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So, since $\epsilon>0$ we can get $(A_i)_{i\in\Bbb N}$ an $\mathcal A-$cover of $E$ such that: $$\sum_{i=1}^\infty \mu(A_i)<\mu^*(E)+\frac{\epsilon}{2}$$ and it's clear that $\mu^*(E)\le \sum_{i=1}^\infty \mu(A_i)$, so we get that: $$\mu^*(E)\le \sum_{i=1}^\infty \mu(A_i)<\mu^*(E)+\frac{\epsilon}{2}$$ thus $$0\le \sum_{i=1}^\infty ... 0 To use the Radon-Nikodym theorem we need that: (1) P_{(X,Y)} \mbox{ is absolutely continuous w.r.t. the Lebesgue measure} \lambda^2 on \mathbb{R}^2 (2) P_{(X,Y)} is \sigma-finite which is satisfied since it is even finite: P_{(X,Y)}(\mathbb{R}^2) = 1 < \infty. (3) The Lebesgue measure \lambda^2 is \sigma-finite, for this we define for ... 0 If an arbitrary centered Gaussian measure is given on a product of a Banach space X with itself, then it is not guaranteed to be the tensor product of two centered Gaussian measures on X. As an example, consider the one-dimensional real Banach space R. On the product space R x R (the Euclidean vector plane) consider the measure with density (up to some ... 0 1-The Lebesgue measure \nu satisfies by definition \nu([a,b])=b-a. In particular \nu(\{a\})=\nu([a,a])=a-a=0. Thus \nu is zero on all unitary set. And now using the aditivity of \nu we have for all countable set A=\{a_n\}_{n=1}^\infty=\bigcup_{n=1}^\infty\{a_n\}: ... 1 Hint: For t\in(0,1):$$P\left(X_{n}\leq t\right)=\sum_{k=0}^{\lfloor nt\rfloor}\frac{1}{n+1}=\frac{\lfloor nt\rfloor+1}{n+1}\in\left(\frac{nt}{n+1},\frac{nt+1}{n+1}\right]$$1 Hint: Convergence in distribution is implied by convergence of the characteristic functions. 3 Let U be a random variable which is uniform on (0,1]. Let X=Y=U^{-1/2}. Then E[X]=E[Y]=2 but E[XY]=+\infty. You can make lots of examples of functions like this, which are not integrable but their square root is integrable, because it diverges "more slowly". On an infinite measure space you can also have the opposite phenomenon with tailing: ... 2 Yeah, your proof is correct. Regarding your first question concerning integrability: If you prefer, you can restate the result as follows. The following statements are equivalent: \mathcal{F} and \mathcal{G} are independent. For all bounded X_F \in m \mathcal{F} and bounded X_G \in m \mathcal{G}, it holds that \mathbb{E}(X_G X_F ) = ... 0 No. If the norm were induced by an inner product it would satisfy the parallelogram law$$||f+g||^2+||f-g||^2=2(||f||^2+||g||^2).$$Simple examples show this is not so (for example, the characteristic functions of two disjoint sets). 0 Hint: if a norm is derived from an inner product in that way, then the inner product is uniquely determined by the norm and there is an explicit algebraic expression for (f,g) in terms of ||f||, ||g||, ||f+g|| and ||f-g||. 1 Hint: take a look at the parallelogram law. 2 Hint:$$ \left|\int f_n - \int f\right| \leq \int |f_n - f| $$and the quantity on the right can be controlled because of uniform convergence and the fact that \mu(X) < \infty. To prove f\in L^1, use something similar such as$$ \int |f| \leq \int |f - f_N| + \int |f_N| $$for some fixed N \in \mathbb{N} large enough. -1 To show that the conclusion of the dominated convergence theorem does not hold, assume it does i.e. assume there exists some dominating function. Then show that there is a contradiction. Related exercise in Probability w/ Martingales: Apply the same logic to g:= \sup_n |n^2xe^{-n^2x^2}|. If we suppose g is integrable, then by the dominated ... 0 This is quite a standard argument in measure theory. You don't need any additional assumptions (if I understand it correctly that you suppose the equality holds for all D in the \sigma-algebra). You can easily reduce the problem to the setting$$\int_D X dP=0 for all D\in \mathcal{D} by taking the difference of X and Y (both sides are finite ... 