# Tag Info

0

Or consider any ascending or descending set. That is, $A_n \subseteq A_{n+1} or A_n \supseteq A_{n+1} \forall n \in \mathbb{N}$. I like @user159813's answer better though.

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Note that the Lebesgue outer measure is usually constructed as follows: Let $m : \mathcal{R}(\mathbb{R}^n) \to \overline{\mathbb{R}}$ where $\mathcal{R}(\mathbb{R}^n)$ is the set of all $n$ dimensional rectangles so that $$[a_1, b_1] \times [a_2, b_2] \times \cdots \times [a_n, b_n] \mapsto \prod_{i=1}^n (b_i - a_i)$$ then we put the Lebesgue outer ...

2

An outer measure is a set function which is not necessarily a measure as it is monotonic and subadditive, but not necessarily additive. Outer measures are usually defined an all subsets of the underlying set. For example, the Lebesgue outer measure is defined on all subsets of the set of reals. However, as I mentioned, it is only a subadditive function of ...

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To quote Durrett, 'the game is "guess and verify."'. Note that if you know $X_n$, then you must have $X_{n+1} = X_n + { 1\over 10^{(n+1)} } d$, where $d \in \{0,...,9\}$ and each $d$ is equally likely. If you know $X_n$, then you must have $X=X_n+ { 1\over 10^{n} } u$, where $u \in [0,1)$, where $u$ is uniformly distributed.

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When covering the rationals bit small intervals you are right: there will be overlap, but that's not a problem. The point is that you will still get stuff less than $\epsilon$ overall, so it's still OK. For example: cover the $n^{th}$ rational by an interval of size $\epsilon 2^{-n}$ then the total sum of all the rational lengths is $$\sum_{n=1}^\infty ... 1 From Example 2.9 in the paper: "Let \mu and \nu be counting measure and Lebesgue measure, respectively." You can decompose \mu as \mu+0 or \mu+\nu or \mu+2\nu, etc... they are all the same measure. -1 Ok to bad, the answer is again: No! For s_n:=n\chi_{[0,\frac{1}{n}]} it follows s_n\to 0 but I(s_n)\to 1. 6 Then can you help me to show that \alpha is in fact stopping time? Well, not very easily I am afraid, since \alpha is not always a stopping time for the filtration (\mathcal F_t)_{t\geqslant0} defined by \mathcal F_t=\sigma(X_s;0\leqslant s\leqslant t). For a counterexample, consider some Bernoulli random variable U such that P(U=1)=\frac12 ... 2 I suggest you take a look at this question. In the accepted answer there is also a link to a very good survey which mentions more about a dozen metrics between probabilities. There are also discussions which metrics are useful for which problems, and results on the comparisons between the metrics (e.g. which one bounds another etc.). The computability ... 2 Byron's answer is perfectly OK. Here is another one. Since the polynomials are uniformly dense in \mathcal C([a,b]) (the space of continuous functions on [a,b]), it follows from the assumption that \int_{[a,b]} g(x)f(x)\, dx=0 for all g\in\mathcal C([a,b]). (This means that the signed measure f(x)dx is the 0 measure). From this, "it is well ... 1 This is a special case of Does scaling lead to weak convergence to the null function? (in the notation of that post, use f=\chi_{[-1,1]}). But since your question does not use the language of weak convergence, I give an answer without it. The idea is any L^p function (for p<\infty) is somewhat localized. Namely, for every \epsilon there exists ... 1 I remember seeing this question here, but I can't find it now. Here's the proof that no such S exists: Suppose such S existed. Define the (\sigma-finite) Borel measure \lambda(A)=\mu(A\cap S). Since \lambda and \mu/2 are equal on the open intervals, which generate the Borel \sigma-algebra, then \lambda=\mu/2. Let \chi_S be the ... 1 Instead of the definition of the upper Minkowski dimension, use the formula from page 78 (this is Mattila's book of course):$$ \overline{\dim}_M B =\limsup_{\epsilon\to0}\frac{\log P(B,\epsilon)}{\log(1/\epsilon)} \tag1 $$As you recalled, the definition of P(B,\epsilon) is that it's the maximal number of disjoint balls of radius \epsilon with centers ... 1 Yes, this follows by unwrapping the definitions. Let's recall that (the restriction of) the completion e(p) is defined as follows: if F'\in\mathcal F'\subset\mathcal F^p, then we can write F'=F\cup N with N\subset N' and F,N'\in\mathcal F and p(N')=0. In this situation, e(p)(F')=p(F). (That this is well defined, is part of the standard theory.) ... 3 I believe this is true without putting any extra assumptions on X: Since \nu is a probability measure, there exists a sequence of finite convex combinations of delta measures \{ \sum_{i=1}^{m_k} a_{k,i} \delta_{x_{k,i}} \}_k converging weakly to \nu, with each x_{k,i} \in X. Therefore if p = \int_X x~\nu(dx), we have$$ p = \lim_{k \to \infty} ...

