hcju/proofwikiqa huggingface dataset:
{{begin-eqn}} {{eqn | l = \Sigma       | o = \ne       | r = \O       | c = {{Defof|Event Space|Event Space: Axiom $(\text {ES} 1)$}} }} {{eqn | ll= \leadsto       | lo= \exists A:        | l = A        | o = \in       | r = \Sigma       | c = {{Defof|Empty Set}} }} {{eqn | ll= \leadsto       | l = A \setminus A       | o = \in       | r = \Sigma       | c = {{Defof|Event Space|Event Space: Axiom $(\text {ES} 2)$}} }} {{eqn | ll= \leadsto       | l = \O       | o = \in       | r = \Sigma       | c = Set Difference with Self is Empty Set }} {{end-eqn}} {{qed}}
Let $\powerset \Omega := \Sigma$. ;Event Space Axiom $(\text {ES} 1)$: From Empty Set is Subset of All Sets we have that $\O \subseteq \Omega$. By the definition of power set: :$\O \in \Sigma$ thus fulfilling axiom $(\text {ES} 1)$. {{qed|lemma}} ;Event Space Axiom $(\text {ES} 2)$: Let $A \in \Sigma$. Then by the definition of power set: :$A \subseteq \Omega$ From Set with Relative Complement forms Partition: :$\Omega \setminus A \subseteq \Omega$ and so by the definition of power set: :$\Omega \setminus A \in \Sigma$ thus fulfilling axiom $(\text {ES} 2)$. {{qed|lemma}} ;Event Space Axiom $(\text {ES} 3)$: Let $\sequence {A_i}$ be a countably infinite sequence of sets in $\Sigma$. Then from Power Set is Closed under Countable Unions: :$\ds \bigcup_{i \mathop \in \N} A_i \in \Sigma$ thus fulfilling axiom $(\text {ES} 3)$. {{qed|lemma}} All the event space axioms are seen to be fulfilled by $\powerset \Omega$. Hence the result. {{qed}}
For $\powerset \Omega$ to be an event space of $\EE$, it needs to fulfil the following properties: :$(1): \quad \powerset \Omega \ne \O$, that is, an event space can not be empty. :$(2): \quad$ If $A \in \powerset \Omega$, then $\relcomp \Omega A \in \powerset \Omega$, that is, the complement of $A$ relative to $\Omega$, is also in $\powerset \Omega$. :$(3): \quad$ If $A_1, A_2, \ldots \in \powerset \Omega$, then $\ds \bigcup_{i \mathop = 1}^\infty A_i \in \powerset \Omega$, that is, the union of any countable collection of elements of $\powerset \Omega$ is also in $\powerset \Omega$. These all follow directly from Power Set is Sigma-Algebra. {{qed}}
;Event Space Axiom $(\text {ES} 1)$: From its definition: :$\Sigma \ne \O$ thus fulfilling axiom $(\text {ES} 1)$. {{qed|lemma}} ;Event Space Axiom $(\text {ES} 2)$: From Set Difference with Empty Set is Self: :$\Omega \setminus \O = \Omega \in \Sigma$ From Set Difference with Self is Empty Set: :$\Omega \setminus \Omega = \O \in \Sigma$ By definition: :$\Omega \setminus A \in \Sigma$ From Relative Complement of Relative Complement: :$\Omega \setminus \paren {\Omega \setminus A} = A \in \Sigma$ Thus axiom $(\text {ES} 2)$ is fulfilled. {{qed|lemma}} ;Event Space Axiom $(\text {ES} 3)$: From Union with Empty Set: :$\forall X \in \Sigma: X \cup \O = X \in \Sigma$ From Union with Superset is Superset: :$\forall X \in \Sigma: X \cup \Sigma = \Sigma \in \Sigma$ From Union is Idempotent: :$\forall X \in \Sigma: X \cup X = X \in \Sigma$ From Union with Relative Complement: :$A \cup \paren {\Omega \setminus A} = \Sigma \in \Sigma$ It follows that axiom $(\text {ES} 3)$ is fulfilled. {{qed}}
