I just had a chat with my real analysis teacher today, we talked about how there is no $\sigma$-algebra such that it has countably many elements, but finitely many.   In the end, he said that I can visualize it by drawing out the $\sigma$-algebra on (0,1) using intervals. But I'm not too sure how to do that, should I draw it out on the real line? Or can I use any type of diagram?
29771
I am going over a tutorial in my real analysis course. There is an proof in which I don't understand some parts of it. The proof relates to the following proposition: ($S$ - infinite $\sigma$-algebra on $X$) $\implies $ $S$ is uncountable. Proof:  Assume: $S=\{A_{i}\}_{i=1}^{+\infty}$. $\forall x\in X: B_{x}:=\cap_{x\in A_{i}}A_{i}$. [Note: $B_{x}\in S$ $\impliedby$ ($B_{x}$ - countable intersection]. Lemma: $B_{x}\cap B_{y}\neq\emptyset \implies B_{x}=B_{y}$. Proof(of lemma): $z\in B_{x}\cap B_{y} \implies B_{z}\subseteq B_{x}\cap B_{y}$. 1.$x\not\in B_{z} \implies x\in B_{x}\setminus B_{z} \wedge B_{x}\setminus B_{z} \subset S \wedge  B_{x}\setminus B_{z} \subset B_{x}$ (contradiction:$\space$ definition of $B_{x}$) $\implies$ $B_{z}=B_{x}$ 2.$y\not\in B_{z} \implies y\in B_{y} \setminus B_{z} \space \wedge \space B_{y}\setminus B_{z} \subset S \space\wedge\space B_{y} \setminus B_{z}\subset B_{y} $(contradiction: definition of $B_{y}$) $\implies$ $B_{z}=B_{y}$ $\implies B_{x}=B_{y} \space \square$ Consider: $\{B_{x}\}_{x\in X}$. If: there are finite sets of the form $B_{x}$ then: $S$ is a union of a finite number of disjoint sets $\implies$ $S$ is finite $\implies$ there is an infinite number of sets of the form $B_{x}$. $\implies$ $|\bigcup\limits_{i\in A \subseteq\mathbb{N}}B_{x_{i}}| \geq \aleph_{0}$.(contradiction) $\square$  There are couple of things I don't understand in this proof:  Why the fact that we found a set ($B_{x}\setminus B_{z}$) in $S$ containing $x$ and is strictly contained in $B_{x}$ a contradiction ?  Why if there are only a finite number of different sets of the form $B_{x}$ then $S$ is a union of a finite number of disjoint sets and is finite ?
21649
This is a question from the book Methods of Real Analysis by R. R. Goldberg.     If $(s_n)$ is a sequence of real numbers and if $$\sigma_n=\frac{s_1+s_2+\cdots+s_n}{n}$$ then prove that: $\operatorname{{lim sup}}\sigma_n \leq \operatorname{lim sup} s_n$.   I don't have any idea how to start working on this problem. Please help. Thanks.
80822
Let $S=\{\{x\}\mid x \in \mathbb{R}\}$. Is $\sigma(S)$ included in the Borel $\sigma$-algebra on $\mathbb{R}$? Is $\sigma(S)$ equivalent to the Borel $\sigma$-algbrea on $\mathbb{R}$?    I think the answer to the first question is yes. Because the Borel sets include singletons, the Borel $\sigma$-algebra must contain the smallest sigma-algebra of the set of all singletons.   Here's my attempt at the second question: my answer would be no. Countable unions (or intersections) of countable sets are countable, so that any $X \in \sigma(S)$ is either countable or a compliment of a countable set. Therefore, the interval $(0,1) \not \in \sigma (S)$ but it is included in the Borel $\sigma$-algebra. My issue here is that I don't know how to formalize this statement, if it is even correct.
68966
Second countability by itself doesn't restrict the cardinality of a topological space, since every set with the trivial topology is a second countable space, but it seems natural to ask whether second countability restricts the cardinality of the Borel $\sigma$-algebra of the space.  Can the cardinality of the Borel $\sigma$-algebra of a second countable space be arbitrarily big? If this is the case is there a simple construction for a second countable space with an arbitrarily big Borel $\sigma$-algebra?
95909
I got this problem from Shiryaev's book "Probability-1". I tried to use technique involving $\pi - \lambda$ system but I failed.  Let $\mathcal{F}_1, \mathcal{F}_2$ be two $\sigma$-algebras of subsets of the set $\Omega$. I want to prove  $$d(\mathcal{F}_1,\mathcal{F}_2)=4 \sup_{A_1 \in \mathcal{F}_1,A_2\in \mathcal{F}_2} |P(A_1 \cap A_2)-P(A_1)P(A_2)|$$  satisfies inequality $0\leq d(\mathcal{F}_1,\mathcal{F}_2) \leq 1$.
