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1

Let's have $p$ the north pole ($p_n=1$, $p_1,\ldots,p_{n-1}=0$) and $A$ the southern hemisphere $A=\{x\in S^{n-1}:x_n\leq0\}$. Then in euclidean metric $d(A_{1/2},p)>1$ thus we know that the pole covers at most the area of $1-\mu(A_{1/2})$. By the MCT we have $1-\mu(A_{1/2})\leq 2e^{-n/8}$ - upper bound on the area covered by one point. It follows that we ...


1

If the sequence $(X_n)$ is independent and $X_n\to X$ almost surely then $X$ is an almost sure constant. Hence the sequence $(X_n-X)$ is independent.


1

Fix $\varepsilon\gt 0$: there exists an integer $N$ such that if $n,m\geqslant N$ and $F\subset A$ is finite, then $$\sum_{\alpha\in F}|f_n(\alpha)-f_m(\alpha)|^2\leqslant\varepsilon.$$ What is important here is that $N$ depends only on $\varepsilon$ but no on the finite set $F$ we are considering. We thus obtain, taking the limit $m \to \infty$, that for ...


1

From your answer I see that it is ok to use standard facts from real analysis, so let's use them! Fubini's theorem on differentiation. Assume $(f_n)_{n\in\mathbb{N}}$ is a sequence of non-decreasing functions on $[a,b]$, and the series $\sum_{n=1}^\infty f_n(x)$ converges for all $x\in [a,b]$, then $$ \left(\sum_{n=1}^\infty ...


1

For the first part of the problem: Suppose $g,h$ are two absolutely continuous functions on $[0,1]$ such that $f$ is equal almost everywhere both to $g$ and $h$ with $g',h'\in L^{2}[0,1]$. We need to show that $g' = h'$ almost everywhere. Suppose for a contradiction that there is a closed interval $[a,b]$ such that $g'(x)\neq h'(x)$ for all $x\in ...


0

For the first part, you need to prove that if a function equals $g$ and another function $g_1$ almost everywhere $A$ and $A_1$ respectively. Then $g$ equals almost everywhere $g_1$ ($A_1\cup A_2$). And two elements equaling almost everywhere in $L^2[0,1]$ are equal. Which gives the well defined part.


0

The first hint can be derived from Parseval's theorem (which is an application / case of Bessel's inequality); note that $\int_A \sin(n_k x) dx$ is the imaginary part of the $n_k$-th Fourier coefficient of $\chi_A$, and $\chi_A \in L^2[-\pi,\pi]$. Thus, $$\|\chi_A\|_{L^2} = \|(a_k)_{\ell^2} \| = \sum_{k \in \mathbb{Z}} |a_k|^2$$ where $a_k$ denotes that ...


2

Assuming that $P_n$ is positive set in the decomposition for $\lambda-n\mu$, note that for each $n$, $$ (\lambda-n\mu)(N_n)\leq 0 $$ Since $0\leq \mu(N_n)<\infty$, this is equivalent to $\lambda(N_n)\leq n\mu(N_n)<\infty$, that is, each $N_n$ has finite measure with respect to $\lambda$. It follows that $N$ is $\sigma$-finite. Next, let $E$ be a ...


1

Write that $B_t - B_s \sim N(0,t-s)$ and the bilinear form as a sum of squares: $$ E\exp (a(B_t - B_s))= \int \frac{dx}{\sqrt{2\pi}} \exp \left(a\sqrt{t-s} x - \frac{x^2}2\right) \\= \int \frac{dx}{\sqrt{2\pi}} \exp \left(-\frac 12\left(x-a\sqrt{t-s}\right)^2\right) \exp\left(\frac 12 a^2\left(t-s\right)\right) = \exp\left(\frac 12 ...


0

Assume that $L^r\subset L^\infty$. Then $L^r\subset L^{r+1}$, hence from the case $s:=r+1<\infty$, we deduce the existence of the wanted $\varepsilon$. Conversely, assume that there is some positive $\varepsilon$ such that for each measurable set $E$, $\mu(E)\in \{0\}\cup [\varepsilon,\infty)$. Let $f$ be an element of $L^r$. Define $E_n:=\{f\gt n\}$. ...


2

Such a measure is called absolutely continuous [with respect to the Lebesgue measure].


