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We need the following result: Proposition 1: Let $(X,\mathcal{M}_{X})$ and $(Y,\mathcal{M}_{Y})$ be measurable spaces. Let $\mathcal{M}$ be the product $\sigma$-algebra of $\mathcal{M}_{X}$ and $\mathcal{M}_{Y}$, i.e., $\mathcal{M}=\sigma\left(\mathcal{M}_{X}\otimes\mathcal{M}_{Y}\right),$ where $\mathcal{M}_{X}\otimes\mathcal{M}_{Y}=\{A\times B\mid A\in\... 0 I think, the simplest way to see this is simply asking about an event for which the probability cannot possibly be given by the probability of the finite dimensional distributions. Assume we're in a topological setting. Let$U$be open and consider the event that there exists a time$t\in [t_1,t_2]$such that$X_t\in U$. Now, if$X$is continuous and$U$... -1 In general union of 2 σ-algebras is not σ-algebra. Looking from the probability perspective. Example - Consider a sample space W = {1,2,3,4}. Two event spaces which are the σ-algebras :- F1 = {W,∅,{1,2},{3,4}} F2 = {W,∅,{1,3},{2,4}} We can check it by the properties, both F1 and F2 satisfy for σ-algebra. But - F1 ∪ F2 = {W,∅,{1,2},{3,4},{1,3},{2,4}} :- ... 0 As an addendum to Noah Schweber's answer, here's a concrete example of a set that is in both$\Sigma_3^0$and$\Pi_3^0$(and hence in$\Delta_3^0$) but neither in$\Sigma_2^0$nor in$\Pi_2^0$. The set of all real numbers in$[0,1]$whose base-$10$representations contain (a) infinitely many$1$'s and (b) finitely many digits that are not in$\{1,2\}$. ... 2 Unfortunately, I don't understand your proof. I can give a proof of my own. We want to show that the$\text{esssup}(f):= \inf\{k\in\mathbb{R}: \mu(|f|>k)=0\}$is always smaller than the actual supremum. To show this let$\sup(f)=M$. This means that for all$x\in \mathbb{R}$we have that$f(x)\leq M$. In particular this means that$\mu(|f|>M) = \mu(\...

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The relevant topic here is descriptive set theory. The standard texts on the subject are Moschovakis and Kechris; I tend to prefer the latter, especially as a first introduction unless you're already dead-set on becoming a logician, but the former is freely available on the author's website. The class of Borel sets is vastly more complicated than that. The ...

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Lebesgue's differentiation theorem does hold for a much larger class of measures; it is not restricted to Lebesgue measure. The following statement is compiled from Measure Theory Vol. 1 by Bogachev (Theorem 5.8.8). Let $\mu$ be a measure on $(\mathbb{R}^n,\mathcal{B}(\mathbb{R}^n))$ which is finite on all balls. If $f \in L^1(\mu)$, then $$f(x) = \lim_{r ... 1$$\int_{Y}f\left(x,y\right)\;d\mu_{2}\left(y\right)=\int_{Y}\mathbf{1}_{\left\{ x\right\} }\left(y\right)-\mathbf{1}_{\left\{ x+1\right\} }\left(y\right)d\mu_{2}\left(y\right)=\mu_{2}\left(\left\{ x\right\} \right)-\mu_{2}\left(\left\{ x+1\right\} \right)=1-1=0$$so that the integral can be calculated by:$$\int_{X}\int_{Y}f\left(x,y\right)\;d\mu_{2}\left(y\...

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It's not true as stated: Let $\Omega$ be uncountable with $\mu(\Omega) =\infty$ and $\mu(\{\omega\})=0$ for each $\omega\in\Omega$. Let $F$ be the smallest $\sigma$-algebra generated by the singletons, i.e. it consists of countable subsets and their complements. Finally, choose $G:=\{\{\omega\} \, :\, \omega\in\Omega\}$. I guess, a similar but more ...

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You don't need DCT for this. Use the fact that $|\frac {\sin t} {1+t^{2}}| \leq \frac 1 {t^{2}}I_{|t|>1}+I_{|t|\leq 1}$.

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$$\sigma(X_0,X_1,...,X_n)=\{(X_0,X_1,...,X_n)^{-1}(A): A \text { is Borel in } \mathbb R^{n+1}\}$$ $$=\{\omega: (X_0(\omega),X_1(\omega),...,X_n(\omega)) \in A: A \text { is Borel in } \mathbb R^{n+1}\}.$$

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This is just a special case of the fact that if $\alpha>1$ then $Lip_\alpha$ contains only constants: Say $f_t(x)=f(x+t)$, and define $F:\Bbb R\to L^1$ by $F(t)=f_t$. Then $$F(t)-F(0)=\sum_{j=1}^n(F(jt/n)-F((j-1)t/n)),$$so if $n$ is large enough that $t/n<\delta$ we have $$||F(t)-F(0)||_1\le\sum_{j=1}^n||F(jt/n)-F((j-1)t/n)||_1\le n(t/n)^2=t^2/n.$$ ...

