Tapio Rajala
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Registered User
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I am a postdoc at the University of Jyväskylä working on different topics in geometric measure theory and geometric function theory.
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Apr 26 |
accepted | On the proof of Modified Vitali Lemma. |
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Apr 22 |
answered | On the proof of Modified Vitali Lemma. |
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Apr 18 |
comment |
Doubling space without Besicovitch covering theorem? Heuristically, problems should arise only when one starts to consider balls of different radii. |
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Apr 18 |
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. |
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Apr 18 |
accepted | Doubling space without Besicovitch covering theorem? |
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Apr 18 |
revised |
Doubling space without Besicovitch covering theorem? added 238 characters in body |
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Apr 18 |
answered | Doubling space without Besicovitch covering theorem? |
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Mar 8 |
accepted | Lebesgue measure of level set of continuous functions |
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Mar 8 |
answered | Lebesgue measure of level set of continuous functions |
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Feb 10 |
accepted | Must the Minkowski sum of a Borel set and a *closed* ball be Borel? |
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Feb 10 |
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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). |
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Feb 9 |
revised |
Must the Minkowski sum of a Borel set and a *closed* ball be Borel? deleted 78 characters in body |
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Feb 9 |
revised |
Must the Minkowski sum of a Borel set and a *closed* ball be Borel? $A$ had two meanings..; added 4 characters in body |
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Feb 9 |
answered | Must the Minkowski sum of a Borel set and a *closed* ball be Borel? |
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Jan 14 |
comment |
Hutchinson’s formula for asymptotically homogeneous Cantor sets With 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. |
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Dec 31 |
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 $\int f = \frac{1}{2}(1+2+4)\sum_{i=1}^\infty 8^{-i} = \frac{1}{2}$. |
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Dec 31 |
accepted | Injective&Intregrable Mapping from $\mathbb R^3$ to $\mathbb R$ |
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Dec 31 |
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. |
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Dec 31 |
revised |
Injective&Intregrable Mapping from $\mathbb R^3$ to $\mathbb R$ added few lines on continuity |
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Dec 30 |
answered | Injective&Intregrable Mapping from $\mathbb R^3$ to $\mathbb R$ |
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Dec 20 |
awarded | ● Yearling |