3

A modification of my answer here. You could use the functional form of the Monotone Class Theorem. Let $\cal H$ be the collection of all bounded, Borel measurable functions $g$ so that $\int g(x) f(x)\, dx=0$. Then $\cal H$ is a monotone vector space (exercise for the reader!). Let $\cal K$ be the set of functions $\{x^k: k\in\mathbb{N}\}$. Then $\cal K$ ...

1

Use the fact that if the $limsup$ is less than infinity for any specific value of $s,$ then the $limsup$ is zero for all strictly smaller values of $s;$ and if the $limsup$ is greater than zero for any specific value of $s,$ then the $limsup$ is infinite for all strictly greater values of $s.$ This is illustrated by the following suggestive computations, ...

2

Weak convergence is defined by the criterion $$\int f(x) d\mu_n \to \int f(x)\,d\mu$$ for all $f \in C_b(X)$. That gives you $\mathcal{P}(X)$ as a subset of the closed unit ball of $C_b(X)^\ast$ (which can be canonically identified with the space of Borel measures with finite variation on the Stone-Čech compactification $\beta X$ of $X$ if $X$ is a ...

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Counterexample:$$X=([0,1]\times\{0\})\cup\bigcup_{n=1}^\infty(\{\frac1n\}\times[0,\frac1n])$$

1

By definition $\mathcal H_\delta(A)^s=\inf\{\sum_id(E_i)^s: A\subset \cup_i E_i, d(E_i)<\delta\}$ Terefore, by definition of infimum, for any positive number $K$, there must exist a family $E_i$ with $d(E_i)<\delta$, $A\subset U_i(E_i)$ and $\sum_i d(E_i)^s<\mathcal H_d^s(A)+K$. Set $K=1$.

0

Well, I found a counter-example... Take $H$ and delete an open arc from each circle (not hitting the point where the circles intersect).

4

For every function $u$, the set $\{u\le k\}$ is well-defined. But the elements of $L^p$ are not functions; they are equivalence classes of functions. Since you consider a Bochner space, let's recall what this means: $u\sim v$ if for almost every $t$, the equality $u(\cdot,t)=v(\cdot, t)$ holds for a.e. $x$. Then $\{u\le k\}$ is really an equivalence class ...

