Let $X\in L^2(\Omega, \mathcal{A}, P)$ and let $\mathcal{G}\subset\mathcal{A}$ be a sub $\sigma$-algebra. Assume that $X\stackrel{d}{=}E[X|\mathcal{G}].$ Prove that $X=E[X|\mathcal{G}]$ a.s.  I have posted and solved a similar question  but I cannot figure out how to make the necessary changes from the assumption that $E[X|\mathcal{G}]\le X$ a.s. to the assumption that $X\stackrel{d}{=}E[X|\mathcal{G}]$ in order to alter this proof to make it work. Any help with that or a full blown new proof would be greatly appreciated.
122536
Suppose $(\Omega,\mathcal F,P)$ is a probability space and that $\mathcal G$ is a sub-sigma-algebra of $\mathcal F$. If $X$ is an integrable, non-negative random variable with the same distribution as $E[X\mid \mathcal G]$, how does one show that $X=E[X\mid \mathcal G]$ a.s.?
69609
A generalized version of the second Borel-Cantelli lemma says     Theorem 5.3.2. Second Borel-Cantelli lemma, II. Let $\mathcal F_n, n \ge 0$ be a filtration   with $F_0 = \{\emptyset, \Omega\}$ and $A_n , n \ge 1$ a sequence of events with $A_n ∈ \mathcal F_n$ . Then   $$ \{A_n \,i.o.\} = \left\{\sum_{n \ge 1} P (A_n |\mathcal F_{n−1}) =\infty \right\}. $$   An exercise of this lemma is     Exercise 5.3.6. Show    $\sum_{n \ge 2} \mathbb P (A_n | \cap_{m=1}^{n−1} A_m^c ) = \infty$ implies $P (∩_{m \ge 1} A_m^c ) = 0$.   I think we can actually prove that $\cap_{m \ge 1} A_m^c = \emptyset$. My proof is like this: Let $B_n = \cup_{m = 1}^n A_n$. Let $\mathcal F_n = \sigma(A_1,\dots,A_n)$, which forms a filtration. It is easy to verify that $$ \mathbb P (A_n | B_{n-1}^c ) = \mathbb P (A_n | \mathcal F_{n-1})(\omega) $$ when $\omega \in B_{n-1}^c$.  Assume there exist $\omega \in \cap_{m \ge 1} A_m^c$. It follows from the previous lemma that $$ \omega \in \left\{\sum_{n \ge 1} P (A_n |\mathcal F_{n−1}) =\infty \right\} = \{A_n \,i.o.\}. $$ In other words, $\omega \in \cap_{m \ge 1} \cup_{n \ge m} A_n$, which contradicts $\omega \in \cap_{m \ge 1} A_m^c$.  Is this the right proof?
52776
It would be great if someone could verify this proof.  Theorem:  $\mathcal{P}(A \cap B) = \mathcal{P}(A) \cap \mathcal{P}(B)$  Proof:  First I prove that $\mathcal{P}(A \cap B) \subseteq \mathcal{P}(A) \cap \mathcal{P}(B)$. Take any subset $X \subseteq A \cap B$. Then $X \in \mathcal{P}(A \cap B)$. Also, $X \subseteq A \wedge X \subseteq B$, meaning that $X \in \mathcal{P}(A) \wedge X \in \mathcal{P}(B)$. This, again, means that $X \in \mathcal{P}(A) \cap \mathcal{P}(B)$, and this proves the statement.  Then I prove that $\mathcal{P}(A) \cap \mathcal{P}(B) \subseteq \mathcal{P}(A \cap B)$. Again,  take any subset $X \subseteq A \cap B$. Using just the same arguments as above, this is also a truth.  The theorem follows.
73041
It is well-known that:     Given a set $X$ and a collection $\cal S$ of subsets of $X$, there exists a $\sigma$-algebra $\cal B$ containing $\cal S$, such that $\cal B$ is the smallest $\sigma$-algebra satisfying this condition.   Certain texts, Lieb and Loss, Analysis, for instance, state that the proof of this assertion requires transfinite induction. On the other hand, one can define $\mathcal B$ to be the intersection of all $\sigma$-algebras containing $\cal S$. Which statement is correct? Or, is there a hidden transfinite induction contained somewhere?  I must confess here that I have only vague ideaos of the rigorous set-theoretic foundations of mathematics.
