# Tag Info

3

$$E_n=\{x\in X\mid |f(x)|\leqslant n\}$$

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 ...

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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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