# Tapio Rajala

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 Name Tapio Rajala Member for 2 years Seen 1 hour ago Website Location Finland Age 31
I am a postdoc at the University of Jyväskylä working on different topics in geometric measure theory and geometric function theory.
 Apr26 accepted On the proof of Modified Vitali Lemma. Apr22 answered On the proof of Modified Vitali Lemma. Apr18 comment Doubling space without Besicovitch covering theorem?Heuristically, problems should arise only when one starts to consider balls of different radii. Apr18 comment Doubling space without Besicovitch covering theorem?Misha: perhaps I am misunderstanding the packing lemma, but it seems to be equivalent with the doubling condition. In particular, the packing lemma would seem to follow directly for example from the existence of a doubling measure on the (completion of the) doubling metric space. Apr18 accepted Doubling space without Besicovitch covering theorem? Apr18 revised Doubling space without Besicovitch covering theorem?added 238 characters in body Apr18 answered Doubling space without Besicovitch covering theorem? Mar8 accepted Lebesgue measure of level set of continuous functions Mar8 answered Lebesgue measure of level set of continuous functions Feb10 accepted Must the Minkowski sum of a Borel set and a *closed* ball be Borel? Feb10 comment Must the Minkowski sum of a Borel set and a *closed* ball be Borel?I don't have an answer (at least yet) for $n=2$. What one can immediately say is that the intersection of the set with any line is Borel (it is the union of the translates of intersections of the set $A$ with the two lines with distance 1 from the line, and a countable number of intervals). Feb9 revised Must the Minkowski sum of a Borel set and a *closed* ball be Borel?deleted 78 characters in body Feb9 revised Must the Minkowski sum of a Borel set and a *closed* ball be Borel?$A$ had two meanings..; added 4 characters in body Feb9 answered Must the Minkowski sum of a Borel set and a *closed* ball be Borel? Jan14 comment Hutchinson’s formula for asymptotically homogeneous Cantor setsWith your fast convergence of $\delta_n$ the measure $m$ has also the property that $m(B(x,r)) \le cr^s$ for all $x \in \mathbb{R}^2$ (with $s = \log 2/ \log 3$). Therefore by Frostman's lemma $\mathcal{H}^s(C_\delta)>0$. For $\dim_\mathcal{H}(C_\delta) = s$ it is enough to assume $\delta_n \to 0$ and $\delta_n \in [0,2/3)$. No fast convergence is needed for this. Dec31 comment Injective&Intregrable Mapping from $\mathbb R^3$ to $\mathbb R$For the moment I can't think of any other way of estimating the integral except by direct calculation using the fact that you can write the integral explicitly as the sum of three geometric sums, giving $\int f = \frac{1}{2}(1+2+4)\sum_{i=1}^\infty 8^{-i} = \frac{1}{2}$. Dec31 accepted Injective&Intregrable Mapping from $\mathbb R^3$ to $\mathbb R$ Dec31 comment Injective&Intregrable Mapping from $\mathbb R^3$ to $\mathbb R$@Gerald: I added a few lines on the continuity outside the boundaries of unit cubes, who form a set of dimension 2. @Ohad: You can define $f$ on whole $\mathbb{R}^3$ by composing with translations and dilations like Gerald suggested. Since the original $f$ has bounded range, you can make the range to be inside your favorite bounded (non-empty and open) interval. Also, defining on $\mathbb{S}^n$ should be easy as well. Dec31 revised Injective&Intregrable Mapping from $\mathbb R^3$ to $\mathbb R$added few lines on continuity Dec30 answered Injective&Intregrable Mapping from $\mathbb R^3$ to $\mathbb R$ Dec20 awarded ● Yearling