56619
Does anyone know how to do this exercise? I couldn't do it.  Let $(X,S,\mu)$ be a $\sigma$ -finite measure space and $g\in M(X,S)$ such that $gs\in\mathcal{L}_1$ for any simple function $s\in\mathcal{L}_p$, $p\in(1,\infty)$. Suppose there is $A\geq0$ such that $\left|\int gs\,d\mu\right|\leq A||s||_p$ for all $s$. Show that $g\in\mathcal{L}_p$ with $\frac1p+\frac1q=1$ and $||g||_q\leq A$.
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 $B$.
109739
Question 2.3.20 in Qing Liu's Algebraic Geometry and Arithmetic Curves:  Let $A$ be a ring, $G$ a finite group of automorphisms of $A$. Let  $$    p: \mathrm{Spec}\ A \to \mathrm{Spec}\ A^{G} $$  be the morphism of affine schemes induced by the obvious inclusion.     Show that, for $x_{1}$, $x_{2} \in \mathrm{Spec}\ A$,    $$ p(x_{1}) = p(x_{2}) \iff \exists \ \sigma \in G : \sigma(x_{1}) = x_{2}.$$   The '$\impliedby$' direction is easy, but I'm having trouble with the other direction (although I have no doubt that it is also simple).   My attempts so far have included assuming that there is no $\sigma \in G$ such that $\sigma(x_{1}) = x_{2}$, so that for each $\sigma$ there is $g_{\sigma} \in x_{1}$ with $\sigma(g_{\sigma}) \notin x_{2}$, and then trying to construct an element of $p(x_{1})\backslash p(x_{2})$ out of these $g_{\sigma}$, for example by taking $$   g = \prod_{\sigma \in G}g_{\sigma} $$ and then  $$   h = \prod_{\sigma}\sigma(g) $$ but this appears to fail as I can't ensure that $\sigma(g_{\tau}) \notin p(x_{2})$ for $\sigma \neq \tau$.  Any hints would be much appreciated!
24105
Let $ A $ be a ring.  Let $ I $ be a preordered set, filtering.  Let $ \Sigma $ a multiplicative subset of $ A $.  Suppose for any given $ i \in I $ a multiplicative subset $ S_i $ of $ A $ contained in $ \Sigma $.  Make the following assumptions:   $ \Sigma = \displaystyle \bigcup_i S_i $ For $ (i, j) \in I^2 $ with $ i \leq j $, elements $ S_i $ become invertible in $ S_ {j}^{-1} A $.   $ \mathcal{D} = ((S_{i}^{-1} A)_{i \in I} \ , \ (S_{i}^{-1} A \to S_{j}^{-1} A)_{i \leq j}) $ is a commutative diagram in the category of filter $ A $ - algebras.  Question :  Show that $ \displaystyle \lim_{\longrightarrow} \mathcal{D} \simeq \Sigma^{-1} A $.  Thank you in advance for your help.
169
Let $f: X \rightarrow Y $ and $g:X \rightarrow Y$ continuous functions and $(X,d_X),(Y,d_Y)$ metric space. Let $E$ by a dense subset in $X$. Prove that $f(E)$ is a dense subset in $f(X)$. If $f(p)=g(p)$ for all $p\in E$, prove that $f(x)=g(x)$ for all $x\in X$.  My attempt: since $E \subset X \Rightarrow f(E)\subset f(X)$ and from the dense property we have $f(X) \subset f(\overline{E})$ and since $f$ is continuous, $f(X) \subset f(\overline{E}) \subset \overline{f(E)}$. Then, $f(E)$ is dense in $f(X)$.  Suppose that $f(p)=g(p), \;  \forall p \in E$. Hence, we have  $$d_{Y}(f(x),g(x)) \leq d_{Y}(f(x),f(p)) + d_{Y}(f(p),g(x)) = d_{Y}(f(x),f(p)) + d_{Y}(g(p),g(x)).$$ By the fact that $f,g$ are continuous, there exist a $\delta_1$ and $\delta_2$ such that $$d_X(x,p) &lt; \delta_1 \qquad d_X(x,p) &lt; \delta_2$$  implies $$d_{Y}(f(x),f(p))&lt;\epsilon /2 \qquad d_{Y}(g(p),g(x)) &lt; \epsilon /2.$$ for all $\epsilon &gt;0.$  Is this correct?
