# Tag Info

0

This is theorem 5.4 page 115 from Theory of the integral, S. Saks. It is proved via Vitali's covering theorem.

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Start by defining $P(\omega)=\bigcap\{F\in\mathscr{F}:\omega\in F\}$ for each $\omega\in\Omega$, and let $\mathscr{P}=\{P(\omega):\omega\in\Omega\}$; you need to show that $\mathscr{P}$ is a partition of $\Omega$. In order to do this, you must show two things: $\bigcup\mathscr{P}=\Omega$, meaning that each element of $\Omega$ belongs to at least one member ...

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Lebesgue measure and its multiples are not the only measures that are invariant under translations. Take, for example, the counting measure over the integers or over the rationals. Sure, they are infinite, but your question doesn't mention finiteness.

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You are having problems because the irrationals are a $G_\delta$ set. A $G_\delta$ set is a set that is realized as a countable intersection of open sets. To see why the irrationals satisfy this, consider the intersection $${\huge\cap}_{q\in \mathbb{Q}}\{q\}^c$$

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The result is true, and has nothing to do with topology or countability. Let ${\cal C}$ be the collection of subsets of $E^I$ of the form $\pi_i^{-1}(E)$ for some $i\in I$ and $E\in{\cal E}$. Then use the fact that $X^{-1}(\sigma({\cal C}))=\sigma(X^{-1}({\cal C}))$.

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1 and 4 are right. 2 and 3 are incorrect.

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By Strong Law of Large Numbers, without any assumptions on the variance of $\epsilon_i$, $$\frac{1}{n}\sum^n_{i=1}\epsilon_i \rightarrow \mathbb E[\epsilon_i]=0 \quad \text{almost surely}$$ So since $\lim_{n \rightarrow \infty} n\beta^n = 0$ and $n^{-1}b_n \rightarrow 0$ almost surely, we have that $$\beta^n b_n\rightarrow 0 \quad \text{almost surely}$$ ...

1

I don't understand why nobody has upvoted Chilango's answer. His answer is correct except an obvious typo and that he supposed $E$ is a normed vector space in which case most applications of measurable functions occur. His proof is also correct when $E$ is a metric space if we replace the norm $|*|$ with a metric $d$. The following proof is just a detailed ...

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For any sequence of sets $(A_n)$ the $\limsup$ is defined by $$\limsup A_n =\bigcap_{n=1}^\infty\left(\bigcup_{j=n}^\infty A_j\right)=\bigcap_{n=1}^\infty B_n.$$ Since the $B_n$ are decreasing, if we assume the measure is finite we have $$\mu(\limsup A_n)=\lim_{n\to\infty}\mu(B_n) =\limsup_{n\to\infty}\mu(B_n).$$But $A_n\subset B_n$, so ...

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I'd like to expand upon David C. Ullrich's answer to question #2. The answer, as he wrote, is: yes, $\phi_1 = \phi_2$. Indeed, let $B$ be a Borel set on the real line. Then for $i \in \{1, 2\}$, $$\phi_i(B) = \phi_i\left(\cup_{n \in \mathbb{Z}}(B \cap (-n, n])\right) = \sum_{n \in \mathbb{Z}} \phi_i(B \cap (-n, n])$$ Therefore, it suffices to show, for any ...

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I think I've figured out a proof. Suppose $\lim\sup_n \mu( A_n) > \mu (\lim\sup_n A_n)$. Then $\exists$ some rational $r$ with $\lim\sup_n \mu( A_n) > r > \mu (\lim\sup_n A_n)$(two real numbers $a, b$ satisfy $a > b$ if and only if there is some rational r with $a > r > b$). So $\exists N ∈ \mathbb N^+$ such that $sup_n \mu (A_n)$ = ...

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Assuming that $Var(\epsilon_1)=\sigma_{\epsilon}^2$ we have $$\mathbb{E}\beta^{2t}b_t^2= \sigma_{\epsilon}^2\frac{\beta^2}{(\beta^2-1)^2}+o(1)$$ Hence, $\beta^tb_t$ is $L^2$-bounded and uniformly integrable so that you can pass the limit inside the expectation... Edit: In the first step, you need to show that $\beta^tb_t$ converges a.s. (to $0$). This ...

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To avoid confusion, let's clarify some notation. Let $\mathcal{M}^{*}$ denote the $\sigma$-algebra of $\mu^{*}$-measurable subsets of $X$. Let $\bar{\mathcal{M}}$ denote the completion of $\mathcal{M}$ with respect to $\mu$. Let $\widetilde{\bar{\mathcal{M}}}$ denote the $\sigma$-algebra of locally measurable subsets of $(X,\bar{\mathcal{M}},\bar{\mu})$. ...