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Your idea using DCT works. To show that $F$ is continuous in any $x_0\in\mathbb{R}$, consider a sequence $(x_n)$ with $x_n\rightarrow x_0$. Then the sequence of functions $h_n(t):=f(t)\,\chi_{(-\infty,x_n)}(t)$ converges pointwise a.e. to $h(t):=f(t)\,\chi_{(-\infty,x_0)}(t)$, and $|h_n(t)|\leq f(t)$. Since $f$ is integrable, DCT implies $$\lim_{n\... 0 It means that \mathbb M is the smallest \sigma-algebra for which I_B is measurable for all B. 8 Consider the Fourier transformation, \hat f(\xi)=\int_\mathbb Rf(x)e^{-2\pi ix\xi}\,dx for \xi\in\mathbb R. Denote f_t(x)=f(x+t), then \hat f_t(\xi)=e^{2\pi it\xi}\hat f(\xi). By hypothesis, \|\hat f(\cdot)(e^{2\pi it\cdot}-1)\|_{L^\infty}=\|\hat f_t-\hat f\|_{L^\infty}\leq \|f_t-f\|_{L^1}\leq |t|^2. So for a.e. \xi\in\mathbb R-\{0\},$$|\hat f(\...

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$|\int_a^{b} f(x)dx-\int_{a+t}^{b+t} f(x)dx| =|\int_a^{a+t} f(x)dx-\int_b^{b+t} f(x)dx|\leq t^{2}$ for $t>0$ sufficiently small. [ Because $\int_a^{b} f(x)dx-\int_{a+t}^{b+t} f(x)dx=\int_a^{b} f(x)dx-\int_a^{b} f(x+t)dx$]. Divide by $t$ and let $t \to 0$. Using Lebesgue's Theorem we get $f(a)-f(b)=0$ whenever $a$ and $b$ are Lebesgue points of $f$. Can ...

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Applying the triangle inequality we have for, each $n\in \mathbb{N}$: $\lVert f(x+n\lceil{\frac{1}{t}}\rceil t)-f(x)\rVert_1\leq n \lceil{\frac{1}{t}}\rceil t^2\leq 2nt$ for $t>0$ sufficiently small. In particular, if $f\neq 0$ a.e. we can, for sufficiently large $n$, set $t=\frac{||f||_1}{2n}$ to conclude that \lVert f(x+n\lceil{\frac{1}{t}}\rceil t)-f(... 1 I read the paper due to Morayne and Solecki, and the idea is the following: - Step 1: Define \begin{align} A_m &= \{\ (k2^{-m},(k+1)2^{-m}]:\ k\in \mathbb{Z}\}, \\ A'_m &= \{\ 1/3+ (k2^{-m},(k+1)2^{-m}]:\ k\in \mathbb{Z}\} \end{align} and notice thatA_m$and$A_m'$are partitions of$\mathbb{R}.$- Step 2: Define$A_m(x)$as the unique element ... 0 I have found a sufficient answer to my problem: Piecewise Deterministic Markov Processes. For more information: https://en.wikipedia.org/wiki/Piecewise-deterministic_Markov_process. 2 As I asked in a comment, Riemann integrable (RI) where? Unless you specify a closed bounded interval, the question doesn't make much sense. (Now you say RI on$\mathbb R,$but what does that mean?) Edit The below is edited from the original, where I misunderstood the hypotheses to imply$\{q_k\}=\mathbb Q.$There are certainly examples where the choice of$...

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The formal question that you asked seems to be: "why is the kernel of the operator $\textrm{id}-\kappa$ equal to the set of constant functions"? And my answer to this question is that it is not, for general reversible Markov kernels - just take $\kappa=\textrm{id}$, which satisfies all your conditions, and then $L^2(\mu)$ is equal to the kernel! To better ...

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There is a typo in the OP. The statement in Ash’s book Probability and Measure Theory is as follows: $\mu_n(A \Delta B_n) = \mu ((A \Delta B_n) \cap A_n)= \mu[(A \Delta(B_n\cap A_n))\cap A_n]=\mu_n(A \Delta (B_n\cap A_n)).$ Proof. By definition, \begin{align*} [A\Delta(B_n\cap A_n)]\cap A_n&=[(A\cap(B_n\cap A_n)^c)\cup(B_n\cap A_n\cap A^c)]\cap A_n\\&...

