Skip to main content

New answers tagged

2 votes

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 $\{f_k\}$...
Beerus's user avatar
  • 1,795
0 votes

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 ...
Hirofumi Shiba's user avatar
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 $A^c$,...
Michael's user avatar
  • 24.3k
3 votes
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 $\epsilon$-$\delta$ ...
John B's user avatar
  • 17.1k
2 votes

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 ...
Lieven's user avatar
  • 1,049
2 votes
Accepted

Sigma sum notation. I just started out with some basic questions and soon stumbled on this.

$$a^3 - b^3 = (a-b)(a^2 + ab + b^2)$$ $$ \sqrt[3]{ k^2 }+\sqrt[3]{ k(k+1) }+\sqrt[3]{ (k+1)^2} = \frac{(\sqrt[3]{ k+1 } - \sqrt[3]{ k} ) ( \sqrt[3]{ k^2 }+\sqrt[3]{ k(k+1) }+\sqrt[3]{ (k+1)^2} )}{\...
Sergei Nikolaev's user avatar
1 vote
Accepted

Using the dominated convergence theorem to evaluate the integral of independent events

Let $X_k=\prod_{n=1}^k 2^{-\chi_{E_n}}$. Then $0 \le X_k \le 1$ and $X_k \to X$ where $X=\prod_{n=1}^\infty 2^{-\chi_{E_n}}$. Hence $\int Xd\mu=\lim \int X_k d\mu$ (by DCT). By indepndence we get $\...
geetha290krm's user avatar
  • 38.2k
0 votes

Show that $\frac{1}{n}X_n\to 0$ a.s.

To show that for any sequence $\left(X_n\right)_{n \in \mathbb{N}} \in\left(L_{\mathbb{P}}^2\right)^{\mathbb{N}}$ of identically distributed random variables it holds that $\frac{1}{n} X_n \rightarrow ...
bruno's user avatar
  • 364
4 votes
Accepted

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

To answer this, the following statement is true: Let $(f_n), (g_n)$ be two sequences in $L^p$ such that $f_n = g_n$ a.e, $f_n \rightarrow f$ a.e, then $g_n \rightarrow f$ a.e Indeed, let $A_n = \{...
Thành Nguyễn's user avatar
0 votes

Space $\mathcal{L}^p(X, \Sigma, \mu)$ is separable iff $(\Sigma, \rho_\Delta)$ is separable

I will give more details to the Fausto di Biase answer. A reference about this question is: "Integration" by Aadrian Cornelis Zaanen, §11, §20, §30. Notation and some preliminaries: Let $(X, ...
Manuel Bonanno's user avatar
1 vote
Accepted

Is $C_c^\infty(\mathbb{R})$ dense in $L^1(\mathbb{R}, (1+x^2)dx)$?

Let $f\in L^1((1+|x|^2)\,\mathrm d x)$. Then $g = (1+|x|^2)\,f \in L^1$ so by the density in $L^1$, there is a sequence of functions $\varphi_n\in C^\infty_c$ such that $\varphi_n\to g$ in $L^1$. But ...
LL 3.14's user avatar
  • 12.7k
1 vote
Accepted

Fractal Geometry: Show that the Hausdorff dimension of a set and its image under $f(x)=x^2$ are the same.

Proof. Let $F_k=F \cap[\frac{1}{k}, k]$. If $x, y \in F_k$, then $$ |f(x)-f(y)|=\left|x^2-y^2\right|=|(x-y)(x+y)|=|x-y||x+y| $$ by the homogeneity of $|\cdot|$. We get $$ \frac{2}{k}|x-y| \leq|f(x)-f(...
calculatormathematical's user avatar
0 votes

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) ...
Arthur's user avatar
  • 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 $(G, K)$ ...
GhostAmarth's user avatar
  • 2,168
4 votes
Accepted

$e^{-\lVert x \rVert ^2}$ is integrable over $\mathbb{R}^n$ and $\int_{\mathbb{R}^n} e^{-\lVert x \rVert ^2} = \pi^{n/2}$

$$\int\dots \int e^{-\sum_i x_i^2} \prod_k \ dx_k =\int \prod _i \ e^{-x_i^2} dx_i = \prod _i\int e^{-x_i^2} dx_i.$$ The integral may be splitted into a power of the square $$\left(\int e^{-x^2-y^2} ...
Roland F's user avatar
  • 2,704
0 votes