35334
I am a bit confused about the cardinality of the Vitali sets.  Just a quick background on what I gather about their construction so far:  We divide the real interval $[0,1]$ into an uncountable number of disjoint classes in such a way that two numbers $x$ and $y$ are in the same class if their difference x-y is a rational number. Therefore the equivalence relation on the real numbers would be: $x\sim y\iff x-y\in\mathbb Q$.   The cardinality of our classes is equivalent to the cardinality of $\mathbb Q$, so the classes have countably infinite elements.  How many of these classes are there? As each only has countably infinite elements, but each element of $[0,1]$ lies in one class, there are uncountably infinite classes.  Now, using the axiom of choice, we take one element out of each class and put it in the set V. This is a Vitali set.  Hopefully this is all good at this point.  So each Vitali set should have uncountably many elements (because we have uncountably many classes), but there should be countably many Vitali sets (because each class has countably many elements) correct?  But then why does  state that "There are uncountably many Vitali sets..."?
57325
Let $\Sigma = \{0,1,*\}$ be a finite alphabet and let $\mathcal{L}\subseteq \Sigma^*$ be a formal language defined as follows  $\{0,1\}\subseteq \mathcal{L}$. If $w_1,\dots,w_n\in\mathcal{L}$ then $*w_1w_2\dots w_n*\in\mathcal{L}$ Nothing more is a an element of $\mathcal{L}$.  I want to prove that  If $*w_1w_2\dots w_n*=*v_1v_2\dots v_m*$ where $w_i, v_j \in \mathcal{L}$ then $n=m$ and $w_i=v_i$ for all $i$.  I was trying to use the fact that every element of $\mathcal{L}$ has an even number of $*$ but maybe I am missing something.
83667
I found a interesting problem that says: Let $(X,S)$ be a measurable space where $X=\mathbb R$ and $S$ is the countable co-countable $\sigma$-algebra in $\mathbb R$, i.e. $S=\{A\subset\mathbb R: A\ \lor\ \mathbb R-A\ \ is\ finite\ or\ countable\}$. So the point is to describe the S-measurable functions $\ f:X\to \Bbb R^*$, where $\Bbb R^*$ are the extended real numbers.
21211
Let $\mathcal A$ be a collection of pairwise disjoint subsets of a $\sigma$-finite measure space, and suppose each set in $\mathcal A$ has strictly positive measure. Show that $\mathcal A$ is at most countable.  I have no clue about this one. Any hints would be appreciated.
83655
Let $(X, \mathcal{J})$ and $(Y, \mathcal{F})$ be two measure spaces. Let us assume that $J$ is a collection of subsets of $X$ which generates $\mathcal{J}$, i.e. $\sigma(J)=\mathcal{J}$. Similarly, assume $\sigma(F)=\mathcal{F}$. Is it always true that $$ \sigma(J)\times\sigma(F)=\sigma(J\times F)? $$ Here, $J\times F$ is the set of all cartesian products of sets in $J$ with sets in $F$.   Please note: in the above expression the two product signs mean different things. On the left I am considering the product sigma algebra $\mathcal{J}\times \mathcal{F}$, while on the right I am considering the $\sigma$-algebra generated by the cartesian products of my ``elementary sets.''
88541
Let $G$ be the Cantor set. It is well known that:   $G$ is perfect and hence closed. $G$ has the cardinality of the continuum. $G$ has measure zero. For any set $S \subset \mathbb{R}$ (I will not keep writing that we are in $\mathbb{R}$) we have $S \text{ is closed} \Leftrightarrow S^c \text{ is open}$. Any open set $O$ can be written as a --- in fact unique --- countable union of disjoint open intervals.   $G^c$ can thus be written as a countable union of disjoint open intervals. We now imagine this union as being superposed on the real line graphically as follows:  R: &lt;&lt;&lt;----(....)---(..)--(.)---------(...)---&gt;&gt;&gt;   where the (...) represents the open disjoint intervals (of differing size) composing $G^c$, and the ---  represents the remaining non-covered real numbers (that is those in $G$). Now we can cover $\mathbb{R}$ in its entirety by "collecting" the --- into disjoint closed intervals. Any one of these closed intervals might of course consist of only a single element. We now obtain:  R: &lt;&lt;&lt;[--](....)[-](..)[](.)[-------](...)[-]&gt;&gt;&gt;   Take the union of these disjoint closed intervals. This must be $(G^c)^c = G$. Now it is not hard to imagine a mapping from the (...)'s to the [...]'s. Just take the next (...) in line for each [...] (and do some trivial fixing at the ends). Therefore we have written $G$ as a countable union of disjoint closed sets. However, $G$ has measure zero and therefore cannot contain any closed sets other than the single element type. Hence $G$ is countable. Contradiction.  Where do I go wrong?
