So I've been going in circles over the last few hours of this seemingly simple problem:  "Let $(X,F,\mu)$ be a probability space. Let $A,B,C$ be sets contained in $F$, with  $\mu(A \cap B) \ge \frac{1}{2}$, $\mu(A \cap C) \ge \frac{1}{2}$, $\mu(B \cap C) \ge \frac{1}{2}$.  Show that $\frac{1}{4} \le \mu(A \cap B \cap C) \le \min\{\mu(A \cap B), \mu(A \cap C), \mu(B \cap C)\}$".  Honestly, I'm plain simply stuck. I've been trying to see if I could get something useful out of De Morgan's identities as well as the distributive laws, but I seem to either be going in circles or end up with something not very useful due to unions I can not identify or the lack of information about $A$, $B$ and $C$. Same thing with the strongly additive property of a measure, which is only stated for the intersection of two sets in my book.  I feel like this should be simple, but I'm staring myself blind at this problem. Can anyone offer some guidance?
122038
Let $\{X,\mathcal{A},\mu\}$ be a probability space. Let $A$, $B$, $C$ be sets in $\mathcal{A}$, which satisfy $$\mu (A\cap B)\geq \frac{1}{2}, \quad \mu (A\cap C)\geq \frac{1}{2}, \quad \mu (B\cap C)\geq \frac{1}{2}.$$ Show that $\frac{1}{4}\leq \mu(A\cap B \cap C)$.  I've managed to show in a previous assignment that $$\mu(A\cap B \cap C)\leq \min\{\mu (A\cap B), \mu (A\cap C), \mu (B\cap C)\},$$ but now I'm stuck with the above inequality. I've tried to draw it, but still no luck. Any hints?
85448
What I've done so far is stated that    $(A \cup A) \cap (A \cup B)$  by distribution $A\cap (A\cup B)$        by 1, definition of $\cup$ $A\cap A$              by 2, definition of $\cup$ $A$                   by 3, definition of $\cap$   I'm not sure if I need to state that $A\cup B = \{x\mid x\in A\lor x\in B\}$ somewhere in there.  I'd really appreciate some help on this one. Thank you!
45022
Question: Let $E_1,E_2,...,E_n$ be measurable subsets of $[0,1]$ with $$ \sum_{k=1}^n m(E_k)  &gt; n-1.$$ Prove that $\bigcap_{k=1}^n E_k$ has positive measure.  This is one of the questions in graduate analysis past year paper. I think we need to assume that the intersection $\bigcap_{k=1}^n E_k$ is nonempty. Otherwise, the question is false. Anyway, let's assume that $\bigcap_{k=1}^n E_k \neq \emptyset.$ If $n=2,$ by inclusion-exclusion principle and assumption, $$m(E_1\cup E_2) + m(E_1 \cap E_2) = m(E_1) + m(E_2) &gt; 2.$$ Since $E_1,E_2 \subseteq [0,1],$ by monotonicity of Lebesgue measure, $$m(E_1) \leq 1, m(E_2) \leq 1.$$ By finite subadditivity of Lebesgue measure, $$m(E_1 \cup E_2) \leq m(E_1) + m(E_2) \leq 2.$$ Hence, $$m(E_1 \cap E_2) &gt; 2 - m(E_1 \cup E_2) \geq 0$$ Therefore, the intersection $\bigcap_{k=1}^n E_k$ has positive measure for $n=2.$ I try prove the general $n$ using induction, but it seems a bit long. Does there exist an efficient method to solve the question? EDIT: Actually I am looking for a direct proof instead of indirect proof or inductive proof. If a direct proof is not possible, then I will accept Marios's answer.
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!
36112
I am trying to show that for $A$, $B \; \subset \mathbb{R}$ $A \cup B$ is Lebesgue measurable such that \begin{align} \infty &gt; \mu(A \cup B) = \mu^{*}(A) + \mu^{*}(B), \end{align} Then $A$ and $B$ are Lebesgue measurable.  Thus far, I have tried reversing the directions in the proof of Caratheodory.  I seem to want to do the following:  If $A \cup B$ is Lebesgue measurable, then, for any $E \subset \mathbb{R}$, \begin{align} \mu^{*}(E) \geq \mu^{*}(E \cap (A \cup B)) + \mu^{*}(E \cap (A \cup B)^{c}). \end{align}  I want to use this and work backwards from here adding and subtracting sets to get  \begin{align} \mu^{*}(E) \geq \mu^{*}(E \cap A) + \mu^{*}(E \cap A^{c}). \end{align} yet I am having trouble.