0

The measurable functions $f: \mathbb{R} \rightarrow \mathbb{R}$ are precisely the ones for which there exists an $a\in\mathbb{R}$ such that $f^{-1}(\mathbb{R}\backslash\{a\})$ is countable. Indeed, such a function is measurable because in this case $f^{-1}((-\infty, b])$ will be countable if $b<a$ and co-countable otherwise. Conversely if an $f: \mathbb{R}... 1 Partial Answer Claim:$f(y)$is Lebesgue integrable. Proof: Note that, for any$q_j$and$q_k,$we have that$m([q_j,q_j+4^{-j}]\cup[q_k,q_k+4^{-k}])\le 4^{-j}+4^{-k},$since the intervals might overlap; if the intervals do not overlap, we get equality. Let $$U=\bigcup\limits_{k=1}^{\infty}\left[q_{k},q_{k}+\frac{1}{4^{k}}\right].$$ So we have $$\int\... 1 Let (\Omega,\mathcal{F},P) be a probability space. Let (A_{n})_n be a sequence in \mathcal{F}. Define A=\limsup_{n}A_{n}:=\cap_{n=1}^{\infty}\cup_{k=n}^{\infty}A_{k}. We go to prove that \lim_{n\rightarrow\infty}P(A_{n}\setminus A)=0. Proof: For each n, let B_{n}=\cup_{k\geq n}A_{k}. Then (B_{n}\cap A^{c})_n is a decreasing sequence and \... 1 If you are looking for textbooks suitable to a beginning graduate student, I would recommend all three of these textbooks Real Analysis: Modern Techniques and Their Applications by Gerald B. Folland. Real and Complex Analysis by Walter by Walter Rudin. Introduction to Measure Theory by Terence Tao. He also maintains lecture notes on his blog. I went ... 1 If \mu_n\to\mu weak* there's not much that can be said about convergence of \mu_n(A). If I recall correctly, assuming of course we're talking about regular Borel measures: If A is compact then \mu(A)\ge\limsup\mu_n(A). If A is open then \mu(A)\le\liminf\mu_n(A), and I think that's about the whole story. Thanks to @Mindlack for pointing out ... 3 No: take X=[0,1], \mu_n=\delta_{1/n}, \mu=\delta_0, and A=\{0\}. 1 There has probably been a mixup in the English translation of the book. In the Russian edition the problem is (correctly) stated as Prove that P(A_n\setminus \limsup_k A_k)\xrightarrow[n\to \infty]{} 0 and P((\liminf_k A_k)\setminus A_n)\xrightarrow[n\to \infty]{} 0. 1 For Q 1: Yes, \|f\|_p \to \|f\|_{\infty} as p \to \infty even for infinite measures. For Q 2: Yes, bounded functions belong to L^{\infty}(\mu) even for infinite measures \mu. For Q 3: \mu(E)=\int_E \frac 1 x d\lambda(x)=0 whenever \lambda (E)=0 so \mu << \lambda so L^{\infty}(\lambda) \subset L^{\infty}(\mu). 0 We give a counter-example which shows that none of the following conditions are sufficient to guarantee that the image of a regular set is regular. More conditions on X,Y, f could presumably be added so as to give sufficiency, but this vague enough as to not be a well-posed question. f:X \to Y is continuous and bounded-to-one, i.e., there exists M \... 1 Let M be in \mathcal{M} and let S be any \sigma-algebra containing \mathcal{A}. If M is countable it's a countable union of singletons, all of which are in S and so M \in S. Otherwise it's the complement of a countable set (which is in S as saw) and so also in S (as S is closed under complements). This show the left to right inclusion. ... 2 ...but the union is our box before is not correct. In the two dimensional case, let B=[0,1]\times[0,1] B_1=[0,1]\times\{0\} B_2=\{0\}\times[0,1] Then B is a square, of which B_1,B_2 are two sides, and B\neq B_1\cup B_2. Higher dimensions are similar. 2 B_1 \cup B_2 \neq B. For example (a_1, \dots, a_b) \notin B_1 and (a_1, \dots, a_b) \notin B_2 while (a_1, \dots, a_b) \in B. 3 No, Z is also a function (\Omega, \mathcal F) \to (\mathbb R, \mathcal B). Specifically$$ Z(\omega) = X(\omega) Y(\omega). $$Expectation is calculated the same as for X or Y, i.e.$$ E[Z] = \int_{\Omega} Z(\omega) \text{d}P(\omega) =\int_{\Omega} X(\omega) Y(\omega) \text{d} P(\omega). $$You can also evaluate things in terms of a measure on \... 4 Consider any A\subset[0,1/2). Let \bar{A}=[0,1/2)\setminus A. (I think the question is strange with the element 1/2, but we are not too much interested in this one point so I ignore) Let X=A\cup(1-\bar{A}). Then, every condition you mentioned is satisfied. However, to make your expectation true, we should be able to say that every subset of [0,1/2)... 0 Just write \lim \inf instead of limit throughout the inequalities. Observe that \lim \inf f_n =\lim f_n because \lim f_n exists by hypothesis. Of course \lim (2+\frac 1 n)=2. 0 Let \{Y_n\}_n be a sequence of simple functions converging pointwise to Y. You know that the result holds for simple functions so for each n let f_n be the corresponding Borel function such that Y_n=f_n(X). Since \{Y_n\}_n converges this tells us that \{f_n\}_n converges on X(\Omega). Now let A\subset \mathbb{R} denote the Borel measurable ... 0 If you mean to ask what is the measure of the intersection, then we can actually observe that the given intersection is empty, so the measure is zero. To see this, say there were an element in this intersection, call it x. There must exist N\in\mathbb{N} so that x<N. Now, note that x\notin (M,\infty) for all M\geq N, a contradiction. 1 Just for reference: An answer, which worked for me in the end, was to use a set of test function g \in C_c(\mathbb R^n) and use the pairing \langle \rho_t, g \rangle = \int g \, \rho_t. For an empirical measure \mu^N = \frac{1}{N} \sum_{i=1}^N \delta_{x_i} we can compute a weak form of the equations$$ \frac{\mathrm d}{\mathrm d t} \langle \rho_t , ... 1 It is false. Choose any$b>a>0$. Define$q$as $$q(x)=\begin{cases}0,x\in(-\infty,0)\cup[2,\infty)\\a,x\in[0,1)\\b,x\in[1,2)\end{cases}$$ Take$\phi$as any function in$C^1_c(\mathbb{R})$that is$1$on$[0,2]$and$0$outside$(-1,3). Assuming the conjecture is true, we should have $$\lim_{\eta\rightarrow 0}\frac{1}{\eta}\int_{\mathbb{R}}(q(x+\... 1 By definition the function \phi maps from \mathbb{R} to \mathbb{R}. Hence there is no ambiguity in taking derivatives: there is only one variable and one direction. Furthermore the notation \phi'(x) is an abbreviation for$$\bigg(\frac{\partial \phi}{\partial t}(t)\bigg) \bigg|_{t=x}.First the function gets differentiated, then you plug in ... 1 Your interpretation of the notation in the integral is correct, \lambda = (\lambda_1, \dots, \lambda_n). This is a completely standard notation for elements of \mathbb{R}^d. For the other part of the question, notice that \begin{align} \sup_{(\lambda_1,\cdots,\lambda_n)\in \partial B_n}\langle (\sum_{i=1}^n \lambda_i T_i)h, (\sum_{i=1}^n \overline{\... 0 The class of all sets A \in \sigma(\mathcal E) such that either x \in A and y \in A or y \notin A and x \notin A is a sigma algebra which contains \mathcal E . Hence it contains \sigma(\mathcal E). 1 Write A = \cup_{\omega \in \Omega} \Pi(\omega) \times \Pi(\omega). A is in fact measurable. To show this define \phi: \Omega \times \Omega \to \Omega by \phi(\omega^1, \omega^2)_i = \begin{cases} 1 & \text{if}\ \omega^1_i = \omega^2_i \\ 0 & \text{if}\ \omega^1_i \neq \omega^2_i. \end{cases}$This is measurable since each coordinate of$\...

