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May
12
comment a interesting question from topological group
I just mean if you take any compact neighbourhood $K$ of the identity and let $H$ be the subgroup of $G$ generated by $K$ then $H$ is an open $\sigma$-compact subgroup of $G$. The Haar measure of $G$ is then just the sum of its restrictions to the cosets of $H$, so by translating $A$ and $B$ we may assume they intersect $H$ in sets of positive measure. In effect we may assume that $G$ is $\sigma$-compact so certainly $\sigma$-finite.
May
12
comment a interesting question from topological group
The integrand is zero for $(x,y)$ outside the compact set $AB\times A$, so such problems do not arise. Alternatively you can always just restrict to a subgroup $H$ of $G$ generated by a compact neighbourhood of the identity: this is an open $\sigma$-compact subgroup.
May
11
answered a interesting question from topological group
May
11
comment a interesting question from topological group
@DavidChan Approximate $\chi_A$ and $\chi_B$ in $L^1$ norm by continuous and compactly supported functions. After convolution the $L^1$ error becomes an $L^\infty$ error, so $\chi_A\ast\chi_B$ is a uniform limit of continuous functions, so continuous. The convolution operation has many fundamental smoothing properties like this one.
May
7
answered Strengthening the intermediate value theorem to an “intermediate component theorem”
May
3
comment a interesting question from topological group
This result is sometimes called Steinhaus's theorem. en.wikipedia.org/wiki/Steinhaus_theorem
May
3
comment a interesting question from topological group
@WilliamCurtis Choose some $x_0$ such that $f(x_0)>0$. There is some neighbourhood $U$ of $x_0$ such that $f>0$ on $U$. That means that for every $x\in U$ there is some $y$ such that $xy^{-1}\in A$ and $y\in B$; in other words $x\in AB$.
May
3
revised a interesting question from topological group
Corrected the definition of convolution.
Apr
26
comment The Lebesgue-Borel measuref the difference between two open balls tends to $0$ as the radii tend to $\infty$
Thanks, yes, I meant the modulus of $y$. It's fixed now.
Apr
26
revised The Lebesgue-Borel measuref the difference between two open balls tends to $0$ as the radii tend to $\infty$
added 2 characters in body
Apr
24
answered The Lebesgue-Borel measuref the difference between two open balls tends to $0$ as the radii tend to $\infty$
Apr
24
comment The Lebesgue-Borel measuref the difference between two open balls tends to $0$ as the radii tend to $\infty$
@G.Sassatelli You're probably right. If so then after rescaling this is the same as the limit of $\lambda_n(B_1(x)\setminus B_1(y))$ as $y\to x$, and now it's clear.
Apr
24
comment The Lebesgue-Borel measuref the difference between two open balls tends to $0$ as the radii tend to $\infty$
I think the statement as written is false. WLOG $x=0$ and $y=e_1$. Then we're measuring the set of points which are a distance $r$ from $0$ but not from $e_1$. This set contains a half ball of radius $1$ centred at $-(r-1)e_1$, so it has measure at least half the measure of the unit ball.
Apr
22
answered Show that there exists an entire function $h$ such that $\lim_{n\to\infty}{h(nz)}=0$ for all $z\ne0$
Apr
21
answered Find $\lim\limits_{n\rightarrow\infty}(n!)^{1/n^2}$
Apr
16
comment Difference between pushforward and differential
@IberêKuntz You can certainly use another map $g$ (instead of $f$, to reduce notation clash) for the identification $T_q\mathbb{R}\cong \mathbb{R}$ if you like, provided that $g'(q)=1$. Then you'll find that, at $q=f(p)$, $f_*(X)(g) = g'(q) X(f) = X(f)$.
Apr
7
answered How do I choose first terms of a Fibonacci sequence?
Mar
12
answered Haar's theorem for the rotation-invariant distribution on the sphere
Mar
10
comment Determinability
If you define $g(x) = \sup\{t\in\mathbb{Q}: x_t=0\}$ then actually $g^{-1}(\{y\})$ is a little nicer than that: whether $x\in g^{-1}(\{y\})$ depends only on $x_t$ for rational $t$.
Mar
10
comment Determinability
Does taking the supremum just over rational $t$ help?