{{begin-eqn}} {{eqn | l = A, B       | o = \in       | r = \Sigma       | c= }} {{eqn | ll= \leadsto       | l = \Omega \setminus A, \ \Omega \setminus B       | o = \in       | r = \Sigma       | c = {{Defof|Event Space|Event Space: Axiom $(\text {ES} 2)$}} }} {{eqn | ll= \leadsto       | l = \paren {\Omega \setminus A} \cup \paren {\Omega \setminus A}       | o = \in       | r = \Sigma       | c = {{Defof|Event Space|Event Space: Axiom $(\text {ES} 3)$}} }} {{eqn | ll= \leadsto       | l = \Omega \setminus \paren {A \cap B}       | o = \in       | r = \Sigma       | c = De Morgan's Laws: Difference with Intersection }} {{eqn | ll= \leadsto       | l = \Omega \setminus \paren {\Omega \setminus \paren {A \cap B} }       | o = \in       | r = \Sigma       | c = {{Defof|Event Space|Event Space: Axiom $(\text {ES} 2)$}} }} {{eqn | ll= \leadsto       | l = A \cap B       | o = \in       | r = \Sigma       | c = Relative Complement of Relative Complement }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = A, B       | o = \in       | r = \Sigma       | c= }} {{eqn | ll= \leadsto       | l = A, \Omega \setminus B       | o = \in       | r = \Sigma       | c = {{Defof|Event Space|Event Space: Axiom $(\text {ES} 2)$}} }} {{eqn | ll= \leadsto       | l = A \cap \paren {\Omega \setminus B}       | o = \in       | r = \Sigma       | c = Intersection of Events is Event }} {{eqn | ll= \leadsto       | l = A \setminus B       | o = \in       | r = \Sigma       | c = Set Difference as Intersection with Relative Complement }} {{end-eqn}} {{qed}}
Let $\NN = \MM \circ \LL : \EE \to \GG$ be the composition. We want to show that for any $p, q \in \EE$ :$\map \GG Q = \map \GG p + \map {M \circ L} {\vec {p q} }$ We find that: {{begin-eqn}} {{eqn | l = \map \GG q       | r = \map {\MM \circ \LL} q }} {{eqn | r = \map \MM {\map \LL p} + \map L {\vec{p q} }       | c = $\LL$ is an Affine Transformation }} {{end-eqn}} Now let: :$p' = \map \LL p$ and: :$q' = \map \LL p + \map L {\vec {p q} }$ so: :$\vec {p' q'} = \map L {\vec {p q} }$ Then: {{begin-eqn}} {{eqn | l = \map \GG q       | r = \map \MM {q'} }} {{eqn | r = \map \MM {p'} + \map M {\vec {p' q'} }       | c = $\MM$ is an Affine Transformation }} {{eqn | r = \map \MM {\map \LL p} + \map M {\map L {\vec {p q} } }       | c = Definitions of $p'$ and $q'$ }} {{eqn | r = \map {\MM \circ \LL} p + \map {M \circ L} {\vec {p q} }       | c =  }} {{end-eqn}} as required. {{Qed}} Category:Affine Geometry k8rj2qi1mssvk7sdmw9vohrdxlizzmu
