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If for $\mathcal{A}$ to be an algebra on a set $\Omega$ you need $\Omega \in \mathcal{A}$, then a subset $\mathcal{A} \subseteq \mathcal{P}(\Omega)$ without $\Omega \in A$ is not an algebra. I'm interpreting 'algebra' to mean a set containing $\Omega$ and closed under complements and finite unions of its elements. If $\mathcal{A} = \{ \{1\}, \{2,3\} \}$ ...

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I don't see a problem. In the example you give, assuming $P = I_1 \cup I_2 \cup I_3$, we would take $P_1 = P_2 = \emptyset$, $P_3 = I_1 \cup I_2$, and $P_4 = I_3$, and then $P_5 = P_6 = \dots = \emptyset$. Maybe you are misinterpreting the sentence "Since each $I_j$ is contained in one of $G_1, G_2, \dots$." What Jones means is "Each $I_j$ is contained in ...

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This is an elaboration of the (slightly simplified version of the) answer suggested by user24142. Let $$\mathscr S\equiv\{F\in\mathscr F\,|\,\forall\varepsilon>0,\exists E\in\mathscr E:|\mathbb P(E)-\mathbb P(F)|<\varepsilon\}.$$ Clearly, $\mathscr E\subseteq\mathscr S$. I claim that $\mathscr S$ is a monotone class. Fix $\varepsilon>0$ and suppose ...

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It's probably good to clarify some definitions. If $\mathscr{A}$ is an algebra of sets, and $\mu \colon \mathscr{A} \to [0,+\infty]$ is a finitely additive set function, then we say that $\mu$ is $\sigma$-additive, if whenever $(A_n)_{n\in \mathbb{N}}$ is a sequence of pairwise disjoint sets in $\mathscr{A}$ such that $\bigcup\limits_{n\in\mathbb{N}} A_n \in ... 1 Your conjecture is correct. The reason is that the set of sets that you can approximate within$\mathscr{F}$is a$\sigma$-algebra. That its closed under complements should be clear, and the countable union condition isn't too hard. It therefore is a$\sigma$-algebra that contains$\mathscr E$and is contained by$\mathscr F$, so it must be$\mathscr F$. 2 Your$\chi_{D}u$can be defined in that way: $$(\chi_{D}u)(x)=\begin{cases} u(x), & x\in D, \\ \\ 0, & x\notin D. \end{cases}$$ Edit: You don't need to mention that$u\in M\subseteq L^p[0,1]$where$M$is a bounded subset of$L^p[0,1]$. This is equivalent ... 0 Since$\{f_i, 1 \le i \le m\}$is a set of Borel measurable functions $$\{x:f_i(x)<c_i\}\subset \mathcal{B}(\mathbb R)$$ for all possible$c_i\in\Bbb{R}$. So $$\{x:f_1(x)<c_1\}\times \{x:f_2(x)<c_2\}\times\cdots\times\{x:f_m(x)<c_m\}\subset \mathcal{B}(\mathbb R)^m=\mathcal{B}(\mathbb R^m)$$ for all possible$c_1\cdots c_m\in\Bbb{R}$. This ... 2 If you are assuming$\mu (E)<\infty $, then I think this will work: Let$n\in \mathbb N$. Then using the definition of the outer measure, there is are open sets$U_{n+1}\subseteq U_n$containing$E$such that$\mu^* (E)=\mu(E)>\mu (U_n)-1/n$. Thus,$\tag1\mu (E)\geq \mu\left ( \bigcap _{n\in \mathbb N} U_n \right )$(because$U_n\subseteq ...

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You're right. you have that the image of an open set under projection is open, then the preimage of that is measurable in $\mathbb{R}^1$. Then, the product of measurable sets is measurable, and you are done.

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For a non-metrizable but locally compact Hausdorff counterexample, consider $\omega_1$ with its order topology. I will show that $\mathcal{S}$ does not even contain all the open sets. Let $\mathcal{B}$ be the collection of all sets $B \subset \omega_1$ such that either $B$ is countable or $B$ contains a club set. (We consider "countable" to include ...

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For the answer to the first question: let $\mathcal S' = \{A \in \mathcal S : A^C \in \mathcal S\}$. See that $\mathcal S'$ contains all open sets, that $\mathcal S'$ is closed under taking complements, and that $\mathcal S'$ is closed under countable unions.

