# Tagged Questions

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234 views

### Can a nowhere differentiable function preserve measurability?

I want to know whether a continuous nowhere differentiable function $f: \mathbf{R} \to \mathbf{R}$ can map Lebesgue measurable sets to Lebesgue measurable sets. More generally I'm interested to know ...
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### Physical meaning of the Lebesgue measure

Question (informal) Is there an empirically verifiable scientific experiment that can empirically confirm that the Lebesgue measure has physical meaning beyond what can be obtained using just the ...
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### Set of General Linear Position with Nonzero Measure

I came to the following question, but I don't have quite a good idea how to approach. Can a set $A\subset \mathbb{R}^n , n\ge 2$ with nonzero measure be in a general linear position? I believe that,...
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### $\forall g\in L^2(\Omega)$ exists $g_n\in H_0^1(\Omega)$ and $\epsilon>0$ s.t $g_n(x)\to g(x),\,a.e$ and $|g_n(x)|\leq |g(x)|+\epsilon$

I asked this question on MSE here some time ago, but I couldn't get an answer. There was a suggestion in the comments for a counterexample using a fat Cantor set, but I couldn't show a contradiction ...
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### Is $f(x,y)=\sum_{n\in\mathbb{Z}\backslash\{0\}}\frac{1}{n}e^{2\pi i(xn+yn^2)}$ essentially bounded?
Let $$f(x,y)=\sum_{n\in\mathbb{Z}\backslash\{0\}}\frac{1}{n}e^{2\pi i(xn+yn^2)}$$ Is it true that $\|f\|_{L^{\infty}(\mathbb{R}^2)}<\infty$? i.e. is $f$ essentially bounded?