{{begin-eqn}} {{eqn | l = A, B       | o = \in       | r = \Sigma       | c= }} {{eqn | ll= \leadsto       | l = A \cup B       | o = \in       | r = \Sigma       | c = {{Defof|Event Space|Event Space: Axiom $(\text {ES} 3)$}} }} {{eqn | lo= \land       | l = A \cap B       | o = \in       | r = \Sigma       | c = Intersection of Events is Event }} {{eqn | ll= \leadsto       | l = \paren {A \cup B} \setminus \paren {A \cap B}       | o = \in       | r = \Sigma       | c = Set Difference of Events is Event }} {{eqn | ll= \leadsto       | l = A \ast B       | o = \in       | r = \Sigma       | c = {{Defof|Symmetric Difference|index = 2}} }} {{end-eqn}} {{qed}}
From the conditions for $\Pr$ to be a probability measure, we have: :$(1): \quad \forall A \in \Sigma: 0 \le \map \Pr A$ :$(2): \quad \map \Pr \Omega = 1$ :$(3): \quad \displaystyle \map \Pr {\bigcup_{i \mathop \ge 1} A_i} = \sum_{i \mathop \ge 1} \map \Pr {A_i}$ where all $A_i$ are pairwise disjoint. From the definition of event space, we have: :$\Omega \in \Sigma$ :$A \in \Sigma \implies \relcomp \Omega A \in \Sigma$ From Intersection with Empty Set: :$\O \cap \Omega = \O$ Therefore $\O$ and $\Omega$ are pairwise disjoint. From Union with Empty Set: :$\O \cup \Omega = \Omega$ Therefore we have: {{begin-eqn}} {{eqn | l = \map \Pr \Omega       | r = \map \Pr {\O \cup \Omega}       | c =  }} {{eqn | r = \map \Pr \O + \map \Pr \Omega       | c =  }} {{end-eqn}} As $\map \Pr \Omega = 1$, it follows that $\map \Pr \O = 0$. {{qed}}
First note that by Successor Set of Ordinal is Ordinal, $x^+$ and $y^+$ are ordinals. By Ordinal Membership is Trichotomy, one of the following must be true: : $x^+ = y^+$ : $y^+ \in x^+$ : $x^+ \in y^+$ We will show that the first two are both false, so that the third must hold. Two preliminary facts: {{begin-axiom}} {{axiom |n = 1         |t = $x \ne y$         |rc = $x \in y$ and Ordinal is not Element of Itself }} {{axiom |n = 2         |t = $y \notin x$         |rc = $x \in y$ and Ordinal Membership is Asymmetric }} {{end-axiom}} By $(1)$ and Equality of Successors: : $x^+ ≠ y^+$ Thus the first of the three possibilities is false. Suppose for the sake of contradiction that $y^+ \in x^+$. Then: {{begin-eqn}} {{eqn |o =       |r = y \in y^+       |c = Definition of successor set }} {{eqn |o = \implies       |r = y \in x^+       |c = $y^+ \in x^+$ and $x^+$ is an ordinal and therefore transitive. }} {{eqn |o = \implies       |r = y \in x \lor y = x       |c = Definition of successor set }} {{end-eqn}} But we already know that $y \notin x$ by $(2)$ and $y \ne x$ by $(1)$, so this is a contradiction, and we conclude that $y^+ \notin x^+$. Thus we have shown that the second possibility is false. Thus the third and final one must hold: $x^+ \in y^+$. {{qed}}
Suppose for the sake of contradiction that $m \in n$ and $n \in m$. Since $m$ is an ordinal, it is transitive. Thus since $m \in n$ and $n \in m$, it follows that $m \in m$. But this contradicts Ordinal is not Element of Itself. {{qed}} Category:Ordinals ldio46x87kr5fy460pxnmroieab85jd
