# Busy days.

Somehow spring break has turned into one of the busier weeks of my year.  Trying to keep up with real life work has not left a ton of time for writing anything thoughtful/reasonable, though at least for continuity I will try to keep a paragraph or so up here each day with my favorite thought of the day.  This also means I can reuse some old graphics!

Today I really enjoyed a particular fact about Sobolev functions.  Recall that these are actually equivalence classes of functions, as they are really defined under an integral sign, which “can’t see” sets of small measure.  However, the following quantifies exactly how small the bad set might have to be:

If $f \in W^{1,p}(\Omega)$ for $\Omega \subset \mathbb{R}^n$, then the limit $\lim_{r \to 0} \frac{1}{\alpha(n)r^n}\int_{B(x,r)}f(y)~dy$ exists for all x outside a set E with $\mathcal{H}^{n-p+\epsilon}(E) = 0$ for all $\epsilon > 0$.

Put another way, every Sobolev function may be “precisely defined” outside a set of small dimension, where the dimension gets smaller as p gets larger.  I suppose a given representative may be worse, but this allows you to require that the member of the equivalence class of Sobolev functions has some nice properties.

The fibers of two functions in a sequence. I was thinking the above argument might imply that the limit was not Sobolev, but the limit is precisely represented outside a set with positive 1-dimensional measure, so the result is silent on this issue.