\(A\) and \(B\) are two sets, we write \(A-B\) for the set of all elements \(x\) such that \(x\in A, x\notin B\). A family \(\mathscr R\) of set is called a ring if \(A\in\mathscr R, B\in\mathscr R\) implies \[A\cup B\in\mathscr R,\quad A-B\in\mathscr R, \quad A\cap B=A-(A-B)\in\mathscr R\] A ring \(\mathscr R\) is called a \(\sigma\)-ring if \[\overset{\infty}{\underset{n=1}{\bigcup}}A_n\in\mathscr R\] whenever \(A_n\in\mathscr R(n=1,2,\cdots)\). And if \(\mathscr R\) is a \(\sigma\)-ring, \[\overset{\infty}{\underset{n=1}{\bigcap}}A_n=A_1-\overset{\infty}{\underset{n=1}{\bigcup}}(A_1-A_n)\in\mathscr R\]
\(\phi\) is a set function defined on \(\mathscr R\) if \(\phi\) assigns to every \(A\in\mathscr R\) a number \(\phi(A)\) of the extended real number system (the range of \(\phi\) does not contain \(\pm\infty\)). \(\phi\) is additive set function if \(A\bigcap B=0\) implies \[\phi(A\bigcup B)=\phi(A)+\phi(B)\] and \(\phi\) is countably additive if \(A_i\bigcap A_j=0\quad(i\ne j)\) implies \[\phi\Bigl(\overset{\infty}{\underset{n=1}{\bigcup}}A_n\Bigr)=\sum_{n=1}^{\infty}\phi(A_n)\]
Let \(R^p\) denote \(p\)-dimensional euclidean space. The set of points \(\mathbf x=(x_1,\cdots,x_p)\) such that \[a_i\leq x_i\leq b_i\quad(i=1,\cdots,p)\] is called an interval in \(R^p\). If \(A\) is the union of a finite number of intervals, \(A\) is said to be an elementary set. If \(I\) is an interval, we define \[m(I)=\prod_{i=1}^{p}(b_i-a_i)\] If \(A=I_1\bigcup\cdots\bigcup I_n\) and if these intervals are pairwise disjoint, we set \[m(A)=m(I_1\bigcup\cdots\bigcup I_n)=m(I_1)+\cdots+m(I_n)\] Let \(\mathscr E\) denote the family of all elementary subsets of \(R^p\), then \(\mathscr E\) is a ring, but not a \(\sigma\)-ring, and \(m\) is additive on \(\mathscr E\).
A nonnegative additive set function \(\phi\) defined on \(\mathscr E\) is said to be regular if: for every \(A\in\mathscr E, \varepsilon>0\), there exist sets \(F\in\mathscr E, G\in\mathscr E\) such that \(F\) is closed, \(G\) is open, \(F\subset A\subset G\) and \[\phi(G)-\varepsilon\leq\phi(A)\leq\phi(F)+\varepsilon\]
Let \(\mu\) be additive, regular, nonnegative and finite set function on \(\mathscr E\). The countable open coverings of any set \(E\subset R^p\) by open elementary sets \(A_n\): \[E\subset\overset{\infty}{\underset{n=1}{\bigcup}}A_n\] Define \[\mu^*(E)=\text{inf}\sum_{n=1}^{\infty}\mu(A_n)\] The \(\text{inf}\) being taken over all countable coverings of \(E\) by open elementary sets. \(\mu^*(E)\) is called outer measure of \(E\), corresponding to \(\mu\). \(\mu^*(E)\ge0\) for all \(E\) and if \(E_1\subset E_2\) then \[\mu^*(E_1)\leq\mu^*(E_2)\]