Why are there no interior points of the positive cone defined by nonnegative functions on $X=L_1[t_1, t_2]$

The statement in the OP is in fact valid in a very general setting: Let $(\Omega,\mathscr{F},\mu)$ be a measure space. If $1\leq p<\infty$, then $L^+_p(\mu)$ has empty interior unless $L_p(\mu)$ ...
Mittens's user avatar
  • 39.5k
2 votes
Accepted

Confused about the Radon-Nikodym theorem applied to the counting measure

The Radon-Nikodym Theorem requires the measure to be $\sigma$-finite. What you found is an example that shows that the hypothesis is crucial, as the counting measure is not $\sigma$-finite on any ...
Martin Argerami's user avatar
3 votes
Accepted

Suppose $1\leq p_1<p_2<+\infty$ and $\mu$ a finite measure, then $\mathscr{L}^{p_2}(X,\mathscr{A},\mu)\subseteq\mathscr{L}^{p_1}(X,\mathscr{A},\mu)$

Thanks to @RyszardSzwarc's comment. Let $f\in\mathscr{L}^{p_2}(X,\mathscr{A},\mu)$, then $\int|f|^{p_2}d\mu<+\infty$. Let $p=\frac{p_2}{p_1}$ and $q=\frac{p_2}{p_2-p_1}$. Then $p,q\in[1,+\infty]$ ...
Beerus's user avatar
  • 1,795
1 vote

Need help understanding the switching of integral limits with Fubini's theorem: $\int_0^cm(\{x:|f(x)|>t\})dt=\int_{\{x:c\geq |f(x)|\geq 0\}}|f(x)|dx$

You are allowed to switch the order of integration when the integrand is nonnegative and both spaces you integrate over are fixed (Tonelli's theorem). To get the integral in that form, use the common ...
Kakashi's user avatar
  • 1,831
1 vote
Accepted

Need help understanding the switching of integral limits with Fubini's theorem: $\int_0^cm(\{x:|f(x)|>t\})dt=\int_{\{x:c\geq |f(x)|\geq 0\}}|f(x)|dx$

The identity you write is incorrect: notice that if $t\in[0,c]$ then $f^{-1}((t,\infty)) = f^{-1}((c,\infty))\cup f^{-1}((t,c])$ hence $m(f(x)>t) = m(f(x)>c)+m(t<f(x)\leq c)$. It follows that ...
krm2233's user avatar
  • 4,791
0 votes

$\mathscr{L}^{p_1}(\mathbb{N},\mathscr{A},\mu)\subseteq\mathscr{L}^{p_2}(\mathbb{N},\mathscr{A},\mu)$: $1\leq p_1<p_2<+\infty$, $\mu$ counting measure

Thanks to @ThànhNguyễn's comment and help! Proposition$\quad$ Let $(\mathbb{N},\mathscr{A},\mu)$ be a measure space, where $\mathscr{A}$ is the $\sigma$-algebra of all subsets of $\mathbb{N}$ and $\...
Beerus's user avatar
  • 1,795
0 votes

Is there an example of a sigma algebra that is not a topology?

I am just a beginner in this domain, so pardon me if my reasoning is wrong. I will be happy hear correct answers from the members. I am going through Measure Theory by John K. Hunter online material (...
Kalidas's user avatar
  • 19
0 votes

Prove independence of events given random variables are iid and have (not absolutely) continuous cdf

If $y\mapsto F_Y(y)$ is continuous and $Y$ has df $F_Y$, then $F_Y(Y)\sim U[0,1]$. Let ${Y}_m^*:=\max(Y_1,...,Y_m)$ where $(Y_n)_{n \in \mathbb{N}}$ are IID with continuous cdf $F_Y$. Then for $m\geq ...
Snoop's user avatar
  • 15.6k
2 votes
Accepted

Let $A:=\{f\in C^1(\mathbb{R}): f, f'\in L^1(\mathbb{R}\}$. Then are the Schwartz functions dense in $A$ w.r.t. $\|f\|=\|f\|_1+\|f'\|_1$?