By Intersection of Ordinals is Ordinal, $m$ is an ordinal. By Intersection is Largest Subset, $m \subseteq a$ for each $a \in A$. It remains to show that $m \in A$. Let $m^+ = m \cup \set m$ be the successor of $m$. By Relation between Two Ordinals: :for each $a \in A$, either $m^+ \subseteq a$ or $a \in m^+$. Let $m^+ \subseteq a$ for all $a \in A$. Then by Intersection is Largest Subset $m^+ \subseteq m$. This contradicts the fact that Ordinal is not Element of Itself. Thus there is an $a \in A$ such that $a \in m^+$. Thus $a \in m$ or $a = m$. Let $a \in m$. Then $a \subsetneqq m$. This contradicts the fact that $m \subseteq a$. Thus $a = m$. Thus it follows that $m \in A$. {{qed}} Category:Ordinals 27ffaqxp1k494b4ndklpntimdi8l86v
{{NotZFC}} By Subset Relation is Ordering, $\subseteq$ is an ordering of any set. Let $A$ be a subclass of $\On$. By Intersection of Ordinals is Smallest, $A$ has a $\subseteq$-smallest element. {{qed}} Category:Ordinal Class qtq8v3lmbjcg3ya0fmf1zooqfb4h4su
{{begin-eqn}} {{eqn|l = x \in y      |o = \implies      |r = x \ne y      |c = No Membership Loops }} {{eqn|o = \implies      |r = x^+ \ne y^+      |c = Equality of Successors }} {{eqn|l = x \in y      |o = \implies      |r = y \notin x      |c = No Membership Loops }} {{eqn|o = \implies      |r = y \notin x^+      |c = $x ≠ y$ and Definition:Successor Set }} {{eqn|l = y^+ \in x^+      |o = \implies      |r = y \in x^+      |c = Successor Set of Ordinal is Ordinal, Ordinals are Transitive, and Set is Element of Successor }} {{end-eqn}} The last part is a contradiction, so $y^+ \notin x^+$. By Ordinal Membership is Trichotomy: : $x^+ \in y^+$ {{qed}}
First note that by Successor Set of Ordinal is Ordinal, $x^+$ and $y^+$ are ordinals. Let $x \in y$. We wish to show that $x^+ \in y^+$. By Ordinal Membership is Trichotomy: :Either $x^+ \in y^+$, $y^+ \in x^+$, or $x^+ = y^+$. Suppose for the sake of contradiction that $y^+ = x^+$. Then $y \in x$ or $y = x$ by the definition of successor set. If $y = x$ then $x \in x$, contradicting the fact that Ordinal is not Element of Itself. If $y \in x$ then since an ordinal is transitive, $y \in y$, again contradicting Ordinal is not Element of Itself. Thus $y^+ ≠ x^+$. Suppose for the sake of contradiction that $y^+ \in x^+$. By the definition of successor set, $y^+ \in x$ or $y^+ = x$. If $y^+ \in x$, then since $y^+$ and $x$ are both ordinals, $y^+ \subsetneqq x$. Then $y \in x$. Since $y$ is transitive, $y \in y$, contradicting Ordinal is not Element of Itself. If instead $y^+ = x$, then $x \in y \in y^+ = x$, so the same contradiction arises because $x$ is transitive. Thus $y^+ \notin x^+$. So the only remaining possibility, that $x^+ \in y^+$, must hold. {{qed}}
Suppose $y^+ \in x^+$. By the definition of successor, $y^+ \in x \lor y^+ = x$. Suppose $y^+ = x$. By Ordinal is Less than Successor, $y \in x$. Suppose $y^+ \in x$. By Ordinal is Less than Successor, $y \in y^+$. By Ordinal is Transitive, $y \in x$.
First note that by Successor Set of Ordinal is Ordinal, $x^+$ and $y^+$ are ordinals. Let $x^+ \in y^+$. Then since $y^+$ is transitive, $x^+ \subseteq y^+$. Thus $x \in y$ or $x = y$. If $x = y$ then $x^+ \in x^+$, contradicting Ordinal is not Element of Itself. Thus $x \in y$.
