Dear MOs,
I am sorry if this problem is too elementary for someone. I just want to get confirmation. Suppose $f\in L^1(R^d)$. Since almost all points are Lebesgue points by the Lebesgue differentiation theorem, can we say that for almost every $x\in R^d$,
$$ \lim_{x'\rightarrow x} f(x') = f(x)\:? $$
I think it is true probably only for $d=1$. Does anyone know some results about this problem?
Thanks a lot!
Since $L^1(\mathbb{R}^d)$ really means equivalence classes of integrable functions, I am interpreting the question as follows. Given $f \in L^1(\mathbb{R}^d)$, does there exist $g$ which is equal to $f$ almost everywhere and such that for almost every $x \in \mathbb{R}^d$, $$\lim_{x' \to x} g(x') = g(x)? $$
Here is a counterexample, even with $d=1$. Let $A \subset [0,1]$ be a measurable set with measure strictly less than $1$, such that for every open subset $U$ of $[0,1]$, the measure of $U \cap A$ is nonzero, and set $f=\chi_A$, the characteristic function of $A$.
(Such $A$ can be constructed, for example, taking an enumeration $\mathbb{Q} \cap (0,1) = \{x_n\}$ and setting $A = \bigcup_{n=1}^\infty \left(x_n-2^{-(n+2)}, x_n + 2^{-(n+2)}\right)$.)
Suppose that $g$ is equal to $f$ almost everywhere. Since the measure of $A$ is less than $1$, $B := g^{-1}(0)$ has positive measure, and we shall now show that $g$ is discontinuous at each $x \in B$.
Fix $x \in B$. For every $\epsilon > 0$, we have that $(x-\epsilon,x+\epsilon) \cap A$ has nonzero measure, and therefore, there exists a point $x'\in (x-\epsilon,x+\epsilon)$ such that $g(x') = 1$. Hence, $g$ is not continuous at $x$.