For every \(A\in\mathscr E\), \(\mu^*(A)=\mu(A)\). The regularity of \(\mu\) shows that \(A\) is contained in an open elementary set \(G\) such that \(\mu(G)\leq\mu(A)+\varepsilon\). Since \(\mu^*(A)\leq\mu(G)\) then \[\mu^*(A)\leq\mu(A)\] The definition of \(\mu^*\) shows that there is a sequence \(\{A_n\}\) of open elementary sets whose union contains \(A\), such that \[\sum_{n=1}^{\infty}\mu(A_n)\leq\mu^*(A)+\varepsilon\] The regularity of \(\mu\) shows that \(A\) contains a closed elementary set \(F\) such that \(\mu(F)\ge\mu(A)-\varepsilon\) and since \(F\) is compact, we have \[F\subset A_1\cup\cdots\cup A_N\] for some \(N\). Hence \[\mu(A)\leq\mu(F)+\varepsilon\leq\mu(A_1\cup\cdots\cup A_N)+\varepsilon\leq\sum_{n=1}^{N}\mu(A_n)+\varepsilon\leq\mu^*(A)+2\varepsilon\] then \[\mu^*(A)=\mu(A)\]
If \(E=\overset{\infty}{\underset{n=1}{\bigcup}}E_n\) then \[\mu^*(E)\leq\sum_{n=1}^{\infty}\mu^*(E_n)\] Assume that \(\mu^*(E_n)<+\infty\) for all \(n\). Given \(\varepsilon>0\), there are coverings \(\{A_{nk}\},k=1,2,\cdots\) of \(E_n\) by open elementary sets such that \[\sum_{k=1}^{\infty}\mu(A_{nk})\leq\mu^*(E_n)+2^{-n}\varepsilon\] Then \[\mu^*(E_n)\leq\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\mu(A_{nk})\leq\sum_{n=1}^{\infty}\mu^*(E_n)+\varepsilon\] then \[\mu^*(E)\leq\sum_{n=1}^{\infty}\mu^*(E_n)\]
For any \(A\subset R^p, B\subset R^p\), we define \[S(A,B)=(A-B)\cup(B-A), \quad d(A,B)=\mu^*(S(A,B))\] We write \(A_n\to A\) if \[\lim_{n\to\infty}d(A,A_n)=0\] If there is a sequence \(\{A_n\}\) of elementary sets such that \(A_n\to A\), we say that \(A\) is finitely \(\mu\)-measurable and write \(A\in\mathfrak M_F(\mu)\). If \(A\) is the union of a countable collection of finitely \(\mu\)-measurable sets, we say that \(A\) is \(\mu\)-measurable and write \(A\in\mathfrak M(\mu)\). \(S(A,B)\) is the symmetric difference of \(A\) and \(B\). We have \[S(A,B)=S(B,A),\quad S(A,A)=0\]
Since \[(A-B)\subset(A-C)\cup(C-B)\] \[(B-A)\subset(B-C)\cup(C-A)\] then \[S(A,B)=(A-B)\cup(B-A)\subset\Bigl((A-C)\cup(C-B)\cup(B-C)\cup(C-A)\Bigr)=S(A,C)\cup S(C,B)\] so \[S(A,B)\subset S(A,C)\cup S(C,B)\]
\[S(A_1\cup A_2,B_1\cup B_2)=\Bigl(((A_1\cup A_2)-(B_1\cup B_2))\cup((B_1\cup B_2)-(A_1\cup A_2))\Bigr)\\ \subset\Bigl((A_1-B_1)\cup(A_2-B_2)\Bigr)\cup\Bigl((B_1-A_1)\cup(B_2-A_2)\Bigr)\\ =S(A_1,B_1)\cup S(A_2,B_2)\] \[\begin{align} S(A_1\cap A_2,B_1\cap B_2)&=S((A_1^c\cup A_2^c)^c,(B_1^c\cup B_2^c)^c)\\ &=S(A_1^c\cup A_2^c,B_1^c\cup B_2^c)\\ &\subset S(A_1^c, B_1^c)\cup S(A_2^c, B_2^c)\\ &=S(A_1, B_1)\cup S(A_2, B_2) \end{align}\] \[\begin{align} S(A_1- A_2,B_1- B_2)&=S(A_1\cap A_2^c,B_1\cap B_2^c)\\ &\subset S(A_1, B_1)\cup S(A_2^c, B_2^c)\\ &=S(A_1, B_1)\cup S(A_2, B_2) \end{align}\] then \[\left. \begin{array}{l} S(A_1\cup A_2,B_1\cup B_2)\\ S(A_1\cap A_2,B_1\cap B_2)\\ S(A_1- A_2,B_1- B_2)\\ \end{array} \right\} \subset S(A_1, B_1)\cup S(A_2, B_2)\]