From the conditions for $\Pr$ to be a probability measure, we have: :$(1): \quad \forall A \in \Sigma: 0 \le \map \Pr A$ :$(2): \quad \map \Pr \Omega = 1$ :$(3): \quad \ds \map \Pr {\bigcup_{i \mathop \ge 1} A_i} = \sum_{i \mathop \ge 1} \map \Pr {A_i}$ where all $A_i$ are pairwise disjoint. Let $A \in \Sigma$ be an event. Then $\paren {\Omega \setminus A} \in \Sigma$ by definition of Event Space: Axiom $(ES \ 2)$. From Intersection with Relative Complement is Empty: :$A \cap \paren {\Omega \setminus A} = \O$ From Union with Relative Complement: :$A \cup \paren {\Omega \setminus A} = \Omega$ So: :$\map \Pr A + \map \Pr {\Omega \setminus A} = 1$ from above, and so: :$\map \Pr {\Omega \setminus A} = 1 - \map \Pr A$ {{qed}}
Let $\mathcal D$ be a $\delta$-algebra whose unit is $\mathbb U$. Let $A_1, A_2, \ldots$ be a countably infinite collection of elements of $\mathcal D$. Then: {{begin-eqn}} {{eqn | l = \forall i: \mathbb U \setminus A_i       | o = \in       | r = \mathcal D       | c = $\mathcal D$ is closed under relative complement with $\mathbb U$ }} {{eqn | ll=\implies       | l = \bigcap_{i \mathop = 1}^\infty \left({\mathbb U \setminus A_i}\right)       | o = \in       | r = \mathcal D       | c = $\mathcal D$ is closed under countable intersections }} {{eqn | ll= \implies       | l = \mathbb U \setminus \bigcup_{i \mathop = 1}^\infty A_i       | o = \in       | r = \mathcal D       | c = De Morgan's Laws: Complement of Union }} {{eqn | ll= \implies       | l = \bigcup_{i \mathop = 1}^\infty A_i       | o = \in       | r = \mathcal D       | c = $\mathcal D$ is closed under relative complement with $\mathbb U$ }} {{end-eqn}} Thus $\mathcal D$ is also a $\sigma$-algebra. {{qed}}
By definition, a probability space $\struct {\Omega, \Sigma, \Pr}$ is a measure space. So, again by definition, an event space $\Sigma$ is a $\sigma$-algebra on $\Omega$. From Sigma-Algebra is Delta-Algebra: :$\ds A_1, A_2, \ldots \in \Sigma \implies \bigcap_{i \mathop = 1}^\infty A_i \in \Sigma$ by definition of $\delta$-algebra. {{qed}}
To establish that $\subseteq$ is an ordering, we need to show that it is reflexive, antisymmetric and transitive. So, checking in turn each of the criteria for an ordering: === Reflexivity === {{begin-eqn}} {{eqn | ll= \forall T \in \mathbb S:       | l = T       | o = \subseteq       | r = T       | c = Set is Subset of Itself }} {{end-eqn}} So $\subseteq$ is reflexive. {{qed|lemma}} === Antisymmetry === {{begin-eqn}} {{eqn | ll= \forall S_1, S_2 \in \mathbb S:       | l = S_1 \subseteq S_2 \land S_2 \subseteq S_1       | o = \iff       | r = S_1 = S_2       | c = {{Defof|Set Equality|index = 2}} }} {{end-eqn}} So $\subseteq$ is antisymmetric. {{qed|lemma}} === Transitivity === {{begin-eqn}} {{eqn | ll= \forall S_1, S_2, S_3 \in \mathbb S:       | l = S_1 \subseteq S_2 \land S_2 \subseteq S_3       | o = \implies       | r = S_1 \subseteq S_3       | c = Subset Relation is Transitive }} {{end-eqn}} That is, $\subseteq$ is transitive. {{qed|lemma}} So we have shown that $\subseteq$ is an ordering on $\mathbb S$. {{qed}}
