MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
In my opinion this is too elementary for MO. However, to give you some direction: for positive results you should consider for example approximate limits. – Tapio Rajala Jul 10 '12 at 9:46
up vote 2 down vote accepted

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 $x$.

share|cite|improve this answer
Dear Aaron Tikuisis, Thank you very much for your counter example. – Anand Jul 10 '12 at 10:56

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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