\[d(A,B)=\mu^*(S(A,B))=\mu^*(S(B,A))=d(B,A)\] \[d(A,A)=\mu^*(S(A,A))=\mu^*(0)=0\] \[d(A,B)=\mu^*(S(A,B))\leq\mu^*(S(A,C)\cup S(C,B))\leq\mu^*(S(A,C))+\mu^*(S(C,B))=d(A,C)+d(C,B)\] then \[\left. \begin{array}{l} d(A_1\cup A_2,B_1\cup B_2)\\ d(A_1\cap A_2,B_1\cap B_2)\\ d(A_1- A_2,B_1- B_2)\\ \end{array} \right\} \leq d(A_1, B_1)+ d(A_2, B_2)\]
If at least one of \(\mu^*(A),\mu^*(B)\) is finite, suppose \(0\leq\mu^*(B)\leq\mu^*(A)\), then \[d(A,0)\leq d(A,B)+d(B,0)\] that is \[\mu^*(A)\leq d(A,B)+\mu^*(B)\] since \(\mu^*(B)\) is finite, it follows that \[\mu^*(A)-\mu^*(B)\leq d(A,B)\] and \[\lvert\mu^*(A)-\mu^*(B)\rvert\leq d(A,B)\]
\(\mathfrak M(\mu)\) is a \(\sigma\)-ring, and \(\mu^*\) is countably additive on \(\mathfrak M(\mu)\). (a) Suppose \(A\in\mathfrak M_F(\mu), B\in\mathfrak M_F(\mu)\), choose elementary subsets of \(R^p\), \(\{A_n\}, \{B_n\}\) such that \(A_n\in\mathscr E, B_n\in\mathscr E, A_n\to A,B_n\to B\), then \[A_n\cup B_n\to A\cup B\] \[A_n\cap B_n\to A\cap B\] \[A_n- B_n\to A- B\] \[\mu^*(A_n)\to\mu^*(A)\] Since \[A_n\cup B_n\to A\cup B\] and \[A_n- B_n\to A- B\] \(\mathfrak M_F(\mu)\) is a ring. Since \[\mu(A_n)+\mu(B_n)=\mu(A_n\cup B_n)+\mu(A_n\cap B_n)\] then \[\mu^*(A_n)+\mu^*(B_n)=\mu^*(A_n\cup B_n)+\mu^*(A_n\cap B_n)\] and since \[\mu^*(A_n)\to\mu^*(A)\] then \[\mu^*(A)+\mu^*(B)=\mu^*(A\cup B)+\mu^*(A\cap B)\] if \(A\cap B=0\) then \[\mu^*(A\cap B)=0\] it follows that \(\mu^*\) is additive on \(\mathfrak M_F(\mu)\). Now let \(A\in\mathfrak M(\mu)\), then \(A\) can be represented as the union of a countable collection of disjoint sets of \(\mathfrak M_F(\mu)\). If \(A=\bigcup A_n'\) with \(A_n'\in\mathfrak M_F(\mu)\), write \(A_1=A_1'\) and \[A_n=(A_1'\cup\cdots\cup A_n')-(A_n'\cup\cdots\cup A_{n-1}')\quad(n=2,3,4,\cdots)\] then \[A=\overset{\infty}{\underset{n=1}{\bigcup}}A_n\] is the required representation. Then \[\mu^*(A)\leq\sum_{n=1}^{\infty}\mu^*(A_n)\] (b) On the other hand, \(A\supset A_1\cup\cdots\cup A_n\) and by the additivity of \(\mu^*\) on \(\mathfrak M_F(\mu)\) we obtain \[\mu^*(A)\ge\mu^*(A_1\cup\cdots\cup A_n)=\mu^*(A_1)+\cdots+\mu^*(A_n)\] then \[\mu^*(A)=\sum_{n=1}^{\infty}\mu^*(A_n)\] Suppose \(\mu^*(A)\) is finite, Put \(B_n=A_1\cup\cdots\cup A_n\) then \[\begin{align} d(A,B_n)&=\mu^*(S(A,B_n))\\ &=\mu^*((A-B_n)\cup(B_n-A))\\ &=\mu^*(\overset{\infty}{\underset{i=n+1}{\bigcup}}A_i)\\ &=\sum_{i=n+1}^{\infty}\mu^*(A_i)\to0 \end{align}\] as \(n\to\infty\) Hence \(B_n\to A\), and since \(B_n\in\mathfrak M_F(\mu)\) then \(A\in\mathfrak M_F(\mu)\). We have thus shown that \(A\in\mathfrak M_F(\mu)\) if \(A\in\mathfrak M(\mu)\) and \(\mu^*(A)<+\infty\) then \(\mu^*\) is countably additive on \(\mathfrak M(\mu)\) If \[A=\bigcup A_n\] where \(\{A_n\}\) is a sequence of disjoint sets of \(\mathfrak M(\mu)\) we ahve shown that \[\mu^*(A)=\sum_{n=1}^{\infty}\mu^*(A_n)\] holds if \(\mu^*(A_n)<+\infty\) for every \(n\). (c) Finally we have to show that \(\mathfrak M(\mu)\) is a \(\sigma\)-ring. If \(A_n\in\mathfrak