1 For both questions, the answer boils down to the triangle inequality. It is more straight forward for the second: if x\in B(2re_i,r), then d(x,0)\leq d(x,2re_i)+d(2re_i,0)<r+2r=3r. For the first question, we first need to calculate the distance between the centers of two circles. Since d(2re_i,2re_j)=2rd(e_i,e_j)=2r\sqrt{2}, a point in our ... 0 If the collection A_1,\ldots,A_k is finite, one can just extend it to a countable collection by padding it with A_n=\emptyset for n>k. This doesn't change the union of the collection. A similar idea goes for the covering collection, that is, one can always assume without loss of generality it is countably infinite purely for notational convenience. 0 Hint: x(t,\xi)=\lim_{n\to\infty}\sum_{k=1}^{2^n}1_{[(k-1)/n,k/n)}(t)\xi(k/n). (Because of right continuity.) Argue that each term in the sum is \mathcal B[0,\infty)\otimes\mathcal F-measurable, hence to is the sum, hence so is the limit. 1 Because \sup_{n\ge 1} n^2 x e^{-n^2x^2}={1\over x\cdot e} for x\in(0,1]. 1 Hint: If |f|\le 1, then |f|^{p_2} \le |f|^{p_1}. 2 Let f \in L^{p_1}(E) bounded. Then there exists M\geq 0 such that \sup_{x \in E} |f(x)| \leq M and \|f\|_{p_1}^{p_1}=\int_E |f(x)|^{p_1}\,dx < \infty. Case 1: p_2<\infty For p_2 > p_1, we have \begin{align} \|f\|_{p_2}^{p_2}& =\int_E |f(x)|^{p_2}\,dx \\ &= \int_E |f(x)|^{p_2-p_1}|f(x)|^{p_1}\,dx \\ & \leq \left(\sup_{x ... 1 Note that if \lambda(N)=0 then R\setminus N is everywhere dense in R. Indeed, if assume the contrary then there will be (a,b) such that (R \setminus N)\cap (a,b)=\emptyset. This means that N contains the set (a,b) and hence N is not of \lambda-measure zero. Since f(x)=g(x) for x \in R \setminus N, \lambda(N)=0 and R\setminus N ... 1 Wikipedia implies that this is not correct even for Hausdorff spaces and Borel measures. However, under the assumptions of \mu being a Radon measure on a Hausdorff space (X, \mathcal{T}), we can prove that it is true: In this case, we have the important property that the complement of the support has measure zero. Let the finite support be given as ... 0 I learnt this solution from terence tao's blog. Since A\times B= \cup_{i\geq 1} (A_i\times B_i) and we have pairwise disjointness, therefore we know that for all a\in A, b\in B we have:1_A(a)1_B(b)=\sum_{i=1}^{\infty}1_{A_i}(a)1_{B_i}(b)$$Thus, for every a\in X we have: ... 1 When X is finite, you can treat \mathcal P as a subset of \Bbb R^X. Total variation is the norm there, and all the norms over finite-dimensional spaces are equivalent, so the induced metric is equivalent to the Eucledian one. Embedded \mathcal P is bounded and closed in Eucledian metric, hence compact. 1 The total variation |\mu| of \mu is defined via$$|\mu|(B) = \sup_{\pi} \sum_{A \in \pi} \|\mu(A)\|_{\mathbb{R}^n}$$where the supremum ranges over all finite partitions of the Borel set B into disjoint Borel sets. In the scalar case, this coincides with the definition via the Hahn decomposition. In the vector-valued case, however, there is no ... 3 It is not too hard to show that \overline{f_n}(x) = \sup_{k \ge n} f_k(x) and \underline{f_n}(x) = \inf_{k \ge n} f_k(x) are measurable functions. Note that these are non increasing, non decreasing sequences respectively, hence they have measurable limits \overline{f}(x) = \lim_n \overline{f_n}(x), \underline{f}(x) = \lim_n \underline{f_n}(x). ... 1 We don't need to show that C is the preimage of a measurable set; we can show it belongs to the sigma algebra directly. It helps translating the conditions for the limit to exist:$$C=\{x\in X: \lim f_n(x) \text{ exists and is finite}\}=\{x\in X: (f_n(x))_{n=1}^\infty \text{ is a Cauchy sequence}\}=\{x\in X: \text{for all }\epsilon>0\text{ ...

