hcju/proofwikiqa huggingface dataset:
=== Necessary Condition === Let $F$ be exact. Then by definition there exists a function whose second partial derivatives of $f$ exist and are continuous: :$\map f {x, y}$ where: :$\dfrac {\partial f} {\partial x} = M$ :$\dfrac {\partial f} {\partial y} = N$ Differentiating $M$ and $N$ partially {{WRT|Differentiation}} $y$ and $x$ respectively: :$\dfrac {\partial M} {\partial y} = \dfrac {\partial^2 f} {\partial x \partial y}$ :$\dfrac {\partial N} {\partial x} = \dfrac {\partial^2 f} {\partial y \partial x}$ From Mixed Second Partial Derivatives are Equal: :$\dfrac {\partial M} {\partial y} = \dfrac {\partial N} {\partial x}$ {{qed|lemma}} === Sufficient Condition === Let $F$ be such that: :$\dfrac {\partial M} {\partial y} = \dfrac {\partial N} {\partial x}$ Take the equation: :$\dfrac {\partial f} {\partial x} = M$ and integrate it {{WRT|Integration}} $x$: :$(1): \quad f = \displaystyle \int M \rd x + \map g y$ Note that the arbitrary constant here is an '''arbitrary function''' of $y$, since when differentiating partially {{WRT|Differentiation}} $x$ it disappears. So, now we need to find such a $\map g y$ so as to make $f$ as defined in $(1)$ satisfy $\dfrac {\partial f} {\partial y} = N$. Differentiating $(1)$ {{WRT|Differentiation}} $y$ and equating it to $N$: :$\displaystyle \dfrac {\partial} {\partial y} \int M \rd x + \map {g'} y = N$ So: :$\displaystyle \map {g'} y = N - \dfrac {\partial} {\partial y} \int M \rd x$ which can be integrated {{WRT|Integration}} $y$ thus: :$\displaystyle \map g y = \int \paren {N - \frac {\partial} {\partial y} \int M \rd x} \rd y$ which works as long as the integrand is a function of $y$ only. This will happen if its derivative {{WRT|Differentiation}} $x$ is equal to zero. We need to make sure of that, so we try it out:  {{begin-eqn}} {{eqn | l = \frac {\partial} {\partial x} \paren {N - \frac {\partial} {\partial y} \int M \rd x}       | r = \frac {\partial N} {\partial x} - \frac {\partial^2} {\partial x \partial y} \int M \rd x       | c = }} {{eqn | r = \frac {\partial N} {\partial x} - \frac {\partial^2} {\partial y \partial x} \int M \rd x       | c = }} {{eqn | r = \frac {\partial N} {\partial x} - \frac {\partial M} {\partial y}       | c = }} {{end-eqn}} This is our initial condition. {{qed|lemma}}
Let us for ease of manipulation express $(1)$ in the form of differentials: :$(2): \quad M \left({x, y}\right) \mathrm d x + N \left({x, y}\right) \mathrm d y = 0$ Suppose that $(2)$ has a general solution: :$(3): \quad f \left({x, y}\right) = C$ where $C$ is some constant. We can eliminate $C$ by differentiating: :$\dfrac {\partial f} {\partial x} \, \mathrm d x + \dfrac {\partial f} {\partial y} \, \mathrm d y = 0$ It follows from $(2)$ and $(3)$ that: :$\dfrac {\mathrm d y} {\mathrm d x} = - \dfrac M N = - \dfrac {\partial f / \partial x} {\partial f / \partial y}$ and so: :$(4): \quad \dfrac {\partial f / \partial x} M = \dfrac {\partial f / \partial y} N$ Let this common ratio in $(4)$ be denoted $\mu \left({x, y}\right)$. Then: :$\dfrac {\partial f}{\partial x} = \mu M$ :$\dfrac {\partial f}{\partial y} = \mu N$ So, if we multiply $(2)$ by $\mu$, we get: :$\mu M \, \mathrm d x + \mu N \, \mathrm d y = 0$ or: :$\dfrac {\partial f} {\partial x} \, \mathrm d x + \dfrac {\partial f} {\partial y} \, \mathrm d y = 0$ which is exact. So, if $(2)$ has a general solution, it has at least one integrating factor $\mu \left({x, y}\right)$. {{qed}}