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.''
21499
Statement: Let $A$, $B$, and $C$ be sets. Then $(A \setminus B) \cap (A \setminus C) = A \setminus (B \cup C)$.   Proof: Let $(x,y) \in ( A \setminus B) \cap (A \setminus C)$, where $ x \in A$ and $y \in (B \cap C$ by definition of intersection. So, $(x,y) \in (A \setminus B \cap C)$ which show that $(A \setminus C) \cap (A \setminus C) \subseteq A \setminus (A \cup C)$. \ Now let $(x,y) \in A \setminus(B \cup C)$, where $ x \in A $. So, $ (x,y) \in (A \setminus B) \cap (A \setminus C)$ by definition of intersection, which shows that $A \setminus (B \cup C) \subseteq (A \setminus B) \cap (A \setminus C)$. Therefore, by the Axiom Set of Equality we know that $(A \setminus B) \cap (A \setminus C) = A \setminus (B \cup C)$ .   Am I on the right track to proving that this is a true statement or is there a better way to show this?
66387
In some cases, we will have to consider the union or the intersection of several, even infinitely many sets, defined in the obvious way. For example, if for every positive integer $n$, we are given a set $S_n$, then   $$\bbox[border:1px solid red]{\bigcup_{n=1}^\infty S_n}=S_1\cup S_2\cup\cdots=\{x\mid x\in S_n\text{ for some }n\},$$   and   $$\bigcap_{n=1}^\infty S_n=S_1\cap S_2\cap\cdots=\{x\mid x\in S_n\text{ for all }n\}.$$   Two sets are said to be disjoint if their intersection is empty. More generally, several sets are said to be disjoint if no two of them have a common element. A collection of sets is said to be a partition of a set $S$ if the sets in the collection are disjoint and their union is $S$.   Normally what I know is that you can make a union or an intersection between only two sets. In this expression, there is a big union of sets. I'm asking about the meaning of this expression – what does it mean? What does the infinity sign do at the top?  Things are even more complicated with De Morgan's laws, which use the same expression:     Two particularly useful properties are given by De Morgan's laws which state that   $$\left(\bigcup_nS_n\right)^c=\bigcap_nS_n^c,\quad\quad\quad\quad\left(\bigcap_nS_n\right)^c=\bigcup_nS_n^c.$$   Anyone who can explain to me the expression or De Morgan's laws would be much appreciated.
25047
The problem statement, all variables and given/known data: Show that if $X$ is a subset of $M$ and $(M,d)$ is separable, then $(X,d)$ is separable. [This may be a little bit trickier than it looks - $E$ may be a countable dense subset of $M$ with $X\cap E = \varnothing$.]  Definitions Per our book:  A metric space $(M,d)$ is separable if there exists a countable dense $E$ contained in $M$.   $E$ contained in $M$ is dense if $\forall m\in M$, $\forall ε&gt;0\in\mathbb R$, $\exists e\in E$ s.t. $d(m,e) &lt; ε$   The attempt at a solution  My best attempt was doomed from the start, because I don't quite understand the hint. My thought process went as follows:  Since $X$ is a subset of $M$, $\forall x\in X, x\in M$. Thus, since $E$ is dense in $M$,  $\forall x\in X$, and ε > 0, $\exists e\in E$ st $d(x,e)&lt;ε $.   At this point, I was done, because the set of $e$'s satisfying the above, is a subset of $E$, a countable set. So a subset of a countable set is dense in $X$, and $X$ is separable. This is incorrect, but I cannot see why.  Any help clearing up the confusion would be greatly appreciated. Thanks!  Edit: I wish I could upvote all of you for your help! I really appreciate the speedy replies and attempts to make this information clear to me.