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The identity $$(a,b) = \bigcup_{n \in \mathbb{N}} [r_n(a),\infty) \cap \bigcup_{n \in \mathbb{N}} [r_n(b),\infty)$$ is not correct. As @uniquesolution pointed out in his (her) comment, you need to assume that $a<r_n(a)<r_n(b)<b$, but even then the identity does not hold true. Instead, it should read $$(a,b) = \bigcup_{n \in \mathbb{N}} ... 1 The argument given in the question is correct and it shows that$$ \{w : w \in A_n \text{ for all $n$ except a finite number}\} \subseteq \bigcup_{n=1}^\infty \bigcap_{k=n}^\infty A_k $$It remains to show that$$ \{w : w \in A_n \text{ for all $n$ except a finite number}\} \supseteq \bigcup_{n=1}^\infty \bigcap_{k=n}^\infty A_k. $$Given that you've written ... 2 Your proof so far is perfectly fine. All you need to do is prove the reverse inclusion and you'll be set. 1 Define for a positive integer n,$$f_n(y):=\sum_{i=1}^n\chi_{A_i}(x) \chi_{B_i}(y).$$For each y, the sequence (f_n(y))_{n\geqslant 1} is non-decreasing and non-negative. Since each function f_n is \Sigma_Y-measurable, we have by the monotone convergence theorem$$\lim_{n\to +\infty }\int_Yf_n(y)d\nu(y)=\int_Y\lim_{n\to +\infty }f_n(y)d\nu(y) ...

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A measure space is always a triple $(\Omega, \mathcal{F}, \mu)$, where the set $\Omega$ is the set with respect to which $\mathcal{F}$ is a sigma-algebra. The value of $\mu(\Omega)$ is called total mass of $\mu$. The measure $\mu$ is a probability measure iff its total mass is $1$. Now observe that $P(I \cap \Omega) = P(I) = P(I) \cdot P(\Omega)$, so both ...

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Given two sequences $x_n$ and $y_n$, we have the relation $$x_n\leq y_n\Rightarrow \sum_n x_n\leq \sum_n y_n.$$ Therefore, if we set $x_n=\sum^{\infty}_{j=1}|I_{n_j}|$ and $y_n=\frac{\epsilon }{2^n}$, applying the above gives $$\sum_n\sum^{\infty}_{j=1}|I_{n_j}|<\sum_n\frac{\epsilon}{2^n}$$

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You really only need to consider your first equation in order to obtain a contradiction. Suppose that there exists $w \in L^{1}_{\mbox{loc}}(\mathbb{R})$ such that $$\int w\varphi dx = \varphi(0),\;\;\;\varphi \in \mathcal{C}_{c}^{\infty}(\mathbb{R}).$$ That equation is inconsistent. To see why, find ...

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There is no such function, as you can see for $t \to 0$. But the point is that differentiability is a local property. Hence, it suffices for each interval $(t_0,t_1)$ with $0<t_0<t_1$ to find a dominating function which is independent of $t$ (but only needs to hold for $t$ in the above interval). This is easy to do, namely $x^2 e^{-t_0 x^2}$ will ...

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We must show that $\mathcal{A}$ contains $\mathbb{R}$ and is closed under pairwise unions and complements, or equivalently (which is what we will do here) under pairwise intersections and complements. Choose any $a \in \mathbb{R}$. Then $\mathbb{R} = (-\infty,a] \cup (a,\infty)$, which shows that $\mathbb{R} \in \mathcal{A}$. To show that $\mathcal{A}$ is ...

