Let $\scr{A}$ be an algebra of subsets. Let $(\Omega, \sigma(\mathscr{A}), P)$ be a probability space. Then for each $B \in \sigma(\scr{A})$ and $\epsilon &gt; 0$, there exists $A \in \scr{A}$ such that $$P(A \Delta B) := P((A \setminus B) \cup (B \setminus A)) &lt; \epsilon.$$  There is a suggestion to consider a collection of such all sets $B$ satisfying the condition. I guess I should look at $$G = \{B \subseteq \Omega :\ \mbox{there exists} \ A \in \mathscr{A} \ \mbox{such that} \ P^*(A \Delta B) &lt; \epsilon \ \forall \epsilon &gt; 0\}$$ (I am not sure here that I should use the measure $P$, or its corresponding outer measure $P^*$ instead).  I find that the hint does not much helpful for me, can anyone suggest more about this problem ?
95285
Theorem. Let $(X,\mathcal B,\mu)$ a finite measure space, where $\mu$ is a positive measure. Let $\mathcal A\subset \mathcal B$ an algebra generating $\cal B$. Then for all $B\in\cal B$ and $\varepsilon&gt;0$, we can find $A\in\cal A$ such that $$\mu(A\Delta B)=\mu(A\cup B)-\mu(A\cap B)&lt;\varepsilon.$$  I don't think there is a proof in this site. It's a useful result for several reasons:  We know what the algebra generated by a collection of sets is, but not what the generated $\sigma$-algebra is. The map $\rho\colon \cal B\times\cal B\to \Bbb R_+$, $\rho(A,A')=\mu(A\Delta A')$ gives a pseudo-metric on $\cal B$. This makes a link between generating for an algebra and dense for the pseudo-metric. We say that a $\sigma$-algebra is separable if it's generated by a countable class of sets. In this case, the algebra generated by this class is countable. An with the mentioned result, we can show that $L^p(\mu)$ is separable for $1\leq p&lt;\infty$, which makes a link between the two notions. In ergodic theory, we have to test mixing conditions only an a generating algebra, not on all the $\sigma$-algebra.
44926
Let $(X, \mathcal B, m)$ be a measure space.  For $1 \leq p &lt; q \leq \infty$, under what condition is it true that $L^q(X, \mathcal B, m) \subset L^p(X, \mathcal B, m)$ and what is a counterexample in the case the condition is not satisfied?
30765
Let $P$ be a probability measure on a Borel $\sigma$-algebra (on some metric space, $\Omega$).  It is called tight if for every $\epsilon &gt;0$, there exists a compact $K$ such that $P(X \in K) \geq 1 - \epsilon$.   It is called inner regular if $P(A) = \sup \{P(K) | K \subset A, K \text{ compact}\}$ for every $A$ in the Borel $\sigma$-algebra.   It is clear that inner regularity implies tightness if we choose $A=\Omega$. For an arbitrary set $A$, using tightness we can obtain a compact $K(\epsilon)$ such that $P(A\cap K(\epsilon)) \geq P(A) - \epsilon$. If $A$ is closed then $K \cap A$ is compact and hence we get inner regularity for closed sets $A$. How can I prove this for general $A$?  Thanks.
33602
My question is the following :   is there a $\sigma$-algebra $\mathcal{T}$ (of subsets of $\mathbb{R^n}$) that contains strictly the $\sigma$-algebra $\mathcal{L}$ of  Lebesgue measurable sets (in $\mathbb{R}^n$), and such that there is a measure on $\mathcal{T}$ that extends the usual Lebesgue measure on $\mathcal{L}$ ?  I guess not, but I did not find a reference.
103551
I am trying to prove that $\mathcal{S} = \{\cup_{n\in K} (n, n+1]: K \subset \mathbb{Z}\}$ is a $\sigma$-algebra on $\mathbb{R}$ using the definition. I want to show two things: (1) $E \in \mathcal{S} \implies E^c \in \mathcal{S}$ and (2) If $E_1, E_2, \dots$ is a sequence of members of $\mathcal{S}$, then $\bigcup_{i=1}^\infty E_i\in\mathcal{S}$. My attempt: Fix $K \subset \mathbb{Z}$. Let $E\in\mathcal{S}$. Then, $E=\bigcup_{n\in K}(n, n+1]$. So, $E^c = \bigcap_{n\in K} (-\infty, n]\cup (n+1, \infty)$. Now, how can I show that $E^c\in\mathcal{S}$? It seems that as $n$ increases, the intervals that constitute $E^c$ get smaller and smaller, so the intersection of all these intervals would get smaller and smaller. Also, how can (2) be proven? Would be have $2$ unions in this part?