We have one of these: : Integrating Factor for First Order ODE: Function of One Variable: $x$ or $y$ only : Integrating Factor for First Order ODE: Function of $x + y$ : Integrating Factor for First Order ODE: Function of $x y$ === Conclusion of Proof === {{:Integrating Factor for First Order ODE/Conclusion}}
Consider the first order ordinary differential equation: :$\map M {x, y} + \map N {x, y} \dfrac {\d y} {\d x} = 0$ We can put our equation: :$(1): \quad \dfrac {\d y} {\d x} + \map P x y = \map Q x$ into this format by identifying: :$\map M {x, y} \equiv \map P x y - \map Q x, \map N {x, y} \equiv 1$ We see that: :$\dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x} = \map P x$ and hence: :$\map P x = \dfrac {\dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x} } N$ is a function of $x$ only. It immediately follows from Integrating Factor for First Order ODE that: :$e^{\int \map P x \rd x}$ is an integrating factor for $(1)$. So, multiplying $(1)$ by this factor: :$e^{\int \map P x \rd x} \dfrac {\d y} {\d x} + e^{\int \map P x \rd x} \map P x y = e^{\int \map P x \rd x} \map Q x$ The result follows by an application of Solution to Exact Differential Equation. {{qed}}
From the Product Rule for Derivatives: {{begin-eqn}} {{eqn | l = \frac \d {\d x} \paren {e^{\int \map P x \rd x} y}       | r = e^{\int \map P x \rd x} \dfrac {\d y} {\d x} + y e^{\int \map P x \rd x} \map P x       | c =  }} {{eqn | r = e^{\int \map P x \rd x} \paren {\dfrac {\d y} {\d x} + \map P x y}       | c =  }} {{end-eqn}} Hence, multiplying $(1)$ all through by $e^{\int \map P x \rd x}$: :$\dfrac \d {\d x} \left({e^{\int \map P x \rd x} y}\right) = \map Q x e^{\int \map P x \rd x}$ Integrating {{WRT|Integration}} $x$ now gives: :$\displaystyle e^{\int \map P x \rd x} y = \int \map Q x e^{\int \map P x \rd x} \rd x + C$ whence we get the result by dividing by $e^{\int \map P x \rd x}$. {{qed}}
By the Series Law for Extremal Length, we have  :$\lambda \left({\Gamma}\right) \ge \lambda \left({\Gamma_1}\right) + \lambda \left({\Gamma_2}\right)$ Hence it remains only to prove the opposite inequality. Let $\rho$ be a metric as in the definition of extremal length, normalized such that $A \left({\rho}\right) = 1$. The claim to be proved is that: :$\left({L \left({\Gamma, \rho}\right)}\right)^2 \le \lambda \left({\Gamma_1}\right) + \lambda \left({\Gamma_2}\right)$   Define $\alpha_j := \sqrt {A \left({\rho {\restriction_{E_j} } }\right)}$ for $j = 1, 2$. Suppose either $\lambda \left({\Gamma_1}\right)$ or $\lambda \left({\Gamma_2}\right)$ is infinite. It trivially follows that both $\alpha_1$ and $\alpha_2$ are positive. Assume both extremal lengths are finite. Let $\alpha_j = 0$. Then: :$L \left({\Gamma_j, \rho}\right) = 0$ Indeed, we can choose a metric $\tilde{\rho}$ that agrees with $\rho$ on $A_j$ and has total area $\varepsilon > 0$. Then: :$\left({L \left({\Gamma_j, \rho}\right)}\right)^2 = \left({L \left({\Gamma_j, \tilde \rho}\right)}\right)^2 \le \varepsilon \cdot \lambda \left({\Gamma_j}\right)$ Because $\varepsilon$ was arbitary, the claim follows. {{WLOG}}, suppose that $\alpha_2 = 0$. Then: :$\left({L \left({\Gamma_j, \rho}\right)}\right)^2 = \left({L \left({\Gamma_1, \rho}\right) + L \left({\Gamma_2, \rho}\right)}\right)^2 = \left({L \left({\Gamma_j, \rho}\right)}\right)^2 \le \lambda \left({\Gamma_1}\right) \le \lambda \left({\Gamma_1}\right) + \lambda \left({\Gamma_2}\right)$ Thus both $\alpha_1$ and $\alpha_2$ are positive Let: :$\rho_j := \dfrac {\rho {\restriction_{E_j} } } {\alpha_j}$ Then $A \left({\rho}\right) = 1$. So: {{begin-eqn}} {{eqn | l = \left({L \left({\Gamma, \rho}\right)}\right)^2       | r = \left({L \left({\Gamma_1, \rho}\right) + L \left({\Gamma_2, \rho}\right)}\right)^2       | c = Definition of $\Gamma$ }} {{eqn | r = \left({L \left({\Gamma_1, \alpha_1 \rho_1}\right) + L \left({\Gamma_2, \alpha_2 \rho_2}\right)}\right)^2       | c = Definition of $\rho_j$ and disjointness of $E_1$ and $E_2$ }} {{eqn | r = \left({\alpha_1 L \left({\Gamma_1, \rho_1}\right) + \alpha_2 L \left({\Gamma_2, \rho_2}\right)}\right)^2       | c = Definition of $L \left({\Gamma_j, \rho_j}\right)$) }} {{eqn | o = \le       | r = \left({\alpha_1^2 + \alpha_2^2}\right) \left({L \left({\Gamma_1, \rho_1}\right)^2 + L \left({\Gamma_2, \rho_2}\right)^2}\right)       | c = Cauchy-Bunyakovsky-Schwarz Inequality }} {{eqn | o = \le       | r = L \left({\Gamma_1, \rho_1}\right)^2 + L \left({\Gamma_2, \rho_2}\right)^2       | c = as $\alpha_1^2 + \alpha_2^2 \le A \left({\rho}\right) = 1$ }} {{eqn | o = \le       | r = \lambda \left({\Gamma_1}\right) + \lambda \left({\Gamma_2}\right)       | c = Definition of Extremal Length }} {{end-eqn}} In conclusion: :$\lambda \left({\Gamma}\right) = \sup_\rho L \left({\Gamma, \rho}\right) \le \lambda \left({\Gamma_1}\right) + \lambda \left({\Gamma_2}\right)$ as claimed. {{qed}} Category:Functional Analysis lqm3iv7tj9ii3k5gv0hwl815klla2wb
If $n$ is even, then at least the last digit of $n^2$ is even. So for $n^2$ to have its digits all odd, $n$ itself must be odd. We can see immediately that $1 = 1^2$ and $9 = 3^2$ fit the criterion. Of the other 1-digit odd integers, we have $5^2 = 25, 7^2 = 49, 9^2 = 81$, all of which have an even digit. Now, let $n > 10$ be an odd integer. There are five cases to consider: * $n = 10 p + 1$: we have $\left({10p + 1}\right)^2 = 100 p^2 + 20 p + 1 = 10 \left({10 p^2 + 2 p}\right) + 1$. * $n = 10 p + 3$: we have $\left({10p + 3}\right)^2 = 100 p^2 + 60 p + 9 = 10 \left({10 p^2 + 6 p}\right) + 9$. * $n = 10 p + 5$: we have $\left({10p + 5}\right)^2 = 100 p^2 + 100 p + 25 = 10 \left({10 p^2 + 10 p + 2}\right)+ 5$. * $n = 10 p + 7$: we have $\left({10p + 7}\right)^2 = 100 p^2 + 140 p + 49 = 10 \left({10 p^2 + 14 p + 4}\right)+ 9$. * $n = 10 p + 9$: we have $\left({10p + 9}\right)^2 = 100 p^2 + 180 p + 81 = 10 \left({10 p^2 + 18 p + 8}\right)+ 1$. It is clear that in all cases the 10's digit is even. So the square of every odd integer greater than $3$ always has at least one even digit. Hence the result. {{qed}}