79302
Given a sequence $(\mu_n)_{n\in \mathbb N}$ of finite measures on the measurable space $(\Omega, \mathcal A)$ such that for every $A \in \mathcal A$ the limit   $$\mu(A) = \lim_{n\to \infty} \mu_n(A)$$   exists. I want to show that $\mu$ is a measure on $\mathcal A$.   What I managed to figure out:   $\mu$ is monotone, additive and - if $\lim_n \mu_n(\Omega)$ is supposed to be taken in $\mathbb R$ - then $\mu$ is also finite (I'll assume this). Also $\mu(\varnothing) = 0$. So we can wlog assume that $\mu(\Omega) = 1$ and that $\mu_n(\Omega) \le 2$ for all $n$.   Now what remains to be shown is that $\mu$ is $\sigma$-additive, or equivalently that for $A_n\downarrow \varnothing$, we have $\mu(A_n) \to 0$ (since $\mu$ is finite).  All attempts at proving this have been futile so far. I don't seem to see the right approach.  If possible, I would like to only receive a hint rather than a full answer. But of course, I'd be quite happy with a full answer, too; if a good hint is hard to find!  Thanks a lot in advance for your help! =)
59173
Let $f\colon\mathbb R \to \mathbb R$ be convex and $X_1,\ldots,X_n$ be i.i.d. random variables (e.g., uniform). Furthermore, suppose $g_1,\ldots,g_n\colon\mathbb R\to\mathbb R$ are increasing. Is it true that $$\mathbb E f(g_1(X_1) + g_2(X_1) + \cdots + g_n(X_1)) \ge \mathbb E f(g_1(X_1) + g_2(X_2) + \cdots + g_n(X_n))? $$ Like the previous version of this question (), which is the special case where the $g_i$ are all the identity, it seems like one approach is to show that $g_1(X_1) + \cdots + g_n(X_1)$ is a mean-preserving spread of $g_1(X_1) + \cdots + g_n(X_n)$. As before, I am also curious if there is a name for this sort of result/inequality.
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.
94631
I am not sure I am using the standard definitions so I will open by defining what I need:        Let $X$ be a set, $\nu:\, P(X)\to[0,\infty]$ will be called an external measure if $\nu(\emptyset)=0$ and for any   $\{A_{i}\}_{i=1}^{\infty}\subseteq P(x)$ (not neccaseraly disjoint) it   holds that   $\nu(\cup_{i=1}^{\infty}A_{i})\leq\sum_{i=1}^{\infty}\nu(A_{i})$   Let $\nu$ be an external measure on a set $X$ then we say that a set $A$ is $\nu$ measurable if for any $E\subseteq X$: $\nu(E)=\nu(E\cap A)+\nu(E\cap A^{c})$      The exercise asks to prove that the set of $\nu$ measurable sets $M\subseteq P(X)$ is an algebra.  I have proved $\emptyset,X\in M$ and that $A\in M\implies A^{c}\in M$ but I am having problems proving closer under union and intersection.  I assume that $A_{1},A_{2}$ are $\nu$ measurable so I get that for any $E$: $$\nu(E)=\nu(E\cap A_{1})+\nu(E\cap A_{1}^{c})$$  $$\nu(E)=\nu(E\cap A_{2})+\nu(E\cap A_{2}^{c})$$  And I need to prove that for any $E'$: $$\nu(E')=\nu(E'\cap A_{1}\cap A_{2})+\nu(E'\cap(A_{1}\cap A_{2})^{c})$$  which is the same as $$\nu(E')=\nu(E'\cap A_{1}\cap A_{2})+\nu((E'\cap A_{1}^{c})\cup(E'\cap A_{2}^{c}))$$  and a similar result to prove closer under union.  I guess that it all have to do with choosing the right $E$'s from knowing that $A_{i}$ are $\nu$ measurable, but I tried different options for an hour now and I don't see this going anywhere.  I need some help in showing closer under union and intersection
79825
I'm taking my first course in Probability and one of my homework problems is to prove that for any two sets:  $$P(A \cup B) = P(A) + P(B) - P(A \cap B) $$  Note that the function $P$ is the probability of an event happening.  From the third axiom, we are given that for any two disjoint sets, that is $ A \cap B = \emptyset$:  $$P(A \cup B) = P(A) + P(B)$$    My thought process follows from this third axiom, but I run into a problem near the end:  Since $A$ and $B$ are disjoint sets, $P(A)=P(A-B)$ and $P(B)=P(B-A)$, so  $$P(A \cup B) = P(A) + P(B)=P(A-B)+P(B-A)$$  From this, it appears that we are essentially taking the symmetric difference of A and B.  This, to me, seems to be equivalent to:  $$P(A-B)+P(B-A)=P(A)+P(B)-P(A \cap B)$$  Hence, by my logic:  $$P(A \cup B) = P(A)+P(B)-P(A \cap B)$$  Which, in essence, doesn't care whether two sets are disjoint or not as we are subtracting the intersection of them.  If the sets are disjoint, then the intersection is merely null.  If they intersect, then we are removing the "extra pieces".  I feel like this is inadequate proof despite my ability to convince myself that it is true..  Where, if anywhere, did I go wrong?  Thanks!
