Pointwise limit at Lebesgue's point - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T09:19:55Z http://mathoverflow.net/feeds/question/101831 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/101831/pointwise-limit-at-lebesgues-point Pointwise limit at Lebesgue's point Anand 2012-07-10T08:06:01Z 2012-07-10T09:54:34Z <p>Dear MOs,</p> <p>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 <a href="http://en.wikipedia.org/wiki/Lebesgue_point" rel="nofollow">Lebesgue points</a> by the <a href="http://en.wikipedia.org/wiki/Lebesgue_differentiation_theorem" rel="nofollow">Lebesgue differentiation theorem</a>, can we say that for almost every $x\in R^d$, </p> <p>$$\lim_{x'\rightarrow x} f(x') = f(x)\:?$$</p> <p>I think it is true probably only for $d=1$. Does anyone know some results about this problem?</p> <p>Thanks a lot!</p> http://mathoverflow.net/questions/101831/pointwise-limit-at-lebesgues-point/101836#101836 Answer by Aaron Tikuisis for Pointwise limit at Lebesgue's point Aaron Tikuisis 2012-07-10T09:54:34Z 2012-07-10T09:54:34Z <p>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 $x \in \mathbb{R}^d$, $$\lim_{x' \to x} g(x') = g(x)?$$</p> <p>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 $A$.</p> <p>(Such $A$ can be constructed, for example, taking an enumeration $\mathbb{Q} \cap (0,1) = {x_n}$ and setting $A = \bigcup_{n=1}^\infty \left(x_n-2^{-(n+2)}, x_n + 2^{-(n+2)}\right)$.)</p> <p>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 $x \in B$.</p> <p>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 $x$.</p>