Let $p$ be an integer. By the Division Theorem, there are unique integers $k$ and $r$ such that $p = 2k + r$ and $0 \le r < 2$. That is, $r = 0$ or $r = 1$, where $r = 0$ corresponds to the case of $p$ being even and $r = 1$ corresponds to the case of $p$ being odd. === Even case === {{:Parity of Integer equals Parity of its Square/Even}} {{qed|lemma}} === Odd case === {{:Parity of Integer equals Parity of its Square/Odd}} {{qed|lemma}} Therefore, if it is not the case that an integer is even, then it is not the case that its square is even. Conversely, if it is the case that an integer is even (and also a perfect square), then it is the case that its square root is even. {{qed}}
By definition of limit superior: :$\displaystyle \limsup_{n \mathop \to \infty} E_n = \bigcap_{i \mathop = 1}^\infty \bigcup_{j \mathop = i}^\infty E_j$ Thus, by Measure is Monotone and Intersection is Subset: :$(1): \quad \displaystyle \map \mu {\limsup_{n \mathop \to \infty} E_n} = \map \mu {\bigcap_{i \mathop = 1}^\infty \bigcup_{j \mathop = i}^\infty E_j} \le \map \mu {\bigcup_{j \mathop = i}^\infty E_j}$ for all $i \in \N$. By Measure is Subadditive: :$\displaystyle \map \mu {\bigcup_{j \mathop = i}^\infty E_j} \le \sum_{j \mathop = i}^\infty \map \mu {E_j}$ However, by assumption $\displaystyle \sum_{n \mathop = 1}^\infty \map \mu {E_n}$ converges. By Tail of Convergent Series tends to Zero this implies: :$\displaystyle \lim_{i \mathop \to \infty} \sum_{n \mathop = i}^\infty \map \mu {E_n} = 0$ Now $(1)$ implies, together with Lower and Upper Bounds for Sequences, that: :$\displaystyle \map \mu {\limsup_{n \mathop \to \infty} E_n} \le 0$ But as $\mu$ is a measure, the converse inequality also holds. Hence: :$\displaystyle \map \mu {\limsup_{n \mathop \to \infty} E_n} = 0$ {{qed}}
Let: :$\displaystyle S_n = \sum_{i \mathop = 1}^n f_i$ Let: :$\displaystyle f = \lim_{n \mathop \to \infty} S_n$ To show the partial sums converge uniformly to $f$, we must show that: :$\displaystyle \lim_{n \mathop \to \infty} \sup_{x \mathop \in D} \size {f - S_n} = 0$ But: {{begin-eqn}} {{eqn | l = \sup_{x \mathop \in D} \size {f - S_n}       | r = \sup_{x \mathop \in D} \size {\paren {f_1 + f_2 + \dotsb} - \paren {f_1 + f_2 + \dotsb + f_n} }       | c =  }} {{eqn | r = \sup_{x \mathop \in D} \size {f_{n + 1} + f_{n + 2} + \dotsc}       | c =  }} {{end-eqn}} By the Triangle Inequality, this value is less than or equal to: :$\displaystyle \sum_{i \mathop = n + 1}^\infty \sup_{x \mathop \in D} \size {\map {f_i} x} \le \sum_{i \mathop = n + 1}^\infty M_i$ We have that: :$\displaystyle 0 \le \sum_{i \mathop = 1}^\infty M_n < \infty$ It follows from Tail of Convergent Series tends to Zero: :$\displaystyle 0 \le \lim_{n \mathop \to \infty} \sum_{i \mathop = n + 1}^\infty \sup_{x \mathop \in D} \size {\map {f_i} x} \le \lim_{n \mathop \to \infty} \sum_{i \mathop = n + 1}^\infty M_i = 0$ So: :$\displaystyle \lim_{n \mathop \to \infty} \sup_{x \mathop \in D} \size {f - S_n} = 0$ Hence the series converges uniformly on the domain. {{qed}} {{expand|Establish how broadly this can be applied - it's defined for real functions but should also apply to complex ones and possibly a general metric space.}}
First, note that $\map {\mathfrak M} {\mu^*}$ is an algebra (of sets). It remains to be shown that $\map {\mathfrak M} {\mu^*}$ is closed under countable union. Because $\map {\mathfrak M} {\mu^*}$ is an algebra (of sets), the union of any two $\mu^*$-measurable sets is $\mu^*$-measurable. Using mathematical induction, it directly follows that the finite union of $\mu^*$-measurable sets is $\mu^*$-measurable. Let $\sequence {S_n}$ be a sequence of $\mu^*$-measurable subsets of $X$. Define $\displaystyle S = \bigcup_{n \mathop = 1}^\infty S_n$. We wish to prove that $S$ is $\mu^*$-measurable. For all $n \in \N$, the set $T_n = S_1 \cup S_2 \cup \cdots \cup S_n$ is $\mu^*$-measurable. By Subset of Union, the sequence $\sequence {T_n}$ is increasing. Also, $T_n \uparrow S$ (as $n \to \infty$) where $\uparrow$ denotes the limit of an increasing sequence of sets. Let $A$ be any subset of $X$. By Set Difference Union Intersection: :$A = \paren {A \cap S} \cup \paren {A \setminus S}$ So by the subadditivity of $\mu^*$, it suffices to prove that $\map {\mu^*} A \ge \map {\mu^*} {A \cap S} + \map {\mu^*} {A \setminus S}$ for any subset $A \subseteq X$. Then: {{begin-eqn}} {{eqn | l = \map {\mu^*} A       | r = \map {\mu^*} {A \cap T_n} + \map {\mu^*} {A \setminus T_n}       | c = Definition of Measurability of $T_n$ }} {{eqn | o = \ge       | r = \map {\mu^*} {A \cap T_n} + \map {\mu^*} {A \setminus S}       | c = Set Complement inverts Subsets and Monotonicity of $\mu^*$ }} {{end-eqn}} Letting $n \to \infty$, the result follows by Outer Measure of Limit of Increasing Sequence of Sets. {{qed}}
Pick any $t$ and define: : $B = \left\{{x \in A: \left\vert{f \left({x}\right)}\right \vert \ge t}\right\}$ Let $\chi_B$ denote the indicator function of $B$ on $A$.  For any $x \in A$, either $x \in B$ or $x \notin B$. In the first case: : $t \chi_B \left({x}\right) = t \cdot 1 = t \le \left\vert{f \left({x}\right)}\right\vert$ In the second case: : $t \chi_B \left({x}\right) = t \cdot 0 = 0 \le \left\vert{f \left({x}\right)}\right\vert$ Hence: : $\forall x \in A: t \chi_B \left({x}\right) \le \left\vert{f \left({x}\right)}\right\vert$ By Integral of Integrable Function is Monotone: : $\displaystyle \int_A t \chi_B \mathrm d \mu \le \int_A \left\vert{f}\right\vert \mathrm d \mu$ But by Integral of Integrable Function is Linear: : $\displaystyle \int_A t \chi_B \mathrm d \mu = t \int_A \chi_B \mathrm d \mu = t \mu \left({B}\right)$ Dividing through by $t$: : $\displaystyle \mu \left({B}\right) \le \frac 1 t \int_A \left\vert{f}\right\vert \mathrm d \mu$ {{qed}}