100865
Good day,  I have the problem of an old exam without solutions. It is the following:    Let $X_1,X_2,...$ be i.i.d. (independent and identically distributed) random variables on the same probability space such stat $$P\left(X_1=-\frac{1}{4}\right)=P\left(X_1=\frac{1}{4}\right)=\frac{1}{2}$$ For all $n\in \mathbb{N}$ define $M_n:=\prod_{k=1}^n (1+X_k)$.   Show that $(M_n)_{n\in\mathbb{N}} $ is a martingale w.r.t. a suitably chosen filtation (This was no problem for me) Use the law of large numbers to show that $(M_n)_{n\in\mathbb{N}}$ converges almost surely and determine the limit. (Here is the problem for me.)     This is the strong law of large numbers (SLLN) we had in the lecture:  Let $X_i$ be uncorrelated random variables, $E(X_i)=\mu, E(X_i^2)&lt;\infty$, $\exists C$ such that $\text{Var}(X_i)\leq C$ for all $i$ and define $S_n=\sum_{i=1}^n X_i$. Then $$\frac{S_n}{n} \to \mu \text{ almost surely.}$$    Therefore I have to get this $M_n$ into the form of $\frac{S_n}{n}$. I had a problems in the class where I had $$ A_n:=\left( \prod_{i=1}^n X_i \right)^{1/n}.$$ This was pretty elegant since we get the wished form via $$\log A_n = \frac{1}{n} \sum_{i=1}^n \log(X_i)=:\frac{1}{n} \sum_{i=1}^n Y_i.$$  I wanted to use this here:   $$\log M_n = \sum_{k=1}^n \log(1+X_k) = \frac{1}{n} \sum_{k} \log(1+X_k)^n $$  But now this "new" random variable $Y_{n,k}:=\log(1+X_k)^n$ is dependent of $n$. I don't know if this is a problem (now I think this is a huge problem, see below.) Let's try and use the SLLN for this: We would get $\log M_n \to E(Y_{n,k})$ if the assumptions are fulfilled but the limit should not be dependent on $n$. $$E(Y_{n,k})=E(\log(1+X_k)^n)=\frac{1}{2} \left( \log(5/4)^n +\log(3/4)^n \right) = \frac{1}{2} \log (15/16)^n \to -\infty$$  I don't know what to do. I think I could use the Martingale Convergence Theorem to get that $M_n$ converges a.s. but via this method I don't get the limit. Further the exam states that I have to use the SLLN explicitly.  I am thankful for every help/hint.
60246
From Rudin book  $\bf 2.17\ $ Theorem $\ $ Suppose $X$ is a locally compact, $\sigma$-compact Hausdorff space. If $\frak M$ and $\mu$ are as described in the statement of Theorem $\it 2.14$, then $\frak M$ and $\mu$ have the following properties: $(a)\ \ $ If $E\in\frak M$ and $\epsilon&gt;0$, there is a closed set $F$ and an open set $V$ such that $F\subset E\subset V$ and $\mu(V-F)&lt;\epsilon$. $(b)\ \ $ $\mu$ is a regular Borel measure on $X$. $(c)\ \ $ If $E\in\frak M$, there are sets $A$ and $B$ such that $A$ is an $F_\sigma$, $B$ is a $G_\delta$, $A\subset E\subset B$, and $\mu(B-A)=0.$  My doubt: In $(c)$ it is written that  that $A$ is an $F_\sigma$ and $B$ is a $G_{\delta}$ My confusion is about symbolic meaning what  is the  meaning of $F_\sigma$ and $G_\delta?$
2105
Let $\mu$ a probabibility measure in $\mathbb R$ with $\mu\{x\}=0\quad \forall x\in\mathbb R$. Show that for all $\alpha \in [0,1]$ there exists $B_\alpha \in\mathcal B(\mathbb R)$ with $\alpha=\mu (B_\alpha)$.  My thoughts: We have to show that the distribution function $F(y)=\mu(-\infty,y)$ is continous. Then we can apply the intermediate value theorem.  I am not sure how to show that this measure is continous. Some help is welcome!