50264
This is a problem in KaiLai Chung's A Course in Probability Theory.     Given a nonnegative random variable $X$ defined on $\Omega$, if $\mathbb{E}(X^2)=1$ and $\mathbb{E}(X)\geq a &gt;0$, prove that $$\mathbb{P}(X\geq \lambda a)\geq (a-\lambda a)^2$$   for $0\leq\lambda \leq 1$.   Let $A=\{x\in \Omega:X(x)\geq \lambda a\}$, we get $$\int_A (X-\lambda a)\geq a-\int_A\lambda a -\int_{A^c}X$$ and $$\int_A (X^2-\lambda^2 a^2)=1-\int_A\lambda^2a^2-\int_{A^c}X^2$$ I want to contrast $\int_A (X-\lambda a)$ and $\int_A (X^2-\lambda^2 a^2)$, but I don't know how to do it, could anyone gives me some hints?
57123
I need help to demonstrate:  Let  $F,G$  two $\sigma$-algebra in $\Omega$, if $F\cup G$ is $\sigma$-algebra in $\Omega$, show $F\subset G$ or $G\subset F$.  Thanks
103441
Let $\Omega$ a set and     $$ C=\left\{A\subset\Omega: A~\text{ is countable or }A^C\text{ is countable}\right\}. $$     Show that $C$ is a $\sigma$-algebra and that     $$ \sigma(\left\{\left\{ x\right\}: x\in\Omega\right\})=C. $$      Hello!  First I have to show that $C$ is a $\sigma$-algebra. Here are my results; I have some problems to finish the proof.  (1) If $\Omega$ is countable, then $\Omega\in C$. If $\Omega$ is not countable, then $\Omega^C=\emptyset$ is. So in any case $\Omega\in C$.  (2) Consider $A\in C$. If $A$ is countable, then $A^C$ is in $C$, because $(A^C)^C=A$ is countable then. If $A$ is not countable, it has to be by definition $A^C$ countable (otherwise $A$ could not be in $C$), so $A^C$ in $C$.  (3) Consider $A_1,A_2,\ldots\in C$. First case: All $A_i, i=1,2,\ldots$ are countable. Then $\bigcup_{i\geq 1}A_i$ is countable as countable union of countable sets. So $\bigcup_{i\geq 1}A_i\in C$. Second case: All $A_i, i=1,2,\ldots$ are not countable. Then all $A_i^C, i=1,2,\ldots$ are countable and for any $i\in\left\{1,2,.\ldots\right\}$ it is $$ \left(\bigcup_{i\geq 1}A_i\right)^C=\bigcap_{i\geq 1}A_i^C\subset A_i^C $$ which means that $\bigcap_{i\geq 1}A_i^C$ is countable. So it follows that $\bigcup_{i\geq 1}A_i$ is in $C$, because its complement is countable.  Now what is with the third case, that some $A_i$ are countable and others not? Do not know how to handle it.  Overmore, I have problems to show $\supset$ of the identity that is to show. The other implikation $\subset$ follows, because $C$ is a $\sigma$-algebra (which I have not completely shown) and $\left\{\left\{ x\right\}: x\in\Omega\right\}\subset C$.    So would you please help me to finish point (3) (case 3) and the implication $\supset$?