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Part 3: Given $\epsilon>0$, there is $h$ such that $$\int_{[|f_k|\ge h]} |f_k|d\mu< \epsilon$$ For this $h$, by post for each $\epsilon >0$ there is a $\delta >0$ such that whenever $m(A)<\delta$, $\int_A f(x)dx <\epsilon$, given $\epsilon$, there is a $\eta$, for any $A$ such that $\mu(A)<\eta$, there is $$\int_A h d\mu < ... 0 This really should be a comment, but it's a bit too long. Part 2 is very similar to part 1.$$\begin{aligned} \int_A|f_k| &= \int_{A \cap [|f_k| \geq h]}|f_k| + \int_{A \setminus [|f_k| \geq h]} |f_k| \\ &\leq \int_{A \cap [|f_k| \geq h]}|f_k| + \int_{A \setminus [|f_k| \geq h]} h \\ &\leq \int_{[|f_k| \geq h]}|f_k| + \int_{A} h \end{aligned}... 2 Take a look further to the front of the book, maybe you can find the definition of the generated sigma-algebra that the author uses. Typically, the generated sigma-algebra is defined just in the way you described. In this case you only need to prove the second assertion of this exercise. However, there are also other definitions. One possible definition is ... 2 First assume f is nonnegative. Define E=\bigcup_{n=1}^\infty E_n, then as f\chi_{E_n}\uparrow f\chi_E so by monotone convergence, \begin{align} \sum_{n=1}^\infty \int_{E_n} f\ \mathsf d\lambda &= \lim_{n\to\infty}\sum_{j=1}^n\int_{E_j} f\ \mathsf d\lambda \\ &=\lim_{n\to\infty} \int_{\bigcup_{j=1}^n E_j} f\ \mathsf d\lambda \\ ... 3 You are thinking too complicated. What we need is that the integral of the mollifier is 1, and that the support of \omega_\epsilon shrinks to the origin as \varepsilon \to 0 (if we took more general mollifiers, where the support need not be compact, we would still have \lim\limits_{\epsilon \to 0} \int_{\lvert x\rvert > \delta} ... 2 This is not exactly an answer to the question as formulated, but it may clarify the precise formulation of the distinction. Convergence a.e. can be written using countable intersections and unions. First defineA_{n,m} = \{ x : |f_n(x) - f(x)| > 1/m \}.$$Now, for the convergence to fail at x, there must be some m such that x \in A_{n,m} for ... 1 Yes, your re-characterization is wrong, becuase it is very well known that convergence in measure does not imply convergence almost everywhere. Here is an example. Consider the intervals E_{i,n}=[{i-1\over n},{i\over n}] and their characteristic functions \chi_{E_{i,n}}, where for each n i runs between 1 and n. So the few first intervals are ... 3 This is not correct. f_n\to f a.e. iff$$\mu(\{x :\lim_{n\to\infty} |f_n(x)-f(x)|\ne0\}) = 0. $$Convergence in measure has the limit outside the measure - one must take care when interchanging limit operations! Now, convergence in measure implies that there is a subsequence that converges a.e., and in a finite measure space (i.e. \mu(X)<\infty), ... 1 Indeed, these cannot be always equal: as you correctly observed, one is a norm and the other is not. And here is an explicit example: f\equiv 1  on the unit interval [0,1] with p_0=2, p_1=4 and t=1/2; subsequently p=8/3. Clearly, \|f\|_{L^{p,\infty}}=1 for every p. Given any \lambda>0, we can write f=f_{0,\lambda}+f_{1 \lambda} with ... 1 Show that \mathcal A = \{ X \subseteq \mathbb R \mid X \text{ is countable or } \mathbb R \setminus X \text{ is countable} \} is a \sigma-algebra. As \mathcal S \subseteq \mathcal A, the claim follows. 3 Computing integral of Gaussian on whole line is possible using standard techniques. The problem is that no elementary antiderivative exists so you can't easily obtain integrals over arbitrary intervals. The function is integrable in the Lebesgue sense, just as it is integrable in "improper" Riemann sense. So Lebesgue theory doesn't introduce anything new ... 1 Every uncountable Borel space is isomorphic to [0,1] with its Borel sets by a result of Kuratowski.So we can always take Borel spaces to be isomorphic to [0,1] (the countable case is usually easy to handle anyways). So if S is a Borel space and \phi:S\to[0,1] is an isomorphism, then the metric d:S\times S\to[0,1] given by ... 2 The "usual" Riemann integral is not defined, but the Lebesgue integral is and is equal to the "improper" Riemann integral: let f(x) = x^{-\frac{1}{2}} on (0,1] and f(0)=0. The f is defined on [0,1]. Now define f_n(x)=f(x)\cdot\chi _{(\frac{1}{n},1]}(x), Then, f_n\nearrow f a.e. and so the Monotone Convergence Theorem applies to say that ... 0 Borel \sigma algebra B[0,1]) of the set [0,1] is infinite but it is not generated by any partition of the interval [0.1]. Assume the contrary and let (X_i)_{i \in I} be such a partition.Since singletons belong to B[0,1]) we deduce that each X_i must be singleton. Then [0,1/2] does not belong to B[0,1]) because it is not presented as ... 1 Note that (why?)$$ \partial B_r (x)\subset \{y \mid d(x,y)=r\}. $$(But in general, equality does not hold). Hence, the sets (\partial B_r (x))_{r>0} are pair wise disjoint. Now it is easy to see that in a collection of pair wise disjoint measurable sets (M_i)_i at most countably many can have positive measure. Otherwise, one of the sets$$ \{i ...

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On finite measure spaces, almost sure convergence is a stronger property than convergence in measure. A sequence converges almost surely if there is a fixed nullset $N$ so that $f_n(x)$ converges to $f(x)$ for all $x \notin N$. However, we don't need such a set for the convergence in measure. We just say that the set on which $f_n$ and $f$ differ by more ...

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This is one of those things that is helpfully studied using an example. A very nice example for this issue is the "wandering block". Informally, the wandering block is the sequence of indicator functions of $[0,1],[0,1/2],[1/2,1],[0,1/4],[1/4,1/2],[1/2,3/4],[3/4,1]$, etc. More explicitly, it is the sequence $g_n(x)$ which comes about from enumerating the ...

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If you omitted condition (b), then nothing in the hypothesis tells you what $f$ is. Remember that a theorem should be correct no matter what particular values you give its variables; as long as the hypotheses are true, the conclusion must be true also. Now suppose you chose some reasonable functions as your $f_n$'s, satisfying hypothesis (a), so they ...