72979
While reading David Williams's "", the following statement caught my fancy:     Every subset of $\mathbb{R}$ which we meet in everyday use is an element of Borel $\sigma$-algebra $\mathcal{B}$; and indeed it is difficult (but possible!) to find a subset of $\mathbb{R}$ constructed explicitly (without the Axiom of Choice) which is not in $\mathcal{B}$.   I am curious to see an example of such a subset.
122265
Let $A\subset\mathbb{R}$ a measurable where $0&lt;m(A)&lt;\infty$. Show that exists for each $0&lt;\alpha&lt;1$ an interval $I$ such that  $$ \frac{m(A\cap I)}{m(I)}&gt;\alpha. $$  MY ATTEMPT: Following a hint a give that:  Let $\varepsilon&gt;0$, exists $G$, a open set such that $m(A)\leq m(G)&lt;m(A)+\varepsilon$. As $G$ is open, we can write as the disjoint sum of open intervals $\dot{\bigcup}_n I_n = G$.So, $$ m(A)\leq m(G)=m\left(\sum \dot{\bigcup_n} I_n \right)\leq \sum m\left(I_n \right)&lt;m(A)+\varepsilon $$ Suppose that $\varepsilon=(\alpha^{-1}-1)m(A)$:  $$ m(A)+\varepsilon=m(A)+m(A)(\alpha^{-1}-1)=m(A)(1+\alpha^{-1}-1)=m(A)\alpha^{-1} $$ and $$ \alpha\sum m(I_n)&lt;m(A)=\sum m(A\cap I_n) $$ $$ \Rightarrow \alpha&lt;\frac{\sum m(A\cap I_n)}{\sum m(I_n)}  $$  But I have a insecurity with that:   (1) I can write that: $m(A)=\sum m(A\cap I_n)$?   (2) How to get the result with this: $$ \alpha&lt;\frac{\sum m(A\cap I_n)}{\sum m(I_n)}  $$
91454
Let $A_1, A_2, \ldots$ and $B_1, B_2, \ldots$ be two sequences of events in some probability triple $(\Omega, \mathcal{F}, \mathbf{P})$. Now, it is true that $\left(\limsup_n A_n\right) \cap \left(\limsup_n B_n\right) \supseteq \limsup_n (A_n \cap B_n)$, and I can think of many different cases where this relation holds true for equality (e.g., let $A_n = B_n$, or $A_n = B_n^c$, or $B_n = \emptyset$, or $B_n = \Omega$). However, I cannot think of any case where $\left(\limsup_n A_n\right) \cap \left(\limsup_n B_n\right) \supset \limsup_n (A_n \cap B_n)$ holds for strict inclusion.  My proof for the relation is simple. Since $A_n \cap B_n \subseteq A_n$ and $A_n \cap B_n \subseteq B_n$, then we can perform some set algebra as follows:  $$\bigcup_{k=n}^\infty (A_k \cap B_k) \subseteq \bigcup_{k=n}^\infty A_k$$ $$\bigcap_{n=1}^\infty \bigcup_{k=n}^\infty (A_k \cap B_k) \subseteq \bigcap_{n=1}^\infty \bigcup_{k=n}^\infty A_k$$  Then we can use the same logic for $B_n$, and come to the conclusion as stated in the beginning.
