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](http://en.wikipedia.org/wiki/Lebesgue_point) by the [Lebesgue differentiation theorem](http://en.wikipedia.org/wiki/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!



Pointwise limit at Lebesgue's point
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 $x$.