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The hypothesis implies that $f_{n}$ is a Cauchy sequence. Observe that $C=\bigcap_{k=1}^{\infty}\bigcup_{N=1}^{\infty}\bigcap_{m=N}^{\infty}\bigcap_{m=N}\{x \in X: |f_{m}- f_{n}| < \frac{1}{k} \}$ The sets $A_{m,n,k} = \{x \in X: |f_{m} - f_{n}| < \frac{1}{k} \}$ $\in \mathcal{A}$. It follows from the properties of $\mathcal{A}$ that $C \in ... 2 It is a quotient space!$\mathbb{R}$here should be thought of as the one-dimensional subspace of$L^2(\Omega)$which consists of the (a.e.) constant functions. The norm given is the canonical norm on the quotient of a Banach space mod a closed subspace. This construction is discussed, for instance, in section III.4 of Conway's A Course in Functional ... 0 Soooo the flips are independent? Consider a probability space$(\Omega, \mathscr F, \mathbb P)$where$\Omega = \{H,T\}^{\mathbb N}$So we have$\omega = (\omega_1, \omega_2, ...)$where$\omega_n \in \{H, T\} \ \forall n \in \mathbb N\mathscr{F} = \sigma(\omega_n = W | \ W \in \{H, T\})$like here (because I guess$\mathscr{F} = 2^{\Omega}$... 2 If$\Bbb E[1_A|\mathscr F_t]=1$, then$\Bbb E[1_A]=1$, which is the same as$1_A=1$. In this case$\Bbb E[1_A|\mathscr F_s]=1$(almost surely) for each$s$. Likewise, if$\Bbb E[1_A|\mathscr F_t]=0$, then$\Bbb E[1_A]=0$, which is the same as$1_A=0$. In this case$\Bbb E[1_A|\mathscr F_s]=0$(almost surely) for each$s$. If$\Bbb E[1_A|\mathscr F_t]=p$, ... 0 Hint: Try to show that there exists$a \in \mathbb{R}$such that $$P[Y \leq a] = 1 \text{ and } P[Y < a] = 0$$ by Kolmogorov$0$-$1$law and$\lim_{x \to \infty} P[Y \leq x] = 1$and$\lim_{x \to -\infty}P[Y \leq x] = 0$. More explicitly, let$F(x) = P[Y \leq x]$be the distribution function of$Y$, then$F$is nondecreasing, right-continuous and ... 1 You can do this change of variables : u =$\log{1/x} \implies x = e^{-u}$So, to calculate the p-norm, consider the following integral$\int_0^{1}\log(\frac{1}{x})^pdx = \int_{0}^{\infty} u^p e^{-u} du$The integral in the r.h.s is the well-known gamma function$\Gamma(p+1)$, which converges for the values of p that we are interested. For the ... 2 Let$q\in\mathbb{Q}$and$n\in\mathbb{N}$be arbitrary. Then, for each$k\in\mathbb{N}$, the event$A_{q,n}^{\left(k\right)}:=\left\{ \left|X_{k}-q\right|<\frac{1}{n}\right\} $has the probability$\mathbb{P}\left(A_{q,n}^{\left(k\right)}\right)=\mathbb{P}\left(A_{q,n}^{\left(1\right)}\right)>0$which is independent of$k\in\mathbb{N}$. Since the ... 3 Notice that your even is a zero-one event. Define the event$A_n:=\{X_n<c\}$for some$c>0$. Then $$\sum_{n\geq 0} P(A_n)=\sum_{n\geq 0} c/n=\infty.$$ So by the reverse Borel Cantelli lemma,$A_n$occurs infinitely often with probability 1. This implies that$P(\omega: X_n(\omega)\rightarrow\infty)<1$, and since it's a zero-one event, its ... 1 For any$a>0,\lim_{x\to 0^+}x^a\ln (1/x) =0.$So given$0<p<\infty,$we have for small positive$x$$$x^{1/(2p)}\ln(1/x) < 1 \implies x^{1/2}(\ln(1/x))^p < 1 \implies (\ln(1/x))^p < x^{-1/2}.$$ Since$\int_0^1 x^{-1/2}\, dx < \infty,$we have the result. 1 The Jordan decomposition of$\mu$is usually given in terms of a Hahn decomposition of the space$X$: there exist measurable sets$P$and$N$with the property that$P \cap N = \emptyset$,$P \cup N = X$,$E \subset P$implies$\nu(E) \ge 0$and$E \subset N$implies$\nu(E) \le 0$. Then by definition$\nu^+(E) = \nu(E \cap P)$and$\nu^-(E) = - \nu(E \cap ...

