Skip to main content
4 votes
Accepted

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 ...
geetha290krm's user avatar
  • 39.2k
2 votes

Missing argument in the proof of the Levy-Khintchine representation .

Hopefully there is a more concise answer. I will prove much more than required, namely that $f_n(x) dx$ converges in distribution towards $\delta_0(dx) + f(x) dx$ with $f$ decreasing, and that $f_n \...
Thomas Lehéricy's user avatar
2 votes

A corollary of Rolle's theorem

This is a specific case of the Darboux theorem. In your case, this is easier, since you are looking for $c$ such that $f'(c)=0$. Since $f$ is differentiable on $[a,b]$, then $f$ is continuous on $[a,b]...
Martigan's user avatar
  • 5,904
1 vote
Accepted

$A= \{(x,y,z) \in \mathbb{R}^3:x^2+2y^2+z^2 < 4z\}$ 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<\...
Sounak's user avatar
  • 66
1 vote
Accepted

Proving existence of certain step functions implies integrability

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 $\...
Bruno B's user avatar
  • 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 ...
Mikhail Katz's user avatar
  • 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$\{0\}=0$
suraj tidke's user avatar

Only top scored, non community-wiki answers of a minimum length are eligible