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If you simply are required to have your $k$s be a subset of $\mathbb N$, then you can take $k=1$, $(x-\frac 1 {3 \epsilon},x+\frac 1 {3 \epsilon}$. If you have to have $I_k$ for every natural number, take $(x-\frac 1 {2^{k+3} \epsilon},x+\frac 1 {2^{k+3} \epsilon})$.

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Why summing them? For each $\varepsilon>0$, take $k\in\mathbb N$ such that $\frac2k<\varepsilon$ and take only the interval $\left(x-\frac1k,x+\frac1k\right)$. That's all.

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To verify that the completion is complete, you have to prove that any subset of a null set of the completion is having measure zero. And any element of the completion can be written as the union of an element of the initial $\sigma$-algebra and an element included in a measure zero set. This is why $\mu(B)$ is supposed to be equal to $0$. Then $\mu(A \cup ... 0 (1) For$\pi_0, observe that \begin{align*} f=c_f+(f-c_f), \end{align*} wherec_f=\int_{\Omega} f d\mu$is the average of$f$(assuming$\mu(\Omega)=1$). To claim$\pi_0(f)=f-c_f$, we have to check that$f-c_f\in L^2_0(\mu)and \begin{align*} f-c_f\perp c_f, \end{align*} which are left to you. (2)(M^{\perp})^{\perp}=\bar{M}$only works when$M$is a ... 1 The importance of absolute continuity comes from the Radon-Nikodym theorem. It states if$\mu \ll \nu$then$\mu$has a density function with respect$\nu\$. A lot of the time it's much easier to prove things if all the distributions we work with have a density function, i.e. are dominated by some common measure. It's also possible phrase absolute continuity ...

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