# Tag Info

### Understanding the proof of $L^{\infty}$ is complete.

I really appreciate @ThànhNguyễn's answer! This answer post is for my own record with extra details worked out. If you find any mistakes, please comment below! Thanks a lot! $\quad\quad$ Let $\{f_k\}$...
• 1,795

### How to conclude the family $\{\mu_T\}_{T>0}$ is tight?

Boundedness of the integrals $$\int_E\phi(V(x))\mu_T(dx),\qquad T>0,$$ and the fact that for any compact subset $C\overset{\text{cpt}}{\subset}\mathbb{R}$, $(\phi\circ V)^{-1}(C)$ is again a ...
1 vote
Accepted

### Given $f \in L^p_{\text{loc}}(\Omega) \setminus L^\infty(\Omega)$, does it follow that $A \cap S(f,K) \neq \emptyset$ for all $K > 0$?

This answer just replicates my comment: For any two sets $A,B$ we can write $B$ as a disjoint union $$B = (B\cap A) \cup (B \cap A^c)$$ If we know the sets $A,B$ are Lebesgue measurable we know $A^c$,...
• 24.3k
Accepted

### Papa Rudin $7.24$ Theorem,

It follows from $(1)$ that $$\lim_{x\to0}\frac{|F(x)-F(0)-Ax|}{|x|}=0,$$ and since $F(0)=0$ and $F'(0)=I$, we get $$\lim_{x\to0}\frac{|F(x)-x|}{|x|}=0.$$ This gives you the $\epsilon$-$\delta$ ...
• 17.1k

### Image of measurable sets under one to one (a.e) functions

That would depend heavily on the measure. In an extreme case, if your measure $\mu$ is concentrated on a singleton then the condition "one-to-one a.e." is satisfied trivially and the ...
• 1,049
Accepted

### A certain distance-related map is well-defined

We have $b=(b_0,b_1,b_2,\ldots)$. Now look at the number $x=d(a,b)$, in base $3$. By construction of $d$, $x$ has ternary expansion consisting of $0$ and $1$, where each trit (ternary digit) ...
• 200k
1 vote

### Are the Lie groups $GL_n(\mathbb R)$ and $GL_n(\mathbb C)$ unimodular?

Here is a proof that doesn't require calculation with the change of variables formula for $\mathbb R^{n\times n}$. Instead, we give a proof that is in flavor of showing that for Gelfand pairs $(G, K)$ ...
• 2,168
Accepted

• 166

• 42.2k

### Proving that restriction of $\sigma$-algebra is a $\sigma$-algebra

Notice that $\Sigma \cap A$ is a $\sigma$-algebra on X is a contradiction whenever $A \subsetneq X$. Since $X \in \Sigma \cap A \implies \exists E\in \Sigma$ such that $X=E \cap A$. This is a ...
1 vote
Accepted

• 130
Accepted

### Is every collection of discrete random variables a function of independent random variables?

Yes, this is always possible. One method would be to take $N = 2^n-1$, and for each $X^i = (X^i_1,X^i_2,\cdots,X^i_n) \in \Omega$ let $p_i := \mu(\{X^i\})$. Let \begin{align*} Y_1 &\sim \text{...
• 13.8k

• 1,092
1 vote
Accepted

### Measurability of $\|f(\cdot, x_{2})\|_{L^\infty(X_{1})}$ (proof of Minkowski's inequality)

Thanks to @PhoemueX for the hint. We may use the $\sigma$-finiteness of $X_{1}$ to write $X_{1} = \bigcup_{n\geq 1}E_{n}$ with $\mu_{1}(E_{n}) < \infty$ and $E_{n}\subseteq E_{n+1}$ for all $n$. ...
• 1,157
1 vote
Accepted

### How to prove that $\left(\frac{r_1}{r_2}\right)^n{\rm vol}(\Omega_{r_1})\leq{\rm vol}(\Omega_{r_2})\leq{\rm vol}(\Omega_{r_1})$ for $0<r_1\leq r_2$?

Please allow me to recycle the notations and rephrase the problem in a way which is, I hope, more intuitive. Let $\Omega \subset \mathbb{R}^n$ be a bounded open set. For $r > 0$, set  \Omega_r = \...
• 5,514

Top 50 recent answers are included