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### Translates of a set of positive Lebesgue measure cover $\mathbb{R}$?

Let $E$ be fat Cantor set (i.e. a Cantor like set of positive measure). Then $E$ has no interior. If $\mathbb R=\bigcup_n (E+x_n)$ then Baire Category Theorem implies that $E+x_n$ has an intetior ...
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• 5,904
1 vote
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### $A= \{(x,y,z) \in \mathbb{R}^3:x^2+2y^2+z^2 < 4z\}$ limit: $\lim_{n \to \infty}\frac{1}{n} \int_A \frac{y^2z}{ln(x^2+2y^2+n) - ln(n)} \ d\lambda_3$

Sketch: On $A$, $x^2+2y^2<4z-z^2$. The LHS is non-negative, so the RHS must be non-negative which gives $z \in [0,4]$. Thus, \begin{align*} A &\subset B := \{(x,y,z): z \in [0,4], x^2+2y^2<\...
• 66
1 vote
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One way to circumvent this problem would be to take the $s_1 = s$ and $s_2 = t$ (you went from one notation to another, that is something to look out for) corresponding to $\varepsilon/2$ instead of $\... • 5,849 2 votes ### Nonstandard Analysis research project ideas There are two approaches to non-standard analysis: (1) the "extension" approach and (2) the axiomatic approach. In the "extension" approach, the real numbers are extended to the ... • 43.8k 1 vote ### Integral of Thomae's function Note that for every partition$P$of$[0,1]$,$L(f,P)=0$. Thus,$\int_0^1f=L(f)$=sup$\{L(f,P):P\; \text{is partition of }[0,1]\}$=sup$\{0\}=0\$

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