Select any $\ds x \in \liminf_{n \mathop \to \infty} E_n$. Then by definition 2 of Limit Inferior of Sequence of Sets, $x$ belongs to $E_i$ for all but finitely many values of $i$. Therefore $x$ belongs to $E_i$ for infinitely many values of $i$. So by the definition 2 of Limit Superior of Sequence of Sets: :$\ds x \in \limsup_{n \mathop \to \infty} E_n$ {{qed}}
We check each of the criteria for an outer measure. From the assumption $\map \gamma \O = 0$: :$\map {\mu^*} \O = 0$ Any cover of a set is also a cover of a subset of it. Thus $\mu^*$ is monotone by Infimum of Subset. It remains to be shown that $\mu^*$ is countably subadditive. Let $\sequence {S_n}$ be a sequence of subsets of $X$. Let $\displaystyle S := \bigcup_{n \mathop = 0}^\infty S_n$. Suppose there does not exist a countable cover for $S_n$ by elements of $\AA$ for some  $n \in \N$. Then there does not exist a countable cover for $S$ by elements of $\AA$. In this case, the result follows immediately. Now suppose that for each $n \in \N$, there exists a countable cover for $S_n$ by elements of $\AA$. Let $\epsilon \in \R_{>0}$ be an arbitrary (strictly) positive real number. By definition of infimum, for each $n \in \N$, we can apply the axiom of countable choice to choose a countable cover $\CC_n \subseteq \AA$ of $S_n$ such that: :$\displaystyle \sum_{x \mathop \in \CC_n} \map \gamma x < \map {\mu^*} {S_n} + \frac \epsilon {2^{n + 1} }$ Let $\displaystyle \CC = \bigcup_{n \mathop = 0}^\infty \CC_n$. Then $\CC$ is a subset of $\AA$ and a cover for $S$. Furthermore, $\CC$ is the countable union of countable sets. By Countable Union of Countable Sets is Countable, $\CC$ is itself countable. Therefore: {{begin-eqn}} {{eqn | l = \map {\mu^*} S       | o = \le       | r = \sum_{x \mathop \in \CC} \map \gamma x       | c = by definition of $\mu^*$ }} {{eqn | o = \le       | r = \sum_{n \mathop = 0}^\infty \sum_{x \mathop \in \CC_n} \map \gamma x       | c = since each $x \in \CC$ belongs to some $\CC_n$ }} {{eqn | o = \le       | r = \sum_{n \mathop = 0}^\infty \paren {\map {\mu^*} {S_n} + \frac \epsilon {2^{n + 1} } } }} {{eqn | r = \epsilon + \sum_{n \mathop = 0}^\infty \map {\mu^*} {S_n}       | c = Sum of Infinite Geometric Sequence }} {{end-eqn}} Since $\epsilon$ was arbitrary, the result follows. {{qed}} {{ACC}}
{{rewrite|Eqn templates would be good, I think}} Let $\left \langle {S_n}\right \rangle_{n \in \N}$ be a sequence of pairwise disjoint elements of $\Sigma$. Let $N \in \N$ be any natural number. By Set is Subset of Union: :$\displaystyle \bigcup_{n \mathop = 0}^N S_n \subseteq \bigcup_{n \mathop = 0}^\infty S_n$ From Non-Negative Additive Function is Monotone: :$\displaystyle f \left({\bigcup_{n \mathop = 0}^N S_n}\right) \le f \left({\bigcup_{n \mathop = 0}^\infty S_n}\right)$ Also, from Finite Union of Sets in Additive Function: :$\displaystyle f \left({\bigcup_{n \mathop = 0}^N S_n}\right) = \sum_{n \mathop = 0}^N f \left({S_n}\right)$ Hence: :$\displaystyle \sum_{n \mathop = 0}^N f \left({S_n}\right) \le f \left({\bigcup_{n \mathop = 0}^\infty S_n}\right)$ {{wtd|Lower and Upper Bounds for Sequences still needs generalization to $\overline \R$ before it can be applied here}} Taking the limit as $N \to \infty$, it follows by Lower and Upper Bounds for Sequences that: :$\displaystyle \sum_{n \mathop = 0}^\infty f \left({S_n}\right) \le f \left({\bigcup_{n \mathop = 0}^\infty S_n}\right)$ The reverse inequality holds because of the countable subadditivity of $f$, and thus equality holds. {{qed}} Category:Measure Theory qg3chhqi8z5mcud6olh2ur2a57cm05i
First, note that $\mathfrak M(\mu^*)$ is a $\sigma$-algebra over $X$. Next, choose $E_1, E_2\in\mathfrak M(\mu^*)$ such that $E_1\cap E_2 = \varnothing$. Thus: {{begin-eqn}} {{eqn | l=\mu(E_1\cup E_2)       | r=\mu^*(E_1\cup E_2)       | c=since $E_1\cup E_2\in\mathfrak M(\mu^*)$ }} {{eqn | r=\mu^*((E_1\cup E_2)\cap E_1) + \mu^*((E_1\cup E_2) - E_1)       | c=by definition of a measurable set }} {{eqn | r=\mu^*(E_1) + \mu^*(E_2)       | c=since $E_1$ and $E_2$ are disjoint }} {{eqn | r=\mu (E_1) + \mu (E_2)       | c=since $E_1, E_2\in\mathfrak M(\mu^*)$ }} {{end-eqn}} Therefore $\mu$ is additive. But because it is constructed from an outer measure, it is also countably subadditive. Since Additive and Countably Subadditive Function is Countably Additive, it follows that $\mu$ is countably additive. Finally, also because it is constructed from an outer measure, $\mu$ is nonnegative. Hence $(X, \mathfrak M(\mu^*), \mu)$ meets all the criteria of a measure space. {{qed}}
Let $\epsilon > 0$ be arbitrary. By definition of convergence a.e., there is a set $E \in \Sigma$ such that $E \subseteq D$, $\map \mu E = 0$ and: :$\displaystyle \lim_{n \mathop \to \infty} \map {f_n} x = \map f x$ for each $x \in A$ where $A = D \setminus E$ is the set difference of $D$ with $E$. For all $n, k \in \N$, define $A_{n, k}$ by: :$A_{n, k} = \set {x \in A: \size {\map {f_n} x - \map f x} \ge \dfrac 1 k}$ Also, define $B_{n, k}$ by: :$B_{n, k} = \displaystyle \bigcup_{i \mathop = n}^\infty A_{i, k}$ Since $f_n$ converges pointwise to $f$ on $A$, it follows by definition of convergence that for each $x \in A$ and $k \in \N$: :$\size {\map {f_i} x - \map f x} < \dfrac 1 k$ for all $i \in \N$ sufficiently large. Thus, when $k$ is fixed, no element of $A$ belongs to $A_{n, k}$ for infinitely many $n$.  Hence, by Definition 2 of Limit Superior of Sequence of Sets: :$\displaystyle \limsup_{n \mathop \to \infty} A_{n, k} = \O$ So we have: {{begin-eqn}} {{eqn | l = 0       | r = \map \mu \O       | c = Measure of Empty Set is Zero }} {{eqn | r = \map \mu {\limsup_{n \mathop \to \infty} A_{n, k} } }} {{eqn | r = \map \mu {\bigcap_{n \mathop = 0}^\infty B_{n, k} }       | c = {{Defof|Limit Superior of Sequence of Sets}} }} {{eqn | r = \map \mu {\limsup_{n \mathop \to \infty} B_{n, k} }       | c = {{Defof|Limit Superior of Sequence of Sets|index = 2}} }} {{eqn | r = \map \mu {\lim_{n \mathop \to \infty} B_{n, k} }       | c = since $B_{n, k}$ is a decreasing sequence of sets }} {{eqn | r = \lim_{n \mathop \to \infty} \map \mu {B_{n, k} }       | c = since $B_{n, k} \subseteq D$, $\map \mu D < \infty$ and by Continuity of Measure }} {{end-eqn}} {{MissingLinks|}} {{handwaving|last two lines}} Thus, since $k$ was arbitrary, we can associate some $n_k \in N$ with each $k$ such that $\map \mu {B_{n_k, k} } < \dfrac \epsilon {2^{k + 1} }$. Setting $B = \displaystyle \bigcup_{k \mathop \in \N} B_{n_k, k}$, we have: :$\displaystyle \map \mu B \le \sum_{k \mathop \in \N} \map \mu { B_{n_k, k} } < \sum_{k \mathop \in \N} \frac \epsilon {2^{k + 1} } = \epsilon$ by the countable additivity of $\mu$ and by Sum of Infinite Geometric Sequence. Also, given any $k$, we have that $x \in A \setminus B$ implies $x \notin B_{n_k, k}$, which means: :$\size {\map {f_i} x - \map f x} < \frac 1 k$  for all $i \ge n_k$. Since this is true for ''all'' $x \in A \setminus B$, it follows that $f_n$ converges to $f$ uniformly on $A \setminus B$. Finally, note that $A \setminus B = D \setminus \paren {E \cup B}$, and: :$\map \mu {E \cup B} \le \map \mu B + \map \mu E = \map \mu B + 0 < \epsilon$ Since $\epsilon$ was arbitrary, it follows that: :$\map \mu {E \cup B} = 0$ Hence $f_n$ converges to $f$ almost uniformly. {{qed}}
By definition of measure, there exists at least one $E \in \Sigma$ such that $\map \mu E$ is finite. So, suppose that $E \in \Sigma$ such that $\map \mu E$ is finite. Let $\map \mu E = x$. Consider the sequence $\sequence {S_n}_{n \mathop \in \N} \subseteq \Sigma$ defined as: :$S_n = \begin {cases} E & : n = 1 \\ \O & : n > 1 \end {cases}$ Then: :$\displaystyle \bigcup_{n \mathop = 1}^\infty S_n = E$ Hence: {{begin-eqn}} {{eqn | l = \map \mu E       | r = x }} {{eqn | ll= \leadsto       | l = \mu \paren {\bigcup_{n \mathop = 1}^\infty S_n}       | r = x }} {{eqn | ll= \leadsto       | l = \sum_{n \mathop = 1}^\infty \map \mu {S_n}       | r = x       | c = Property $(2)$ }} {{eqn | ll= \leadsto       | l = \map \mu E + \sum_{n \mathop = 2}^\infty \map \mu \O       | r = x }} {{eqn | ll= \leadsto       | l = \sum_{n \mathop = 2}^\infty \map \mu \O       | r = 0 }} {{end-eqn}} It follows directly that $\map \mu \O = 0$. {{qed}}
Assume $f_n$ converge a.u. to $f$ on $D$. Then for each $\epsilon > 0$ there is a $B_\epsilon \subseteq D$ with $\map \mu {B_\epsilon} < \epsilon$ outside of which $f_n$ converges uniformly to $f$. Thus, $f_n$ converges pointwise to $f$ outside of each $B_\epsilon$. Next, define $\displaystyle B \equiv \bigcap_{n \mathop \in \N} B_{\frac 1 n}$. First, $B \in \Sigma$ since $\Sigma$ is a sigma-algebra, so that $\map \mu B$ is defined. Second, note that since the measure is monotonic: :$\forall n \in \N: \map \mu B \le \map \mu {B_{\frac 1 n} } < \frac 1 n$ Hence: :$\map \mu B = 0$ But $x \in D - B$ only if $x \notin B_{\frac 1 k}$ for some $k \in \N$. So by the above: :$\map {f_n} x \to \map f x$ It follows by definition that $f_n$ converges a.e. to $f$. {{qed}}