0

The Cantor-Lebesgue function $f: \mathbb{R} \to \mathbb{R}$ is a standard counterexample to this (with Lebesgue measure). To show it is a counterexample, we need to find a set $S$ such that $S$ is measurable, yet $f(S)$ is not measurable. Let $V$ be a Vitali set, and let $v_0 \in V$ be the unique rational element of the vitali set. Take $S = f^{-1}(V ...


4

You're right that $E$ shouldn't be closed, but it doesn't follow that it should be open. The only open set of measure 0 is the empty set. You could take $E$ to be the set of rational numbers (in $\mathbb R$).


2

Consider $\Bbb Q$. The boundary and the closure of $\Bbb Q$ is $\Bbb R$.


1

First of all, $f_n\to f$ in $L^2$ implies $f_n\to f$ in measure, so the second assumption is redundant. There is a standard counterexample to show that convergence in $L^p$, $1\le p<\infty$, does not imply convergence a.e., much less "almost uniformly". Namely, enumerate dyadic subintervals of $[0,1]$ as $I_1,I_2,\dots$ (order does not matter), and let ...


1

For $p<1$, we say that $f_n\to f$ in $L^p$ if $\lim_{n\to\infty}\int |f_n-f|^p=0$. Define $g_n:=|f_n-f|^p$: it converges to $0$ in $\mathbb L^1$ by assumption and $g_n\to |f-g|^p$ almost everywhere. We deduce from the case $p=1$ that $|f-g|^p=0$ a.e. hence $f=g$ a.e.


1

The only outer measures for which your condition holds are the so-called regular outer measures $^{[1]}$, that is, the ones for which every set $A\subset X$ admits a measurable set $E$ such that $A\subset E$ and $m^\star(A)=m(E)$. (In Munroe's book Introduction to measure and integration, such a set $E$ is called a measurable cover for $A$.) To prove this ...


3

This is false. Let $\mu$ be Gaussian measure on $\mathbb{R}$ and $\lambda$ Lebesgue measure.


2

In fact $\lim_N Y_N$ exists almost surely is equivalent to $E(|X_1|) < +\infty$, which is not the case for Cauchy variable. One direction of above assertion is the famous law of large number. To prove the other direction, note that if $\lim_N Y_N$ exsits alomst surely, then $\lim_N\frac{X_N}{N} = 0$ alomst surely, which we will see is impossible if ...


1

Hint: a function is Lebesgue measurable if the inverse image of every open set is Lebesgue measurable.


0

Meanwhile I was working the problem came user141421's answer which shows that, for $p \gt 1$, $$I(p)=\int_2^\infty \left(\frac1{x\log^2x}\right)^p\mathrm dx$$ converges. Just for your curiosity, I give you a few values of this integral $$I(1)=\frac{1}{\log (2)}\simeq 1.4427$$ $$I(2)=\frac{\log (2) \left(2 \log ^2(2) \text{Ei}(-\log (2))-1+\log ...


1

For $p>1$, the integral converges, since $$\dfrac1{x\log^2(x)} < \dfrac1x$$ and $$\int_2^{\infty} \dfrac{dx}{x^p} = \dfrac1{(p-1)2^{p-1}}$$


1

We need two ingredients: If a sequence $(f_n)_n$ converges in some $L^p$ then there exists a subsequence $(f_{n_k})_k$ that converges almost everywhere. A subsequence of a convergent sequence converges to the same limit. Now, since $(f_n)_n$ converges to $g$ in $L^1$ then, (by 1.), there there exists a subsequence $(f_{n_k})_k$ that is simply ...


2

Yes, it is true. Define a (complex-valued) measure on $\mathbb{R}$ by $$\mu(B) := \sum_{k \in \mathbb{Z}} c_k \delta_k(B), \qquad B \in \mathcal{B}(\mathbb{R}). \tag{1}$$ Since $\sum_{k \in \mathbb{Z}} |c_k|<\infty$, $\mu$ is a finite measure on $\mathbb{R}$. From $(1)$ we see that the formula $$\int f(x)\left( \sum_{k \in \mathbb{Z}} c_k \delta_k(dx) ...


0

I would do it as follows; say for $\overline D_\nu\mu$. First, observe that for each fixed $r>0$, the functions $x\mapsto \mu(B_r(x))$ and $x\mapsto \nu(B_r(x))$ are Borel. There is a reference for this (not completely trivial) fact in the link you give. It follows that the set $\Omega=\{ x\in\mathbb R^N;\; \nu(B_r(x))>0\;\hbox{for all}\;r>0\}$ is ...