The Schwartz space is dense in $W^{k,p}(\mathbb R^n)$ (1), so it is in particular dense in $W^{1,1}(\mathbb R^n)$ which has the norm $$\lVert u\rVert=\sum_{|\alpha|\leq 1}\lVert D^\alpha u\rVert_1 = \...
Tina 's user avatar
  • 166
2 votes

For every measurable $E$ with $0<m(E)< \infty$, there is $[a,b]$ such that $m(E \cap [a,b]) > \frac{1}{2} m(E)$

In this question, we have that, if $0 < m(E) < \infty$, for any $\epsilon > 0$, there exists $M > 0$ such that $$ m(E \setminus [-M, M]) < \epsilon $$ Now, choose $\epsilon = \...
Thành Nguyễn's user avatar
0 votes

For every measurable $E$ with $0<m(E)< \infty$, there is $[a,b]$ such that $m(E \cap [a,b]) > \frac{1}{2} m(E)$

Hint: Consider a function $\phi: \mathbb{R} \to [0, \infty]$ defined by $\phi(x) = m(E \cap B(0,x))$ where $B(0,x)$ is the ball centred at $0$ with radius $x$. First show that $\phi$ is continuous, ...
approximate-identity's user avatar
0 votes

What is the motivation of Measure Theory when there is probability theory?

Maybe I don't understand your question but... Unless your question is actually a duplicate of Probability and measure theory ... It sounds like you're asking why measure theory exists in general for ...
BCLC's user avatar
  • 13.6k
3 votes
Accepted

Proof: If $f\in\mathscr{L}^{\infty}(X,\mathscr{A},\mu)$, then $\{x\in X:|f(x)|>\|f\|_{\infty}\}$ is locally $\mu$-null.

How can we make sure that such a sequence $\mathbf{\{M_n\}}$ exists? If $t$ is a real number and $A \subset \mathbb{R}$ such that $\inf A = t$, then there is a sequence $a_1 \ge a_2 \ge a_3 \ge \...
kobe's user avatar
  • 42.2k
0 votes

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 ...
mathgoo's user avatar
1 vote
Accepted

Radon Nikodym derivative and distribution function

Since $\mu$ is a Borel Measure, and $\varphi$ is an increasing function, let $\beta_{\varphi}$ denote the Lebesgue-Stieltjes measure correspoding to $\varphi$ on $\mathbb{R}$. That is, a function $\...
unconscious statistician's user avatar
3 votes

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

As it has been pointed out in an earlier comment, $\Sigma \cap A$ is not a $\sigma$-algebra on $X$, but on $A$. Your observation, then, shows either that $\Sigma \cap A$ is not a $\sigma$-algebra on $...
Emilio Mora's user avatar
2 votes
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{...
user6247850's user avatar
  • 13.8k
0 votes

Proof in measure theory

To be honest, that is one of the results that everyone uses but (almost) no one knows how to prove it without checking a reference. The main idea is to use the spherical coordinates which is $(x_1,\...
Raoní Cabral Ponciano's user avatar
0 votes
Accepted

Stokes theorem for currents on manifolds with corners

This question has been largely answered by Huber 2023: Proposition 6.3: Let $X$ be a definable $C^p$-manifold with corners for $2\leq p<\infty$. Let $G\subset X$ be a pseudo-oriented compact ...
Derivative's user avatar
  • 1,748
0 votes

$\sigma$-finite measure $L^p$ space is isometric to a finite measure $L^p$ space

This is a (very) late answer. The measure $λ$ given by $λ(Α) = \int_A h \, d\mu $ is finite and the operator $A \colon L^p(\mu) \to L^p(λ)$ given by $Af = h^{-1/p} f$ is an linear isometry since $$ \|...
Evangelopoulos Foivos's user avatar
5 votes
Accepted

If $E \in \mathcal{A}$ satisfies $\mu(E)>0$, then there exists $F \subset E, F \in \mathcal{A}$ with $0 < \mu(F) < \infty$

Version 1: $F\subset E$ means $F\subsetneq E$: This is in general false. We can have "atoms". For example, let $X$ be any nonempty set, $x_0\in X$, $\mathcal{A}=\mathcal{P}(X)$ (power set of ...
Severin Schraven's user avatar
2 votes
Accepted