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Suppose that exist $A$ invariant set with $0<\mu(A)<1$. Let $\epsilon>0$ take $x\in A$ and $y\in A^c$ density points. Then there is $R>0$ such that $\mu(B(x,r)\cap A)>(1-\epsilon)2r$ and $\mu(B(y,r)\cap A^c)>(1-\epsilon)2r$ for all $r\leq R$. Take $r\leq R$, then there exist $k\geq 1$ such that $d(f^k(x), y)<\frac{1}{3}r$. So $B(y, ... 0 You're correct that the you can only apply the rule$\mu(A) = \mu_1(A_1) \mu_2(A_2)$when$A = A_1\times A_2.$A triangle does not meet this criteria. However, you can create sets of rectangles that tell us what the . So, for the sake of convenience, let$a=1$and cut the square$[1,0] \times [1,0]$up into$n^2$equal squares in the usual way. [Perhaps ... 1 Actually no, a function has at most one Jordan decomposition. You have two decompositions, but at most one is the Jordan decomposition. Anyway, finiteness for one decomposition does not imply finiteness for the other. Consider $$0=0-0=x^+-x^+.$$ If$f$has bounded variation and all your measures happen to be finite then yes,$\phi_1=\phi_2$. Because for ... 1 Note that $$\int\frac{h}{f}\,{\rm d}P=\int\frac{h^{2}}{f}\,{\rm d}\lambda,$$ where$\lambda$is the Lebesgue measure on$\mathbb{R}. But by Cauchy Schwarz, we have $$1=\int h\,{\rm d}\lambda=\int\frac{h}{\sqrt{f}}\cdot\sqrt{f}\,{\rm d}\lambda\leq\sqrt{\int\frac{h^{2}}{f}\,{\rm d}\lambda}\cdot\sqrt{\int f\,{\rm d}\lambda}=\sqrt{\int\frac{h^{2}}{f}\,{\rm ... 2 The spectrum of the C^*-algebra C_b(\mathbb R) of bounded continuous functions on \mathbb R corresponds to the Stone-Čech compactification \beta \mathbb R of \mathbb R. Any finite positive Borel measure on \beta \mathbb R gives you a positive linear functional on C_b(\mathbb R). However, the points of \beta \mathbb R \backslash \mathbb R ... 0 The method I present in this answer easily generalizes to \mathbb R^d. It is outlined in exercise 5 of chapter 7 (differentiation) in Rudin's Real and Complex Analysis. 0 A proof using metric density is outlined in Exercise 5 of Chapter 7 (Differentiation) of Rudin's Real and Complex Analysis, 3rd edition. I present my version. We generalize to possibly distinct sets A and B of positive measure. The set A has a point a of metric density where$$m(A\cap (a-\delta, a +\delta ))/2\delta > 3/4,$$and it suffices to ... 3 We have$$f(X)\subseteq\{0\}\cup f(\mathrm{supp}(f)),$$which is compact in \mathbb{R} (since \mathrm{supp}(f) is compact and f is continuous), hence bounded. 1 If you do know the Weierstraß theorem, then you can prove it like that: Let f \in \mathcal{K}(X) and denote by K the support of f. Then f|_{K^c}=0 by the very definition of the support. Moreover, by the Weierstraß theorem, f|_K is bounded. Combining both facts, proves that f is bounded. If you do not know the Weierstraß theorem, then have e.g. ... 0 If \varphi is C^1 and injective (not quite enough, since the derivative may be 0, otherwise the measure falls apart). We want to compute$$ m(\varphi(I)) =\int_{\varphi(I)} dy $$Since \varphi is C^1 and injective, we have the inverse function theorem. I.e. we have \varphi^{-1} : \mathbb{R}^n \to \mathbb{R}^n. Thus it makes sense to consider x = ... 1 A probability density function exists if and only if corresponding cumulative distribution function is absolutely continuous. A probability measure (or simply probability) is not a probability density function, but a measure (if a cumulative distribution function is specified, it is the corresponding Lebesgue-Stieltjes measure). Any random variable's ... 3 I think I can confirm your suspicion that this doesn't necessarily hold if the target space is non-Hausdorff, assuming I haven't made a mistake somewhere... Let \mathbb{R} be the real line in its standard topology. Let \mathbb{R}_0 be the real line with the topology whose non-empty open sets U are precisely the standard open sets U \subseteq ... 0 I think if you use \ln(1-x)<-x, then you can show that % %TCIMACRO{\dprod }% %BeginExpansion {\displaystyle\prod} %EndExpansion \left( 1-a_{i}\right) >0\rightarrow% %TCIMACRO{\dsum }% %BeginExpansion {\displaystyle\sum} %EndExpansion \ln\left( 1-a_{i}\right) >-\infty\rightarrow% %TCIMACRO{\dsum }% %BeginExpansion {\displaystyle\sum} ... 1 A measure on a product space need not be a product measure. Suppose$$ X = \left. \begin{cases} 0 & \text{with probability }1/4, \\ 1 & \text{with probability }3/4, \end{cases} \right\}\text{ and }Y = \left. \begin{cases} 0 & \text{with probability }1/3, \\ 1 & \text{with probability }2/3. \end{cases} \right\} \tag 1 $$The following two ... 