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Observe that : $\left|\sin\left(\dfrac{x}{n}\right)\right| < \dfrac{x}{n} < x, x > 0$ and you can prove this fact. Thus it follows that : $|f_n(x)| < g(x) = \dfrac{x}{e^x-1}$. Next consider $f(x) = \sin\left(\dfrac{x}{n}\right) - \dfrac{x}{n} \to f'(x) = \dfrac{\cos\left(\dfrac{x}{n}\right)}{n} - \dfrac{1}{n} < 0\to f(x) < f(0) = 0 \to ... 2 These are just the pointwise max and min of$f$and$g$, respectively. That is, $$(f\vee g)(x)=\max(f(x),g(x))$$ and $$(f\wedge g)(x)=\min(f(x),g(x)).$$ More generally, if$f$and$g$are elements of a poset,$f\vee g$refers to the least upper bound of$f$and$g$and$f\wedge g$refers to the greatest lower bound of$f$and$g$(if such elements exist). ... 2 Typically, we have $$(f\vee g)(x)=\max\bigl(f(x),g(x)\bigr)$$ and $$(f\wedge g)(x)=\min\bigl(f(x),g(x)\bigr).$$ This accords with a typical notion in boolean algebras that is often generalized to partially ordered sets, where$a\vee b$and$a\wedge b$indicate (respectively) the least upper bound and greatest lower bound of$a,b.$1 According to this study by Jean-Philippe Villeneuve (in French): http://culturemath.ens.fr/histoire%20des%20maths/htm/villeneuve2009/theories-de-la-mesure.html the essential property of sigma-additivity that is used to define sigma-algebra was first used both by Émile Borel in 1898 for his work on the theory of functions in which there is a chapter on the ... 1 As Cameron Buie pointed out, the claim is not true in general. However, it is true if$\tau$is second-countable and$\mathcal B$is a countable topological basis (in particular, it is true in separable metric spaces). To see this, suppose that$U\in\mathcal \tau$is any open set. Then, there exists a subcollection$\mathcal B_{U}\subseteq\mathcal B$, ... 6 Not necessarily. For example, consider the discrete topology on$\Bbb R,$which has $$\mathcal B=\bigl\{\{x\}:x\in\Bbb R\bigr\}$$ as a base. The$\sigma$-algebra generated by$\mathcal B$is the set of all subsets$A$of$\Bbb R$such that either (1)$A$is finite or countably infinite, or (2)$\Bbb R\setminus A$is finite or countably infinite. However, the ... 1 The concrete example I like to think of is where$\Omega = [0,1]^2$with Lebesgue measure, and$\mathcal F_0 = \{A \times [0,1]: \text{$A$ is a Lebesgue measurable subset of $[0,1]$}\}$. Then$ E(f\mid\mathcal F_0) = g $, where $$g(x,y) = \int_0^1 f(x,\eta) \, d\eta .$$ In fact, every conditional expectation can be thought of in this form. I have a paper ... 1 The conditional expectation$Y = E[X\mid\mathcal{F}]$can be interpreted as the best$\mathcal{F}$-measurable approximation for$X$. More specifically: If$X$is square-integrable, then the conditional expectation is the orthogonal projection of$X$to the subspace of$L^2$that consists of$\mathcal{F}$-measurable functions. Note that the conditional ... 0 For a discrete random variable,$X$, the probability mass function, denoted$f_X$satisfies the following $$\mathbb{P}[X\in\{a_1,a_2,\dots\}]=\sum_{n=1}^\infty f_X(a_n).$$ For a continuous random variable,$Y$, the analogue to PMF is called the probability density function, denoted$f_Y$. And it satisfies for any measurable set$U$, $$\mathbb{P}[Y\in ... 2 An atom in a measurable space (M,\mathcal{M}) is a nonempty set A\in\mathcal{M} such that B\in\mathcal{M} and B\subseteq A implies that either B=\emptyset or B=A. If \mathcal{M} is generated by a countable family of sets, so it is countably generated, every point in M is in a unique atom and the atoms form a partition of M. A countably ... 1 The answer is no: Consider X=\{a,b,c\} and \Pi=\{\emptyset,\{a\},\{b\},\{c\},X\} with$$\lambda(\{a\})=\lambda(\{b\})=\lambda(\{c\})=\lambda(X)=1\,.$$The cheat is that \Pi doesn't contain any nontrivial triple of sets with P=Q_1\cup^* Q_2, so the 'additivity' (rather 2-additivity) holds vacuously. 1 Turns out we're talking about \ell_\infty[0,1] with the uniform norm, making the bit about the discrete topology on [0,1] irrelevant. Anyway, no f is not measurable. If E is any subset of [0,1] there is an open set V with E=f^{-1}(V), namely$$V=\bigcup_{t\in E}B(f(t),1/3).$\$ Similarly for any function into any metric space with ...

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