99412
Consider a topological group $G$. I want to prove that      if $\cal{U}$ is the set of all neighborhoods of $e$ (the identity element), then $\overline{\{e\}}=\cap_{U\in \cal{U}}U$.   What I thought was:  i) If $x\in \overline{\{e\}}$, then every neighborhood $V$ of $x$ intersects $\{e\}$, hence every $V$ is a neighborhood of $e$ as well, i.e. $V\in \cal{U}$.  ii) On the other hand, if $x\in \cap_{U\in \cal{U}}U$, then $x\in U$ for all $U \in \cal{U}$, meaning every $U$ is a neighborhood of $x$ intersecting $\{e\}$.  Now, I think that the missing part is an argument justifying that there is no neighborhood of $e$ that is not a neighborhood of $x$ (part i)), and that there is no neighborhood of $x$ that is not a neighborhood of $e$ (part ii)). My idea for this is to argue that there is a homeomorphism taking $e$ to $x$, therefore the neighborhood system of one cannot have more elements that the neighborhood system of the other.  Is this correct? If not, is there a way I can fix it?
46835
How to prove that $E[Y]=E[E[E[Y|X_1, X_2]]]$ ?  PS. I don't see how $E[E(Y|X_{1},X_{2})|X_{1}]=Y[Y|X_{1}]$ and $E[Y]=E[E(Y|X_{1})]$ can be used here. But it feels close. Please help, I'm stuck PPS. Thanks @Deep North. Let's consider the case when X1 is independent of X2
31013
Let $\mathcal{G} = \{A_1, \ldots, A_n\}$ be a partition of a set $\Omega$, $\mathcal{F} = \sigma(\mathcal{G})$. Prove that $X : \Omega\to\mathbb{R}$ is a random variable if and only if it is constant on each of the partition elements $A_j$.  I have attempted both directions but faced some difficulties to continue writing the proof. I have also consulted some sources that have a not so clear proof.  Here is the proof: We know that $$\sigma(\mathcal{G}) = \left\{\sum_{i\in I}A_i : I \subset \{1,\ldots,n\}\right\}.$$ ($\Leftarrow$) Let $B\in \mathcal{B}(\mathbb{R})$. In this case $X(\omega) = a_i$, $\forall\omega\in A_i$ for some constants $a_i$, (then I am not sure how to proceed to show that it is a random variable), the proof says that $X = \sum_{i\in I}^n a_i \mathcal{{1_A}_i}$ and $\{X \in B\} = \sum_{i : a_i\in \mathcal{B}}A_i\in \mathcal{F}$, so $X$ is a random variable.  My question is why are we concluding or defining $X = \sum_{i\in I}^n a_i \mathcal{{1_A}_i}$ (what is the purpose), in other words I don't understand it's flow of reasoning.  ($\Rightarrow$) We use proof by contradiction: If $X$ is not constant on $A_i$, then $\exists i_0$ such that for some $\omega_1, \omega_2 \in A_{i_0}$ one has $a:= X(\omega_1) \not= X(\omega_2)$. Then I am not sure how to proceed from here.  Many thanks in advance.
12378
$\mathcal{C}_1$, $\mathcal{C}_2$ are collections of subsets of $X$,then Im having hard time seeing why the  following is true. Can someone explain them to me?  $\mathcal{C}_1 \subseteq \mathcal{C}_2 \implies \sigma( \mathcal{C}_1) \subseteq \sigma( \mathcal{C}_2) $  $\sigma(\sigma( \mathcal{C} )) = \sigma( \mathcal{C} ) $  where $\sigma( \mathcal{F} ) $ is the sigma field generated by the collection $\mathcal{F}$ of subsets of $X$  thanks
20849
My teacher gave me an example of performing the subject:     Example      Let $\Omega = \Bbb R$ and $\mathcal R = \{(-\infty,-1),(1,+\infty)\}$.   Then $\sigma(\mathcal R) = \{\emptyset, \Bbb R, (-\infty,-1),  (1,+\infty), [-1, \infty), (-\infty,1],  (-\infty,-1)\cup(1,+\infty),[-1,1]\}$.   There was a different example, where she also generated the smallest  $\sigma$-algebra for the family of sets $\mathcal A = \{A,B\} \subset 2^\Omega$ in the same way: $\sigma(\mathcal R) =  \{\emptyset, \Omega, A, B, A^C,B^C,A\cup B, (A\cup B)^C\}$.   I certainly understand why $\emptyset$, $\Omega $ and a family of sets itself are there, and I certainly know that $\sigma$-algebra is closed under the operations of complement and union. What I don't understand about the other elements of generated $\sigma$-algebras: why we are taking exactly them? Does this work in a general case?