3 That "Wikipedia" definition is for the special case where the covariance matrix is \sigma^2 I. In that case they are equivalent. A measure \mu on \mathbb R^n has density \rho with respect to Lebesgue measure \lambda^n iff \mu(A) = \int_A \rho(x) \; d\lambda^n(x) for all Lebesgue measurable sets A. 1 Let A=\{(x,y)\in\mathbb{R}^2:x=y\}. You could use that B\otimes B are the Borel sets on \mathbb{R}^2 so a closed subset of \mathbb{R}^2 is measurable. If you don't know that yet you can go from first principles: Divide A into a countable union of A_n = \{(x,y)\in A: -n\leq x\leq n\}. Show that A_n is measurable by starting with the square ... 0 Hmm. Writing integrals as integrals: Say \epsilon>0. You need to show there exists \delta>0 so P(A)<\delta implies \int_A|X_n|\,dP<\epsilon. You know that \int f(|X_n|)\,dP\le K for some fixed K. Hmm. Ok. Say L>0 is a large number. Choose M so$$\frac{f(t)}t>L\quad(t>M).Now write A=B\cup C, where B=\{t\in ... 0 In (A), assuming you're talking about that normal variable: Saying \mu(B)=P(X^{-1}(B)) is the definition of "\mu is the distribution of X". And now the definition of "X is normal with mean 0 and variance 1" is "if we let \mu be the distribution of X, as defined above, then \mu(B)=\int_B f\,d\lambda, where f is as above". Regarding the ... 3 Suppose that S \cup N is measurable. Since N is measurable, it follows that N^c is measurable. Therefore (S \cup N) \cap N^c = S \backslash N is measurable. Also, since N has measure zero and since S \cap N \subset N, it follows by completeness of Lebesgue measure that S \cap N is measurable. Therefore S = (S \cap N) \cup (S \backslash N) ... 1 Assuming a correct version of the standard definition of an outer measure \mu^* given a measure \mu, and assuming that we're supposed to simply assuume that \Omega_0\subset\Omega satisfies \mu^*(\Omega_0)=\mu(\Omega) (that word "let" threw me off): Suppose A\in F. If we let A_1=A and A_n=\emptyset for n>1 then A_n\in F and ... 2 A lot of times these two mean the same thing, but it is important to consider the superset of which this is an interval. Sometimes, (especially in measure theory, which is why I mention it) it is useful to work in the extended reals, which includes a point at \infty, so (a,\infty) means every number greater than a accept infinity and (a,\infty] ... 4 \begin{align*} (a,\infty)=&\,\{x\in\mathbb R\,|\,x>a\},\\ (a,\infty]=&\,\{x\in\mathbb R\,|\,x>a\}\cup\{\infty\}. \end{align*} The latter set includes an extra point termed “positive infinity.” Note that it is not a real number, but in certain areas of mathematics, especially in measure theory, it is useful to extend \mathbb R by this single ... 1 Let n be a large even number. Take n equally spaced points along the unit interval. A probability distribution which gives equal weight to all these points converges weakly to the uniform distribution. The product of two copies of this distribution converges weakly to uniform on the unit square. Since this distribution is equal to the product of it's ... 2 Well, if the book doesn't include any finiteness hypotheses at all (look carefully!) then the book is simply wrong. If \mu^*(A)=\mu^*(B)=\infty then \mu^*(A)-\mu^*(B) is undefined. It's true if \mu^*(A)=\infty and \mu^*(B)<\infty, by the way. In that case \mu^*(A)\le\mu^*(B)+\mu^*(A\Delta B) shows that \mu^*(A\Delta B)=\infty. (It sounds ... 0 Alternate proof: Fix \epsilon > 0. Since the sequence \{ X_n \}_{n=1}^{\infty} is monotone and X_ n \to X in probability, then the events \{ |X_n - X| \geq \epsilon \} are monotone decreasing. Therefore, \begin{align} P(\{ |X_n - X| \geq \epsilon \} \,\, \text{i.o.}) &= P(\cap_{n=1}^{\infty} \cup_{i=n}^{\infty} \{|X_i - X| \geq \epsilon ... 0 I think it is praiseworthy that you are interested in these infinite sums and are thinking of ways to combine them in order derive new results. However there is a problem: all these sums are not strictly convergent, but exist only if one extends a correct result to a case where they are not really properly defined. Unfortunately this means that the ... -1 There is no official way to add these divergent alternating series, but\sum (-1)^n$is known as Grandi's series: $$\sum_{n=0}^\infty (-1)^n = \begin{cases} 1 & \text{ if }n\text{ is odd} \\ 0 & \text{ if }n\text{ is even} \\ \end{cases} \approx \frac{1}{2}$$ And for your second series$(1 - 1 + 1 - 1 + \dots)^2 = 1 - 2 + 3 - 4 + \dots = ...