0

Given $ \varepsilon> 0 $ is $ k \in \ N $ such that $\varepsilon >1/k$ consider a 'homogeneous' partition of $ [0,1] \times [0,1] $, i.e., all sub- rectangles of partition has area $1/k^2$, we know that $ m_B = 0$ and $ m_B = 1$ if $B$ intection the line $ y = x $ is not empty set, so $ S (f, P)-s (f, P) = \sum_ {B \ in P} (m_B-m_B) vol (B) = \sum_ ...


3

The german word for pre-measure is Prämaß, sweet and simple, and pre-measures is Prämaße. Maß simply means measure, and prä is the german version of the english pre. Maßnahme is only then a correct translation of measure if measure is used as a synonym for action. So for example "Measures haven been taken" translates to "Maßnahmen wurden ergriffen", and ...


-1

I don't speak much German, but here's my attempt: $$\text{Pre-Measure}_{\text{English}}=\text{VorMaß}_{\text{German}}$$ To show that my attempt is "right", I found an article that uses $\text{Maß}$, by Oliver Deiser - Ordinalzahlen in der Analysis und Maßtheorie.


-1

Vorgemaßnahme? Not sure, as thats a direct translation rather than a Mathematical translation, but that means premeasure root for room. Vorge- = pre- maßnahme=measure


1

Well your definition of being bounded in $L^1(\Omega)$ is strange, but with this one, and if $|\Omega|$ refers to the measure of $\Omega$, then you just have $\int_{\Omega}|f_n|dx \leq \int_{\Omega}Cdx \leq |\Omega|C$ so $\delta=\frac{\eta}{C}$ works well... If your definition of being bounded in $L^1$ is the usual one, id est ...


0

The set of measurable sets form a $\sigma - algebra$. So they are closed under countable unions and intersections. Subset of a measurable set need not be measurable in general.See the http://en.wikipedia.org/wiki/Vitali_set. But if the measurable set has measure zero and if the space is complete then all its subsets are measurable. If $E$ is a measurable ...


1

Let $P$ be a non-measurable subset of the real axis $R$. The following conditions are equivalent: (i) for each measurable subset $E \subset P$ the condition $m(E)=0$ holds; (ii) the inner $m$ measure of $P$ is equal to zero.


0

To get pairwise disjoint sets let $\tilde{U}_1 = U_1$ and $\tilde{U}_j = U_j - \overline{U_{j-1}}$ Then, $\sum_{j=1}^m l(\tilde{U_j}) \leq \sum_{j=1}^m l(U_j) < \epsilon$ for all $m$ and $X \subset \bigcup_{j=1}^m \tilde{U}_j$ .


0

You can reduce this to the Fourier basis problem, if that problem is known to you. More explicity, if $f \in L^{2}[0,1]$, then $$ f = 0\;\;a.e. \iff \int_{0}^{1}f(x)e^{i2\pi nx}\,dx =0,\;\; n=0,\pm 1,\pm 2,\pm 3\cdots \;. $$ If $\int_{0}^{1}f(x)x^{n}\,dx=0$ for $n=0,1,2,3,\cdots$, then expand the exponentials in a power series to see that ...


1

Since $\nu$ is absolutely continuous with respect to $|\nu|$, there exists by (Radon-Nikodym) $f\in L^{1}(X,\mathfrak{M},|\nu|)$ such that $d\nu = fd|\nu|$. Equivalently, $\nu(E) = \int_{E}fd|\nu|$ for all $E\in\mathfrak{M}$. To verify the other condition, suppose there were an set $U$ of non-zero measure such that $|f(x)|\neq 1$ for all $x\in U$. ...


0

Hint: consider $\phi = 1_{f <0}$ in the definition of the weak convergence. details: you know that $$ \lim \int f_n \phi = \int f\phi $$ $LHS = \lim \int f_n \phi \ge 0$ because $f_n \phi \ge 0$ $RHS = \int_{f<0} f \le 0$ so both of then are $0$. In particular, with $\lambda$ the Lebesgue mesure, $$ \lambda\{ f<0\} = 0 $$ hence $f \ge 0$ ae. ...