Prove that $\lim_{k \to \infty} \int_{\mathbb{R}} g_k fdm_1 = f(0)$

Via the change of variables $u=kx$ we find $$k\int_{\mathbb{R}} g(kx) f(x) \ dm_1 =\int_{\mathbb{R}} g(u) f\left(\dfrac{u}{k}\right) \ dm_1 $$ Given that $f$ is bounded by some constant $M$ and $g$ is ...
Julio Puerta's user avatar
  • 8,177
5 votes
Accepted

Measure of projection of a set is zero

This is essentially the same question as "can the Minkowski sum of two null sets be a non-null set?" That's because if $S = A \times B$, then the projection of $S$ onto the diagonal is $\{\...
Izaak van Dongen's user avatar
2 votes

A conjecture about measurable functions

Your conjecture is false. Let $\mathcal H$ be the $\sigma$-algebra of Lebesgue measurable sets in $\mathbb R$. Knowing whether each half-line $(-\infty, x)$ occurs, i.e. whether $X < x$, uniquely ...
Robert Israel's user avatar
0 votes

Inner measure doesn't care whether finite or countable

I was thinking about this today while reading the Tao's book and have this proof which I'm not sure whether it's correct or not (also this proof is not related to subadditivity of elementary measure ...
InfSuplife's user avatar
2 votes

Why do we require intervals in Jensen's inequality?

I assume by interval the author allows for unbounded intervals, e.g. $\mathbb{R}$ is an interval. As you can see there are no problems in the proof if $I$ is interchanged with $\mathbb{R}$ or $[0,\...
LostStatistician18's user avatar
3 votes
Accepted

Why do we require intervals in Jensen's inequality?

The theorem remains valid when $I= \mathbb R,$ in fact that is the more usual form in which it is stated. Your more general form comes in handy when you want to consider functions $f$ that are only ...
Lieven's user avatar
  • 1,049
2 votes

Finite linear combination of characteristic functions measurable implies each set is measurable?

The claim is not true. Suppose $X=[0,1]$ and $\mathcal{F}$ is the Lebesgue $\sigma$-algebra on $[0,1]$. Let $V$ be the Vitali set on $[0,1]$. Define $A=X\setminus V$. Then $$\chi_{[0,1]} = \chi_V + \...
Joseph Basford's user avatar
1 vote

"Leibniz's rule" for $t\in\mathbb{R}^n$

Partial derivatives are derivatives of single-variable functions obtained by fixing all other points, so the proof is easily reduced to the case $n=1$. If you want a reference, look at Amann Escher, ...
peek-a-boo's user avatar
  • 57.3k
0 votes

If $\mathcal{A'}$ is an algebra then $\mathcal{A}=\{A \cap E : A \in \mathcal{A'}\}$ is an algebra on $E.$

In this case $B^c=E-B$ since $B\subset E$ and you want to show that $\mathcal{A}$ is an algebra on $E$. To show that $B^c \in \mathcal{A}$ you can observe that there exists $B'\in\mathcal{A'}$ so that ...
pschz's user avatar
  • 1
1 vote
Accepted

The Medians of Lipschitz Functions on $(X,d,\mu)$ (Existence and Uniqueness)

Short answer: Existence and uniqueness of a median (of a continuous function $\varphi : X \to \mathbb{R}$) follow under the additional assumptions (i) $\mathrm{supp}(\mu) = X$, (ii) $X$ is connected ...
Jordan Payette's user avatar
1 vote

Equivalence of definitions of "standard Borel space"

I think these are equivalent but I agree that books in descriptive set theory do not make these things clear. From the Srivastava book A course on Borel sets we have: "A standard Borel space is a ...
Michael's user avatar
  • 24.3k
4 votes

$\left |\int _{X} log(|f|)d\mu \right | < \infty$ if $\int_{X} |f| d\mu <\infty $?

The integral may diverge for two reasons. The first one concerns the measure by which we integrate. If the measure is large, the integral may also be large (like $\int_{\mathbb{R}}1 \,\mathrm{d}\...
Marek Kryspin's user avatar
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 $n$. ...
Karthik Kannan's user avatar
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 = \...
Jordan Payette's user avatar

Top 50 recent answers are included