Let $\mathcal F$ be the set of ideals of $R$ of the form $x R$, with $x$ not a unit and such that $x$ cannot be decomposed in the form: :$x = u p_1 \dotsm p_r$ where $u$ is a unit and $p_1, \dotsc, p_r$ irreducible. We show by contradiction that $\mathcal F = \O$. {{AimForCont}} $\mathcal F \ne \O$. Since $R$ is noetherian, we can choose a maximal element $a R \in \mathcal F$. By construction, $a$ is not irreducible, so we can write: : $a = b c$ with $b, c$ non-units and not associates. By the definition of associates in a commutative ring this means that $b R \subsetneq a R$ and $a R \subsetneq b R$. Since $a R$ is assumed maximal, this means that $b R$ and $c R$ do not belong to $\mathcal F$. Therefore there exist units $u, v$ and irreducible elements $p_1, \dotsc, p_r, q_1, \dotsc, q_s$ such that: :$b = u p_1 \dotsm p_r$ and: :$c = v q_1 \dotsm q_s$ But this implies that: :$a = b c = \paren {u v} p_1 \cdots p_r \cdot q_1 \dotsm q_s$ This is a contradiction, since by hypothesis $a$ cannot be written in this form. {{Qed}} Category:Ring Theory Category:Factorization qtrvz0wcnqlgi83sxmiqiv3zduptpbv
<onlyinclude> Suppose for the sake of contradiction that $\neg \On \subseteq A$. Then: :$\paren {\On \setminus A} \ne \O$ From Set Difference is Subset, $\On \setminus A$ is a subclass of the ordinals. By Ordinal Class is Strongly Well-Ordered by Subset, $\On \setminus A$ must have a smallest element $y$. Then every strict predecessor of $y$ must lie in $A$, so by the premise, $y$ must also be an element of $A$. This contradicts the fact that $y$ is an element of $\On \setminus A$. Therefore $\On \subseteq A$. {{qed}} {{LEM|Reductio ad Absurdum|4}} Category:Ordinals hisui4zkmig66hqob794i8gni2513om
Suppose for the sake of contradiction that $x$ has another immediate successor, $z ≠ y$. Then by definition: : $x \prec y$ : $x \prec z$ Since $\preceq$ is a total ordering and $y ≠ z$: : $y \prec z$ or $z \prec y$. If $y \prec z$ then $x \prec y \prec z$, contradicting the fact that $z$ is an immediate successor of $x$. If $z \prec y$ then $x \prec z \prec y$, contradicting the fact that $y$ is an immediate successor of $x$. Thus we conclude that there can be no such $z$. {{qed}}
{{NotZFC}} Let $I$ be the image of $f$. By Union of Subset of Ordinals is Ordinal: Corollary, $\bigcup I$ is an ordinal. But by Union is Smallest Superset, each element of $I$ is a subset of $\bigcup I$. Thus $\bigcup I$ is an upper bound of $f$. {{qed}} Category:Ordinal Class Category:Class Mappings iuql11k1dffyyyz7fk5x1kix4oeapzo
By the Axiom of Replacement, $\map F y$ is a set. Thus by the Axiom of Unions, $\bigcup \map F y$ is a set. By Union of Subset of Ordinals is Ordinal, $\bigcup \map F y$ is transitive. By the epsilon relation $\bigcup \map F y$ is well-ordered. Thus $\bigcup \map F y$ is a member of $\On$, the ordinal class. {{qed}} Category:Ordinals 7gktcr8xvoxtlsybhbswt68kntcp9ua
Proof by induction: For all $n \in \N: n \ge 2$, let $\map P n$ be the proposition: :$A_1, A_2, \ldots, A_n$ are independent {{iff}} $\Omega \setminus A_1, \Omega \setminus A_2, \ldots, \Omega \setminus A_n$ are independent. === Basis for the Induction === $\map P 2$ is the case: :$A_1$ and $A_2$ are independent {{iff}} $\Omega \setminus A_1$ and $\Omega \setminus A_2$ are independent. This is proved in Independent Events are Independent of Complement: Corollary. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :$A_1, A_2, \ldots, A_k$ are independent {{iff}} $\Omega \setminus A_1, \Omega \setminus A_2, \ldots, \Omega \setminus A_k$ are independent. Then we need to show: :$A_1, A_2, \ldots, A_{k + 1}$ are independent {{iff}} $\Omega \setminus A_1, \Omega \setminus A_2, \ldots, \Omega \setminus A_{k + 1}$ are independent. === Induction Step === This is our induction step. Suppose $A_1, A_2, \ldots, A_{k + 1}$ are independent. Then: {{begin-eqn}} {{eqn | l = \map \Pr {\bigcap_{i \mathop = 1}^{k + 1} A_i}       | r = \map \Pr {\bigcap_{i \mathop = 1}^k A_i \cap A_{k + 1} }       | c =  }} {{eqn | r = \prod_{i \mathop = 1}^k \map \Pr {A_i} \times \map \Pr {A_{k + 1} }       | c = as all of the $A_1, A_2, \ldots, A_k, A_{k + 1}$ are independent }} {{eqn | r = \map \Pr {\bigcap_{i \mathop = 1}^k A_i} \times \map \Pr {A_{k + 1} }       | c =  }} {{end-eqn}} So we see that $\displaystyle \bigcap_{i \mathop = 1}^k A_i$ and $A_{k + 1}$ are independent. So $\displaystyle \bigcap_{i \mathop = 1}^k A_i$ and $\Omega \setminus A_{k + 1}$ are independent. So, from the above results, we can see that $\Omega \setminus A_1, \Omega \setminus A_2, \ldots, \Omega \setminus A_{k + 1}$ are independent. So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. The reverse implication follows directly. Therefore: :$A_1, A_2, \ldots, A_n$ are independent {{iff}} $\Omega \setminus A_1, \Omega \setminus A_2, \ldots, \Omega \setminus A_n$ are independent. {{qed}} {{proofread}}
This is an instance of Probability of Independent Events Not Happening with all of $A_1, A_2, \ldots, A_m$ being instances of $A$. The result follows directly. {{qed}} Category:Probability Theory jfor173v5hwwv46csmqgmacah73w0e4
Let $\displaystyle B_i = A_i \setminus A_{i-1}$ for $i \in \N: i > 0$. Then: :$A = A_0 \cup B_1 \cup B_2 \cup \cdots$ is the union of disjoint events in $\Sigma$. By definition of probability measure: {{begin-eqn}} {{eqn | l = \Pr \left({A}\right)       | r = \Pr \left({A_0}\right) + \Pr \left({B_1}\right) + \Pr \left({B_2}\right) + \cdots       | c =  }} {{eqn | r = \Pr \left({A_0}\right) + \lim_{n \to \infty} \sum_{k \mathop = 1}^n \Pr \left({B_k}\right)       | c =  }} {{end-eqn}} But we have: :$\Pr \left({B_i}\right) = \Pr \left({A_i}\right) - \Pr \left({A_{i-1}}\right)$ for $i \in \N: i > 0$. So: :$\displaystyle \Pr \left({A}\right) = \Pr \left({A_0}\right) + \lim_{n \to \infty} \sum_{k \mathop = 1}^n \left({\Pr \left({A_i}\right) - \Pr \left({A_{i-1}}\right)}\right)$ The last sum telescopes. Hence the result: :$\displaystyle \Pr \left({A}\right) = \lim_{n \to \infty} \Pr \left({A_n}\right)$ {{qed}}
{{expand|The below needs to be done properly.}} Set $A_i = \Omega \setminus B_i$ and then apply De Morgan's laws and the result for an increasing sequence of events. {{qed}}
To show that $U$ is discrete random variable on $\struct {\Omega, \Sigma, \Pr}$, we need to show that: :$(1): \quad$ The image of $U$ is a countable subset of $\R$; :$(2): \quad \forall x \in \R: \set {\omega \in \Omega: \map U \omega = x} \in \Sigma$. First we consider any $U_u = \set {\omega \in \Omega: \map U \omega = u}$ such that $U_u \ne \O$. We have that $U_u = \set {\omega \in \Omega: \map X \omega + \map Y \omega = u}$. Consider any $\omega \in U_u$. Then: :$\omega \in X_x \cap Y_x$ where: :$X_x = \set{\omega \in \Omega: \map X \omega = x}, Y_x = \set {\omega \in \Omega: \map Y \omega = u - x}$ Because $X$ and $Y$ are discrete random variables, both $X_x \in \Sigma$ and $Y_x \in \Sigma$. As $\struct {\Omega, \Sigma, \Pr}$ is a probability space, then $X_x \cap Y_x \in \Sigma$. Now note that: :$\displaystyle U_u = \bigcup_{x \mathop \in \R} \paren {X_x \cap Y_x}$ That is, it is the union of all such intersections of sets whose discrete random variables add up to $u$. As $X_x$ is a countable set it follows that $U_u$ is a countable union of countable sets. From Countable Union of Countable Sets is Countable it follows that $X_x$ is a countable set. And, by dint of $\struct {\Omega, \Sigma, \Pr}$ being a probability space, $U_u \in \Sigma$. Thus $U$ is a discrete random variables on $\struct {\Omega, \Sigma, \Pr}$. {{qed}} {{explain|Might need to be more rigorous on the boundary instances.}} {{proofread}}