29826
For some collection of sets $A$, let $\sigma(A)$ denote the $\sigma$-algebra generated by $A$.  Let $C$ be some collection of subsets of a set $Y$, and let $f$ be a function from some set $X$ to $Y$. I want to prove:  $$f^{-1}(\sigma(C))=\sigma(f^{-1}(C))$$  I could prove that  $$\sigma(f^{-1}(C)) \subset f^{-1}(\sigma(C))$$  since complements and unions are 'preserved' by function inverse. But how do I go the other way?   EDIT: One way to go the other way would be to argue that any set in $\sigma(C)$ must be built by repeatedly applying the complement, union and intersection operations to elements of $C$ and all these operations are preserved when taking the inverse. The problem I am facing with the approach is formalizing the word "repeatedly".  [not-homework]
65000
I'm really struggling to grasp why $\sigma$-fields are not necessarily closed under finite union.  If the above statement is true then;  Let $(\Omega, \mathcal{F},P) $ be our probability space such that $(A,B \in \mathcal{F} \Rightarrow A \cup B \not\in \mathcal{F})$.  Surely $P(A \cup B) $ then cannot be evaluated.  Any advice on how I could move forward would be greatly appreciated.
89249
Suppose I have a sigma algebra which is countably generated. I want to find a random variable such that it is generated by that random variable. If the countable class is just a single set then my random variable is indicator function. But, how to do in this case ? If we sum up all the indicators to define our random variable then the sum may diverge. How to make sure it converge ?
49287
If $\mathcal{F}_1 \subset \mathcal{F}_2 \subset \dotsb$ are sigma algebras, what is wrong with claiming that $\cup_i\mathcal{F}_i$ is a sigma algebra?  It seems closed under complement since for all $x$ in the union, $x$ has to belong to some $\mathcal{F}_i$, and so must its complement.  It seems closed under countable union, since for any countable unions of $x_i$ within it, each of the $x_i$ must be in some $\mathcal{F}_j$, and so we can stop the sequence at any point and take the highest $j$ and we know that all the $x_i$'s up to that point are in $\mathcal{F}_j$, and thus so must be their union. There must be some counterexample, but I don't see it.
68767
Let $(a_n)_{n≥0}$ and $(b_n)_{n≥0}$ be two sequences of a nomed algebra such that $\sum{\| a_n\|}$ and $\sum{\| b_n\|}$ converge, and$ $$$\forall n, \  a_n, b_n \neq 0$$  Show that $(\forall  k \in \mathbb N^*,  \ \sum_{n=0}^{\infty}{a_n}^k = \sum_{n=0}^{\infty}{b_n}^k) \implies \exists\ \sigma \in{\mathfrak S}_{\mathbb N} \ ,\ (a_n) = (b_{\sigma(n)})$  The vector-space should be a finite-dimensional $\mathbb R$-vector-space.   I do not see how to deal with the hypothesis. If you have ideas/hints...  It would be relevant to deal with complex sequences even if it is less general. But I don't think one can generalize using real and complex cases.
95444
I'm stuck on the following problem (Source: Real Analysis for Graduate Students; Exercise 2.6; ):     Suppose $\mathcal A$ is a $\sigma$-algebra with the property that whenever $A \in \mathcal A$, there exists $B, C \in \mathcal A$ with $B \cap C = \emptyset$, $B \cup C = A$, and neither $B$ nor $C$ is empty. Prove that $\mathcal A$ is uncountable.   I think there is the added assumption that this is only true for $A \in \mathcal A$ having at least two elements, so it doesn't hold for singletons and the emptyset?  I tried to show this by way of contradiction and say that $\mathcal A = \{A_k\}$ is countable, but I didn't see how this could get me to a contradiction.  The other approach I tried was to look at $X = B_1 \cup C_1$ where $B_1 \cap C_1 = \emptyset$, then look at $B_1 = B_2 \cup C_2$ where $B_2 \cap C_2 = \emptyset$ and so on. If I look at the $C_k$'s they are pairwise disjoint and I either have a finite number of them or an infinite number of them in which case I've created a countable sequence of pairwise disjoint nonempty elements of $\mathcal A$. This path looked promising, but I couldn't see what to do next. Any ideas?  As a secondary question: I think this result is supposed to be used to show that if $\mathcal A$ is a $\sigma$-algebra with infinitely many elements, then it's uncountable, but I wasn't able to show that the property mentioned above (the one I'm trying to show) was satisfied in this case.