1

Let $A_n = f_n^{-1} (M,\infty)$. Let $B=A \cap [a,b]$ and $B_n = A_n \cap [a,b]$ where $0 \le a<b$. Since $f_n(x) \to f(x)$ for ae. $x$, if $x \in B$, then $x \in B_n$ for all $n$ sufficiently large, that is $1_{B_n}(x) \to 1$. Hence $\mu(B \cap B_n) \to \mu B$. We see that $f_n(x) \to f(x)$ for ae. $x \in B$, hence $\liminf_n f_n(x) 1_{B_n \cap B}(x) ... 1 Here is a slightly more general approach for comparison. If$f$is integrable on [a,b], then for any$\epsilon > 0$there exists$\delta > 0$such that for any measurable set$D \subset [a,b]$with measure$\mu(D) < \delta$, then $$\left|\int_{D}f(x)\,dx\right|< \epsilon.$$ Since$f$is measurable , then for any$\epsilon>0$it can be ... 3 I don't see any use of any Hardy-Littlewood inequality here. You have a measure space$(Q,\delta^\alpha \,dx\,dt)$, which has finite total measure. I will denote the measure by$\mu$for simplicity. The assumption (2.7) says that$u$is in the weak$L^{\hat q}( d\mu)$space. Then it's just a matter of interpolation to get that$u\in L^q( d\mu)$for every ... 1 If we set $$I_n = \int_{0}^{n}\left(1-\frac{x}{n}\right)^n\cos\frac{x}{\sqrt n}e^{x/2}dx$$ we obviously have: $$I_n \leq \int_{0}^{n}\cos\frac{x}{\sqrt{n}}e^{-x/2}dx\leq \int_{0}^{\sqrt{n}}\cos\frac{x}{\sqrt n}e^{-x/2}dx+(n-\sqrt{n}) e^{-\sqrt{n}/2}\leq 2+o(1),$$ while: $$\left|I_n-\int_{0}^{n}\cos\frac{x}{\sqrt{n}}e^{-x/2}dx\right|\leq ... 0 The supremum of a countable family of measurable functions is measurable. So is the infimum. Both follow from the fact that sub-level sets are countable unions (resp. intersections). So, how to express D^+f by means of such operations? Introduce$$ f_{h} = \frac{f(x+h)-f(x)}{h} $$Take the supremum over all rational h such that 0<h<1/n. By ... 0 The "only if" part only holds for real-valued f. For any f \in L^2(\mathbb{R}), be it real- or complex-valued, we have the continuous differentiability of$$g(t) = \int_\mathbb{R} f(x)^2 \exp (-tx^2)\,dx$$and the formula$$g'(t) = \int_\mathbb{R} \left(-x^2\right)f(x)^2\exp (-tx^2)\,dx$$on (0,+\infty) by the dominated convergence theorem, since ... 3 "Locally" is ambiguous here f is locally Lipschitz in \Omega if and only if f \in W^{1,\infty}_{loc}(\Omega) The validity of this claim depends on interpretation of "locally Lipschitz". Does it mean every point of \Omega has a neighborhood in which f is Lipschitz, or there is L such that every point of \Omega has a neighborhood in ... 0 Looks fine. Just a small comment on your string of inequalities following "n > N implies ...": I would only keep the first two inequalities, the latter two are not necessary. 1 since f is positive . the function F(x)=\int_{0}^{x}f(t)dt is an increasing funcion . and so since 0\leq F(x)\leq \int_{0}^{1}f(t)dt. If I=0 then necesarly F(x)=0 =>... f=0. 3 In the big Montel theorem, and often otherwise, a normal family is a family \mathcal{F} of meromorphic or holomorphic functions, such that every sequence (f_n) of functions in \mathcal{F} has a subsequence that converges uniformly on compact sets to a meromorphic/holomorphic function, or to the constant function z \mapsto \infty. Your (f_n) ... 2 Well, we assume that you're considering Riemann integral, which is equivalent to Darboux integral. Suppose that \int_a^bf=0 and f\ge0, we'd show that \exists x\in[a,b],f(x)=0. Suppose I is a closed interval, and \int_I f=0. By definition of Darboux integral, given \epsilon>0, there's a partition x_0<x_1<\dotsb<x_n such that ... 0 The following is not a complete answer. In the following let f be some candidate for a counterexample. First observation: f cannot be monotone. If it were, say, increasing, then we would have$$\int_0^x |f'| dx \leq f(x)-f(0)$$by the Lebesgue decomposition theorem. Similarly \int_0^x |f'(x)| dx \leq f(0) - f(x) if f is decreasing. Second ... 1 Maybe I am missing something, but the following seems to suffice. By the definition of the product measure (or Tonelli's Theorem if you prefer),$$ (\mu\times\nu)(E)=\int 1_Ed(\mu\times\nu)=\int \left(\int 1_E(x,y)d\mu(x) \right) d\nu(y). $$But 1_E(x,y)=1_{E^y}(x), so$$ \int \left(\int 1_E(x,y)d\mu(x) \right) d\nu(y)=\int ... 2 For your example intervals, suppose that$f$is$0$on$[0,1/2)$and$1$on$[1/2,1]$. 