Can $L^\infty$ function be redefined so that every point is a Lebesgue point?

Question

Consider a reasonable subset $\Omega$ of an Euclidian space with a scalar-valued $f \in L^\infty(\Omega)$. (I'll define reasonability later.)

It is well-known that almost every point of $\Omega$ is a Lebesgue point; that is, for almost all $x$ we have $$ f(x) = \lim_{\varepsilon \to 0} \frac{1}{|\Omega \cap B(x,\varepsilon)|}\int_{\Omega \cap B(x,\varepsilon)} f(y) dy. $$

(The integral is the Lebesgue integral and the the absolute values around a set indicate Lebesgue measure of that set.)

Questions

1. Does the limit in the definition of Lebesgue point exist everywhere; that is, can we redefine the function on a set of measure zero so that every point is a Lebesgue point?

2. Supposing the limit does not exist everywhere, is there some characterisation or canonical set of examples for the exceptional points? I suppose there must be some sort of oscillation happening, but more intuition would be nice.

Reasonable subset

For the purposes of this question, let us say that $\Omega$ is reasonable if and only if there is an open set $U \neq \emptyset$ with $U \subseteq \Omega \subseteq \overline U$. ($U$ is the interior of $\Omega$.)

Answer 1

Actually what you defined is a "weak" Lebesgue point. A Lebesgue point is a point $x$ such that

$$\lim_{\varepsilon \to 0} \frac{1}{|\Omega \cap B(x,\varepsilon)|}\int_{\Omega \cap B(x,\varepsilon)} |f(y)-f(x)|\, dy =0.$$

Every Lebesgue point is a weak Lebesgue point. So we can show that the answer to 1. is no by producing an $f$ and a point $x$ that is not a weak Lebesgue point for $g,$ for every $g=f$ a.e.

We can do this on $\mathbb {R}.$ Define $a_n = 1/n!.$ Set

$$f = \sum_{n=1}^{\infty}\chi_{(a_{2n}, a_{2n-1})}.$$

Then for $n$ odd,

$$\frac{1}{2a_n}\int_{-a_n}^{a_n} f > \frac{1}{2a_n}\int_{(a_{n+1}, a_n)}f = \frac{1/n!-1/(n+1)!}{2/n!}= \frac{1-1/(n+1)}{2} \to 1/2.$$

For $n$ even,

$$\frac{1}{2a_n}\int_{-a_n}^{a_n} f < \frac{a_{n+1}}{2a_n} = 1/[2(n+1)] \to 0.$$

Thus $\lim_{h\to 0^+}\int_{-h}^h f$ does not exist. If you redefine $f$ on a set of measure $0,$ none of these quantities change. So if $g=f$ a.e., $g$ fails to have a weak Lebesgue point at $0.$