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Make a simple change of variables, $u=x^{-1}$ so that $du = -x^{-2}dx\iff -u^{-2}du = dx$ Then you get $$\int_\infty^1 -{\log u\over u^2}\,du=\int_1^\infty {\log u\over u^2}\,du$$ And this is easily verified to be integrable by directly comparing with $$\int_1^\infty {du\over u^{3/2}}$$ Since we know $$\lim_{x\to\infty} {\log x\over \sqrt{x}}=0$$ So ...

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The answer is negative. If $(r_n)$ is lacunary sequences (that is $r_{n+1}>ar_n$ for some $a>1$), then for any probability preserving transformation $T\colon X\to X$ and any $\delta>0$ there is $A\subset X$ with $0<\mu(A)<\delta$ such that $$\limsup_n \frac{1}{n} \sum_{k=0}^{n-1} \chi_A (T^{r_k}x)=1 \quad \text{ for a.e. x\in X.}$$ Akcoglu ...

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Let $\phi(x,y) = 1_A(y-x)$. We need to ensure that $\phi$ is $\mu \times \lambda$ measurable. The function $(x,y) \to x-y$ is measurable, hence the set $\{ (x,y) | x-y \in A \}$ is measurable, and so is the complement. Then the set $\{(x,y) | \phi(x,y) < \beta \}$ is either empty, the entire space of the complement of the set $\{ (x,y) | x-y \in A \}$ ...

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Lemma 7.1.2. (p. 68) of Measure Theory, volume 1, Vladimir I. Bogachev: If two finite signed Borel measures on any topological space coincide on all open sets, they coincide on all Borel sets. Its simple proof uses: Lemma 1.9.4. If two probability measures on a measurable space $(X,A)$ coincide on some class $E\subset A$ that is closed with respect to ...

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Try this: compute $$\int_{\mathbb R} \mu(A + x) d\lambda(x) = \int_{\mathbb R} \int_{\mathbb R} \chi_{A + x}(y) \, d\mu(y) d\lambda(x).$$ Measurability issues aside, Tonelli's theorem implies that this equals $$\int_{\mathbb R} \int_{\mathbb R} \chi_{A +x}(y) d\lambda(x) d\mu(y).$$ Since \chi_{A + x}(y) = 1 \iff y \in A + x \iff x \in -A + y \iff \chi_{-A ... 1 Here is an alternative, perhaps slightly easier proof using Dynkin's π−λπ−λ Theorem: http://math.stackexchange.com/a/813414/283164 Although it is sketched for \mathbb R, it works more generally. BTW, Lemma 7.1.2. (p. 68) of Measure Theory, volume 1, Vladimir I. Bogachev: If two finite signed Borel measures on any topological space coincide on all open ... 0 I think I found what I was looking for. A little thought should be enough to realize that the k-th digit of the dyadic expansion is zero on the set $$B_k=\bigcup_{i=1}^{2^{k-1}}\left\lbrack \frac{2(i-1)}{2^k},\frac{2\left(i-1\right)+1}{2^k} \right).$$ Now, what I need is the intersection, namely \begin{align} A_n=& ... 1 If one weakens the condition as said, that is considers sets \Lambda \subseteq \def\P{\mathfrak P}\P(S) that have (1) S \in \Lambda(2) If A, B \in \Lambda, then A \setminus B \in \Lambda(3) If (A_n) \in {}^{\mathbf N}\Lambda is increasing, then \bigcup_n A_n \in \Lambda. then every \Lambda satisfying (1)-(3) is - of course - a ... 1 In fact, in the general case, it is enough to check the condition against all sets Q in the \sigma-algebra generated by \mathcal H, such that \mu^*(Q)<\infty. Proof: The condition for A to be Caratheodory-measurable by \mu^\ast is:\mu^\ast(Q) = \mu^\ast(Q \cap A) + \mu^\ast(Q \cap A^c)$$for all sets Q \subseteq X. Note that this ... 0 No. Let \Gamma be the collection of rational points inside the unit ball and suppose that u = 1 on a neighborhood of the unit ball. Since \Gamma is countable it is trivially N-1 rectifiable, and since \Gamma_\epsilon contains the unit ball for \epsilon > 0 you have \displaystyle \int_{\Gamma_\epsilon} u(x) \, dx \ge 1. The limit on the left ... 1 Yes, \mathcal{S}(\mathbb{R}^n) is dense in C_0(\mathbb{R}^n). Let f be in C_0(\mathbb{R}^n). Recall that this implies that f is uniformly continuous. Your claim is easy using mollifiers. Let me show you how this works. Pick a smooth positive function \psi with support contained in the open unit ball B_1(0) such that \int_{\mathbb{R}^n} \psi ... 1 (Might be still trivial) Let \mu_1, \mu_2 be any measure on X with \mu_1 (X) = \mu_2(X)<\infty  and \gamma :D \to Y be any map. Then$$f : D\times X \to Y, \ \ F(t, x) = \gamma (t)$$satisfies that f(t, \cdot)_\sharp \mu_1 = f(t, \cdot)_\sharp \mu_2 = \mu_i(X) \delta_{\gamma(t)} for all t\in D but \mu_1 might not be \mu_2. For a somewhat ... 2 The Fourier transform is a linear map, you only have to check that$$\forall f\in L^1(\mathbb R), \mathcal F(f)=0 \Rightarrow f=0.$$Let f\in L^1(\mathbb R) such that \mathcal F (f)=0. Hence \mathcal F(f) is L^1(\mathbb R) since its the zero function and therefore its Fourier inverse exists. So$$f=\mathcal F^{-1}(\mathcal F(f))=\mathcal ...

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