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As the other answers have said, your reasoning is very off: these are divergent sequences, and don't behave the way you think they do. On the other hand, a reasonable question to ask is: "Can we make sense of these divergent series?" The answer is yes, although it is not easy: there are several methods for summing (some) classically divergent series. See ...

3

Every claim you started with is wrong. Your $S_{1}, S_{2}, S_{3}$ are all divergent series. It looks like you declared these series to be equal to their Ramanujan summation assignment, (which is akin to setting a matrix equal to its own determinant), and then proceeded as if you actually meant it converged to that value.

1

You start with wrong conditions. For example, the infinite sum $\sum_n^{\infty} (-1)^n$ is not equal to $1/2$.

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For any sequence $f_n \to f''$ in $L^p$, there exists a subsequence such that $f_{n(k)} \to f''$ almost everywhere as $k \to \infty$. There are (at least) two possibilites to prove this statement: In the proof of the Riesz-Fischer theorem (which states that $L^p$ is a complete space), one usually constructs such a sequence (see e.g. René Schilling: ...

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Note that if $\mu(X)=\infty$, then your series condition doesn't necessarily imply $f\in L_{\mu}^{p}(X)$. For example, take $X=\mathbb{R}$ with Borel $\sigma$-algebra and Lebesgue measure. If $f=1/2\chi_{[0,\infty)}$, then clearly $f\notin L^{p}$; however, $\lambda_{f}(n)=0$ for $n\geq 1$. Assume that $\lambda_{f}(0)<\infty$. Observe that for $p>0$, ...

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One example is a version of the continuous mapping theorem which states that if $X_n \rightsquigarrow X$ then $f(X_n) \rightsquigarrow f(X)$ for a continuous function $f$. Using the a.s. representation (Skorohod's Representation theorem) there is a sequence of random variables $Y_n$ and a random variable $Y$ defined on a common probability space having the ...

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Define $$f(t)=\sum_{n=-\infty}^\infty \hat\mu(n)e^{int}.$$The series converges uniformly so $f$ is continuous. The uniform convergence also shows that $$\hat f(n)=\hat\mu(n).$$So uniqueness (for complex measures) shows that $\mu=f$, or more carefully $d\mu=f\,dt$. If you don't buy the uniqueness for complex measures bit: Lemma: If $\nu$ is a complex ...

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This answer is Community Wiki! Construction Denote compact sets: $$\mathcal{K}(\mathbb{C}):=\{K\subseteq\mathbb{C}:K\text{ compact}\}$$ Suppose it is positive: $$f\in\mathcal{B}(\mathbb{C}):\quad f\geq0$$ And it is bounded: $$\Delta_n:=2^{-n}\|f\|_\infty<\infty$$ Also compact support: $$K:=\overline{\{f\neq0\}}\in\mathcal{K}(\mathbb{C})$$ By local ...

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Counterexample. In one dimension, let $K_k=\{0,1/k,2/k,3/k,\dots,1\}$. Each $K_k$ has zero measure. The limit of $K_k$ in the Hausdorff metric is the interval $[0,1]$, of measure $1$. You can also attach the same interval to the sets $K_k$ to make their measure positive. The above counterexample could be ruled out by requiring $K_k = ... 1 It depends on how one defines$L^2(X)$. One way of defining$L^2(X)$is as the set of all measurable functions$f:X\rightarrow\mathbb{R}$such that$|f|^2$is integrable over$X$. It is, however, more standard to view$L^2(X)$to be the set of equivalence classes of functions whose domain is$X$and are square-integrable over$X$. Two functions$f\$ and ...

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