu$ that is finite on bounded sets and any $f: X \rightarrow \mathbb{R}$ locally $\mu$-integrable, there is $A \subseteq X$ s.t. $\mu(X \setminus A)=0$ and

$$\forall x \in A: \lim_{r \rightarrow 0} \frac{1}{\mu(B_r(x))} \int_{B_r(x)} f(y) \mu(dy) = f(x)$$

Here, $B_r(x)$ is the ball of radius $r$ with center $x$.

This holds for $X$ a Riemannian manifold or $X$ a locally compact separable ultrametric space. I'm interested to understand how it fails on somewhat more general spaces. In particular:

  • Can the theorem fail on $X$ a compact separable metric space? Can you provide a counterexample (i.e. $X$, $\mu$ and $f$ s.t. the identity fails)?

  • Is there a compact Polish space $X$ s.t. the theorem can fail for any metrization of $X$ (i.e. for any metrization there are $\mu$ and $f$ s.t. the identity fails)?

  • 1
    $\begingroup$ These questions are relevant: mathoverflow.net/questions/244665/… $\endgroup$ Jan 29 '17 at 20:42
  • $\begingroup$ and mathoverflow.net/questions/218457/… $\endgroup$ Jan 29 '17 at 20:43
  • $\begingroup$ Could it be that you misquoted something in your opening paragraph? If I compare it with the wikipedia article for example, then the corresponding claims are only made about the Lebesgue density theorem there. $\endgroup$ Jan 30 '17 at 22:03
  • $\begingroup$ Which part do you find dubious? For Riemannian manifolds I think that the theorem follows quite easily from the Euclidean case by considering a local diffeomorphism with $\mathbb{R}^n$ (resulting eccentricity will be bounded on compact sets). $\endgroup$
    – Vanessa
    Jan 30 '17 at 22:46
  • $\begingroup$ @Squark: I think we are having some basic misunderstandings here and in the comments to my answer below. The point I'm trying to make (here and below) is that already on $\mathbb R^2$, it is certainly not true that $|A|^{-1} \int_A f(x-t)\, dt \to f(x)$ for a.e. $x$ as $\textrm{diam}(A)\to 0$, $0\in A$, if the sets $A$ get too general, and eccentricity is the problem. For example, you can't allow general rectangles. Maybe it would be helpful if you stated precisely what statement exactly you're interested in rather than just refer to it as the Lebesgue differentiation theorem. $\endgroup$ Jan 31 '17 at 1:42

I assume by Lebesgue differentiation theorem you mean the statement that $|B(x)|^{-1}\int_{B(x)} f(y)\, dy \to f(x)$.

Then it's not clear to me what your set-up on $X=[0,1]^{\mathbb N}$ is (what's the measure?), but in any event, for an arbitrary metric, this already fails on $\mathbb R^2$. You can take a metric that gives you wide thin rectangles as small balls, for example $$ d(x,y)=\max (|x_2-x_1|, |y_2^{1/3}-y_1^{1/3}|) $$ (if $y<0$, then $y^{1/3}$ just means $-|y|^{1/3}$).

Update: This answer was originally based on my recollection of the "standard fact" that the higher-dimensional Lebesgue differentiation theorem fails for rectangles if the eccentricity is not restricted. This much is true if arbitrary rectangles are allowed, but of course the situation here is different, and I'm not sure now what the situation is (and in fact I'm not even sure it's not an open question).

  • 1
    $\begingroup$ Thanks for answering, Christian! Regarding $[0,1]^\omega$: I meant that I want a counterexample with any measure. Regarding $\mathbb{R}^2$: the equation you wrote is not a metric since the "distance" between $(-1,0)$ and $(+1,0)$ vanishes. I guess you meant to e.g. take the half-plane $x \geq 0$? But this is isomorphic to the usual $L^1$ metric (transform by $x' = x^2$)? $\endgroup$
    – Vanessa
    Jan 30 '17 at 20:44
  • $\begingroup$ (also, I realized that what I really need to know is "Is there a compact Polish space s.t. for any metrization there is a counterexample?", but I guess it's too late to rephrase the question) $\endgroup$
    – Vanessa
    Jan 30 '17 at 20:47
  • $\begingroup$ @Squark: No, I wanted a metric whose balls are rectangles with unbounded eccentricity, see my new version please. $\endgroup$ Jan 30 '17 at 20:49
  • $\begingroup$ OK, and what is $f(x,y)$? $\endgroup$
    – Vanessa
    Jan 30 '17 at 21:08
  • 1
    $\begingroup$ Your new construction is still isomorphic to the usual $L^1$ metric via the transformation $y' = y^{\frac{1}{3}}$. Moreover, any construction that has bounded eccentricity in some neighborhood of any point except for a set of zero measure cannot work (of course we can also start playing with measure, but I don't know how). $\endgroup$
    – Vanessa
    Jan 30 '17 at 22:41

The classic reference book is:

Hayes & Pauc, Derivation and Martingales


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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