57303
I'm studying measure theory, and I came across a theorem that says that, given two $\sigma$-finite measure spaces $\left(X,\mathcal M,\mu\right)$ and $\left(Y,\mathcal N,\nu\right)$, and a set $E\in \mathcal M\otimes\mathcal N$, the function $x\mapsto \nu\left(E_x\right)$ is measurable.  The definition of the product-$\sigma$-algebra $\mathcal M\otimes\mathcal N$ is the $\sigma$-algebra generated by the rectangles $A\times B$ with $A\in\mathcal M$ and $B\in\mathcal N$. Also, the $x$-section $E_x$ is defined as $E_x=\left\{y\in Y:\left(x,y\right)\in E\right\}$.  My problem is that I can't even convince myself that, given $E\in\mathcal M\otimes \mathcal N$, the sections $E_x$ are in $\mathcal N$ for all $x$. It looks like it has something to do with unitary sets being measurable (which shouldn't be necessary, I think), but I can't prove it even assuming that $\left\{x\right\}\in\mathcal M$ for all $x$.
81310
Suppose $Pr$ is non-atomic that is $Pr(A)$ implies the existence of $B$ with $B\subset A$ such that $0&lt;Pr(B)&lt;Pr(A) $ and $A\in\mathcal{F}$ with $Pr(A)&gt;0$. (a)show that for every $\epsilon&gt;0$, we have $B\subset A$ such that $0&lt;Pr(B)&lt;\epsilon$. (b)Prove that if $0&lt;a&lt;Pr(A)$ then there exists $B\subset A$ such that $Pr(B)=a$.  Here is a hint: fix $\epsilon_n \to 0$ and define inductively numbers $x_n$ and sets $G\in \mathcal{F}$ with $H_0=\emptyset$, $H_n=\cup_{k&lt;n} G_k$, $x_n=\sup\{Pr(G): G\subset A\setminus H_n, Pr(H_n\cup G)\leq a\}$ and $G_n\subset A\setminus H_n$ such that $Pr(H_n\cup G_n)\leq a$ and $Pr(G_n)\geq (1-\epsilon_n)x_n$. Let $B=\cup_{k} G_k$.
103506
I want to consider a space of probability measures on some set $X$, such that the space of measures is complete, not in the sense of  (though probably that too), but as in the standard analysis meaning. I haven't studied far enough in probability measures, so I can't tell -- what conditions do I need on my space such that it is complete? And how should I metrize it?   Thanks!  Edit: A reference/citation to a text would also be super helpful!
101067
This is exercise 1.2.19 from Chang &amp; Keisler's Model Theory, which has been giving me a headache for some time now.  Let $\mathscr{S}$ be a given propositional language of any cardinality (i.e. the set of sentence symbols is allowed to have any cardinality). Call a theory simply any set $\Gamma$ of sentences from $\mathscr{S}$. A set $\Delta$ of sentences is said to be a set of axioms for $\Gamma$ iff $\Gamma$ and $\Delta$ have the same consequences (i.e. they are equivalent). Finally, a set $\Sigma$ of sentences is said to be independent iff for each $\sigma \in \Sigma$, $\sigma$ is not a consequence of $\Sigma - \{\sigma\}$. Exercise 1.2.19 asks us to prove that a theory $\Gamma$ always has an independent set of axioms.  Now, if $|\Gamma| \leq \omega$, the proof is more or less simple; for the finite case, cf. , whereas, if $\Gamma = \{\gamma_0, \gamma_1, \dots\}$, one first constructs a set $\Gamma' = \{\psi_0, \psi_1, \dots\}$ such that $\psi_0 = \sigma_0$ and $\psi_n = \bigwedge_{i \leq n} \sigma_n$ (i.e. the conjunction of all $\sigma_i$ such that $i \leq n$). Then one constructs another set $\Delta = \{\delta_0, \delta_1, \delta_2, \dots\}$ such that $\delta_0 = \psi_0$ and $\delta_{n+1} = \psi_n \rightarrow \psi_{n+1}$. It's not difficult to show that $\Gamma'$ is a set of axioms for $\Gamma$ and that $\Delta$ is an independent set of axioms for $\Gamma'$.  But I'm completely lost about how to proceed in the transfinite case. It's clear to me that the above construction will not work directly; any ideas about how to adapt that strategy? Or should I look for an entirely different strategy?  EDIT: Notice that this question is different from ; the other question (and answer) is only asking about the countable case, the solution of which I have already provided a rough sketch in my own question. The general case, however, for theories of any cardinality, is, as C&amp;K remark, much harder.