1 Trivially we observe that the maximum of the list can only increase as we add more variables to it (for any given instance of the list). $$\newcommand{\Max}{\operatorname{Max}} \forall n, 0\leq \Max(n)\leq 1, \Max(n+1)\geq \Max(n)$$ However, as each$X_i$is a random variable, so to is$\Max(n)$. So we need to examine the behaviour of the probability ... 0 Let$Y_n = \max \{ X_1,\dots,X_n \}$. Then$Y_{n+1} = \max \{ Y_n,X_{n+1} \}$. Consider$E[Y_{n+1} | Y_n]$. Note that$Y_{n+1} - Y_n \geq 0$, so $$0 \leq E[Y_{n+1} - Y_n | Y_n] = E[Y_{n+1} | Y_n] - Y_n$$ So$Y_n$is a submartingale. It is also nonnegative, so it converges a.s. by Doob's martingale convergence theorem. This does not state what the limit ... 2 If I understand you correctly, you can assume that there exists$\phi_n$monotonically increasing to$f$. Since$(X, \mathcal{M},\mu)$is$\sigma$-finite, you can consider$\{X_i\}_{i \in \mathbb{N}}$measureable so that$\cup_iX_i =X$and$\mu(X_i)< \infty$for each$X_i$. Now consider$Y_i = \cup_{i=1}^n X_i$. (I'll let you deduce a convenient ... 0 Presumably you're proved this for a finite measure$\mu$? Sigma-finiteness means that there is some countable collection of open sets each of finite measure, whose union is the whole space. You can use the finite-case construction on each of these open sets. Then simply join these constructions together, one by one. 2 Your guess is correct : the answer is "No" in general; For example, let$\mu_1$be the counting measure on$\mathbb R$, and let$\mu_2$be the measure defined by$\mu_2(\emptyset)=0$and$\mu_2(A)=\infty$if$A\neq\emptyset$. On the other hand, if the space$X$is the union of an increasing sequence of open sets on which the two measures are finite, the ... 2 Hint: what's the collection of all set where the 2 measures agree? Does that collection form a monotone class? Does it contains all open set? 4 $$\mathbf 1_{(x,x+a]}(y)=1\iff x\lt y\leqslant x+a\iff y-a\leqslant x\lt y\iff \mathbf 1_{[y-a,y)}(x)=1$$ 2 If$p\gt 1$, we define the$L^{p,\infty}$semi-norm by $$\lVert f\rVert_{p,\infty}^p:=\sup_{t\gt 0}t^p\lambda\{s, |f(s)|\gt t\}$$ (this is equivalent to a norm, namely,$\sup_{A,\lambda(A)\in (0,\infty)}\mu(A)^{1/p-1}\int_A|f|\mathrm d\lambda$). If we define$x:=k^{1/m}$and if we use the inequality, we obtain $$x^{2m\frac{N+1}N}\lambda\{|u|\lt ... 1 Let x \in E be a point with Lebesgue density 1 (by the Lebesgue Density Theorem, such a point exists). Fix \varepsilon > 0 such that$$\frac{m(E\cap(x - \varepsilon, x + \varepsilon))}{m((x - \varepsilon, x+ \varepsilon))} > \frac{1}{2}.$$Now suppose there is no y \in (0, \varepsilon) such that x - y, x + y \in E. Then A:= -(E\cap(x - ... 2 For sequences, there is the following simple characterization: Proposition. Suppose f_n, f \in C_0(X). The following are equivalent: f_n \to f weakly; f_n \to f pointwise and \sup_n \|f_n\|_\infty < \infty. Proof. 1 \implies 2: The evaluation maps f \mapsto f(x) are continuous linear functionals, so f_n \to f pointwise, ... 0 I'm not sure, if this sufficies for an answer..anyhow, I think it's a neat corollary. I believe this is due to Mackey: Suppose you have two locally compact, Hausdorff, second countable groups G, H and a measurable homomorphism \phi \colon G \to H . Then \phi is actually continuous. To prove this, start with open neighborhoods U,V of e_{H} ... 1 Yes: This is true for functions which are smooth enough, and in fact (if f is smooth) we have$$\int |\cos kx| f(x) dx \to A \int f(x) dx$$where A is the average of |\cos x| over a single cycle. Now given f \in L^1, choose a sequence of smooth enough f_n converging to f in the L^1 norm, and then note$$\left|\int |\cos kx| f(x) dx - \int ... 0 The answer to user133834's question is no. Let$\Omega=[0,1] \cap Q$($Q$denotes a set of all rational numbers) and$\tau$be defined as follows: $$\tau=\{\emptyset\}\cup\{ q:q \in [0,1]\cap Q\} \cup \{ [a,b]\cap Q :0\le a<b\le 1\}\cup$$ $$\{ [a,b[\cap Q :0\le a<b\le 1\}\cup\{ ]a,b]\cap Q : 0 \le a<b \le 1\}\cup$$$$\{ ]a,b[\cap Q :0\le a<b\le ... 1 I guess you mean infinite sums. The following statement holds true, which you may think of more general (or not). If$A$is a countable set, then any positive, say$\sigma$-finite measure defined on$2^A$is a sum of Dirac measure. This follows directly from$\sigma$-additivity of the measure. Consider any such measure$\mu$on$A\$, and define ...

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