1

Let $E_n = \{x : f_n(x) \ne g_n(x)\}$, and let $f, g$ denote the respective suprema. If $$f(x) \ne g(x)$$ then $x \in E_n$ for some $n$ (do you see why?). Thus $$\{x : f(x) \ne g(x)\} \subseteq \bigcup_{n = 1}^{\infty} E_n$$ which has what measure?


3

We know that continuous functions are dense in $L^p([a,b])$ (where $-\infty<a<b<\infty$) and also that polynomials are dense in $C([a,b])$ and so we see that polynomials are dense in $L^p([a,b])$ by standard arguments. Thus we can see that your statement is effectively equivalent to asking if a function is orthogonal to everything in $L^2([a,b])$. ...


0

In fact, one could take any function $f\in L^2([0,1])$ such that $\mu\{x \, | \, f(x) \neq 0\} = 0$. Then $\int_0^1 x^n f(x)\, \text{d}\mu(x) = 0$, but $f(x) \neq 0$. This is, of course, not a counterexample, but it does give an idea of the best possible result one could expect. Namely, if $\int_0^1 f(x) x^n \,\text{d}x = 0$ for every $n\geq 0$, then $f = ...


1

For the first part, note that if $\mu^*(Y)=\infty$, then every set $A\in\mathcal{M}$ such that $Y\subset A$ satisfies $\mu(A)=\infty$. In particular, $\mu(X)=\infty$ and you can take $X$ as the set you're looking for. If $\mu^*(Y)<\infty$, for each $n\in\mathbb{N}$, let $A_n$ be a set in $\mathcal{M}$ such that $\mu(A_n)<\mu^*(Y)+\frac{1}{n}$ and ...


1

I'm going to use the definition of Radon measure as in Federer's book, Geometric Measure Theory (2.2.5) as follows: By a Radon measure we mean a measure $\phi$, over a locally compact Hausdorff space $X$, with the following three properties: If $K$ is a compact subset of $X$, then $\phi(K)<\infty$. If $V$ is an open subset of $X$, then $V$ ...


0

Your idea is good, and you must only take a little more care for it to work. As you said, if $K$ is a compact metric space, then $C(K)$ is separable. Now, suppose that $X$ is is a locally compact, $\sigma$-compact metric space. You can find a sequence $\left\{K_n\right\}_{n\in\mathbb{N}}$ of compact subsets of $X$ satisfying: ...


0

Since the function lies above its tangents, we have $$f(x)\ge f(a)+f'(a)(x-a)$$ As $x\to 0$ we get $$0\ge f(a)+-af'(a)\implies af'(a)-f(a)\ge 0$$ And $$\frac{d}{dx}\left({f(x)\over x}\right)=\frac{f'(x)x-f(x)}{x^2}\ge0$$ which establishes that $g$ is increasing (I'm not sure if you should perhaps use strict inequalities instead).


5

There is a very precise sense in which the answer to your question is "$0$". Let us denote by $ND$ the set of all continuous nowhere differentiable functions on, say, the interval $[0,1]$, and by $\mathcal C([0,1]$ the Banach space of all continuous functions on $[0,1]$. Then, it can be shown that $SD:=\mathcal C([0,1])\setminus ND$ (the set of all ...


1

If $(\Omega,\Sigma,\mu)$ is a probability space, it is possible to prove that one can choose representatives from elements of $L_\infty(\Omega,\Sigma,\mu)$ such that all finite algebraic operations are preserved. Such a choice of representaives is known as a Lifting. Liftings exist by the, terribly advanced, von Neumann-Maharam Lifting Theorem. But if we ...


1

The convergence is uniform when taking out $[1-\epsilon, 1]$ is because for each $x\in [0,1-\epsilon)$ and for each $c>0$, there exists an $N$ such that for every $n\geq N$ we have $$x^n < (1-\epsilon)^n <c.$$ The reason that the convergence is not uniform when only taking out the end point $1$ is because for each $n\in \mathbb{N}$ and for each ...


2

$$ f(x,y) \ge \begin{cases} e^{-1}, &\text{if } y\le 1/x, \\ 0, &\text{otherwise}; \end{cases} $$ Since the set $\{ x\ge1,\, 0\le y\le1/x\}$ has infinite measure, the right-hand side is not integrable, hence neither is the left-hand side.



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