70467
$(\Omega, \mathcal{F}, P)$ is a probability space. $X$ is a r.v. defined on it, and $\mathcal{G}_1, \mathcal{G}_2$ are two $\sigma$-algebra, can we claim the following: $$ \mathbb{E}\{\mathbb{E}[X|\mathcal{G}_1]|\mathcal{G}_2\}=\mathbb{E}\{\mathbb{E}[X|\mathcal{G}_2]|\mathcal{G}_1\}=\mathbb{E}[X|\mathcal{G}_1\cap\mathcal{G}_2]. $$ If this is not true, why the following is true: $X_1,X_2,\ldots,X_n$ are independent r.v. on the same probability space, $Z=f(X_1,X_2,\ldots,X_n)$ is another r.v. And we have: $$ \mathbb{E}\{\mathbb{E}[Z|X_1,\ldots,X_{i-1},X_{i+1},\ldots,X_n]|X_1,\ldots,X_i\}=\mathbb{E}[Z|X_1,\ldots,X_{i-1}]. $$
96222
Suppose $\mu$ is a measure in a ring $R$ of a set $X$. So for any mutually disjoint subsets $A_1,A_2,\ldots$ such that $\sum_{i=1}^\infty A_i\in R$, we have that $\mu(\sum_{i=1}^\infty A_i) = \mu(A_1)+\mu(A_2)+\cdots$. (i.e. countable additivity)  I'm wondering: Does that imply countable subadditivity? I.e. for any subsets $A_1,A_2,\ldots$ such that $\sum_{i=1}^\infty A_i\in R$, is it always true that $\mu(\sum_{i=1}^\infty A_i)\le \mu(A_1)+\mu(A_2)+\cdots$?  I believe it should be true, because you can take the mutually disjoint subsets $A_1,A_2-A_1,A_3-(A_1\cup A_2),A_4-(A_1\cup A_2\cup A_3),\cdots$, each of which is in $R$ (because unions and differences of sets in $R$ are again in $R$), then apply countable additivity, then use monotonicity on the fact that $A_n-(A_1\cup\cdots\cup A_{n-1})\subseteq A_n$.  Is that right? I just want to make sure I didn't make any mistake in my logic.
116611
Let $D$ be a $\sigma$-algebra of subsets of a set $X$, and let $Y$ be an arbitrary subset of $X$. Let $B= \{ A\cap Y | A \in D \}$. Show that $B$ is a $\sigma$-algebra of subsets of $Y$.  If we let $A=X$, then clearly $Y \in B$.  Suppose $G \in B$. Show $G^c \in B$.  So $G = A\cap Y$ for some $A \in D$.  So $G^c = A^c \cup Y^c$  I'm stuck here! Not sure how to show $G^c \in B$  Let $G1,G2,G3,.... \in B$. Not sure how to show that the infinite union of the $G_i$'s is an element of $B$.
53129
Let $\Omega$ be a set.  If $\Omega$ is finite, then any $\sigma$-algebra on $\Omega$ is finite.  If $\Omega$ is infinite and countable, a $\sigma$-algebra on $\Omega$  cannot be infinite and countable.     What about if $\Omega$ is not countable ? Is it possible to find an uncountable $\Omega$ with a $\sigma$-algebra that is infinite and countable ?
15881
My Question: Let $(\Omega,\mathcal{F},\textbf{P})$ be a probability triple such that $\Omega$ is ${countable}$. Prove that it is impossible for there to exist a sequence $A_1,A_2,\ldots \in \mathcal{F}$ which is ${independent}$ such that $\textbf{P}(A_i)=\frac{1}{2}$ for each $i$. Hint: First prove that for each $\omega\in \Omega$, and for each $n\in \mathbb{N}$ we have $P(\{\omega\})\leq \frac{1}{2^n}$. Then derive a contradiction.   My Work: Let $\omega \in \Omega$ be arbitrary. Then since $\textbf{P}(\Omega)=\textbf{P}(\bigcup \{\omega\})=1$, and $(\Omega,\mathcal{F},\textbf{P})$ is a valid probability triple, then $\textbf{P}$ is countably additive, so that $\textbf{P}(\bigcup \{\omega\})=\sum\limits_{n=1}^\infty \textbf{P}(A_n)$. Here I am not sure where to go.  My problem: I am really not seeing how to go about this problem. Any help is appreciated.