M(\mu), n=1,2,3,\cdots\), since the sum of countable sets is countable, then \(\bigcup A_n\in\mathfrak M(\mu)\). Suppose \(A\in\mathfrak M(\mu), B\in\mathfrak M(\mu)\) and \[A=\overset{\infty}{\underset{n=1}{\bigcup}}A_n,\quad B=\overset{\infty}{\underset{n=1}{\bigcup}}B_n\] where \(A_n, B_n\in\mathfrak M_F(\mu)\), then \[A_n\cap B=\overset{\infty}{\underset{i=1}{\bigcup}}(A_n\cap B_i)\in\mathfrak M(\mu)\] and since \[\mu^*(A_n\cap B)\leq\mu^*(A_n)<+\infty\] \[A_n\cap B\in\mathfrak M_F(\mu)\] Hence \(A_n-B\in\mathfrak M_F(\mu)\) and since \[A-B=\overset{\infty}{\underset{n=1}{\bigcup}}(A_n-B)\] then \(A-B\in\mathfrak M(\mu)\) Then \(\mathfrak M(\mu)\) is a \(\sigma\)-ring.
We replace \(\mu^*(A)\) by \(\mu(A)\) if \(A\in\mathfrak M(\mu)\). Thus \(\mu\), originally only defined on \(\mathscr E\), is extended to a countably additive set function on the \(\sigma\)-ring \(\mathfrak M(\mu)\). This extended set function is called a measure. The special case \(\mu=m=\prod_{i=1}^{p}(b_i-a_i)\) is called the Lebesgue measure on \(R^p\).
Set \(E\) is called a Borel set if \(E\) can be obtained by a countable number of operations, starting from open sets, each operation consisting in taking unions, intersections or complements. The collection of all Borel sets \(\mathscr B\) in \(R^p\) is a \(\sigma\)-ring and if \(E\in\mathscr B\) then \(E\in\mathfrak M(\mu)\).
A set \(X\) is said to be a measure space if there exists a \(\sigma\)-ring \(\mathfrak M\) of subsets of \(X\) (which are called measurable sets) and a non-negative countably additive set function \(\mu\) (which is called a measure), defined on \(\mathfrak M\). If \(X\in\mathfrak M\), then \(X\) is said to be a measurable space.
Let \(f\) be a function defined on the measurable space \(X\), with values in the extended real number system. The function \(f\) is said to be measurable if the set \[\{x|f(x)>a\}\] is measurable for every real \(a\).
Let \(s\) be a real-valued function defined on \(X\), if the range of \(s\) is finite, we say that \(s\) is a simple function. Let \(E\subset X\) and put \[K_E(x)=\begin{cases} 1, & (x\in E) \\[2ex] 0, & (x\notin E) \end{cases}\] \(K_E\) is called the characteristic function of \(E\).
Every real function can be approximated by simple functions Let \(f\) be a real function on \(X\), there exists a sequence \(\{s_n\}\) of simple functions such that \(s_n\to f(x)\) as \(n\to+\infty\) for every \(x\in X\). If \(f\) is measurable, \(\{s_n\}\) may be chosen to be a sequence of measurable functions. If \(f\ge 0\), \(\{s_n\}\) may be chosen to be a monotonically increasing sequence. If \(f\ge0\), define \[E_{ni}=\Bigl\{x\Bigl|\frac{i-1}{2^n}\leq f(x)<\frac{i}{2^n}\Bigr\},\quad F_n=\{x|f(x)\ge n\}\] put \[s_n=\sum_{i=1}^{n2^n}\frac{i-1}{2^n}K_{E_{ni}}+nK_{F_n}\] In the general case let \[f=f^+-f^-=\text{max}(f,0)-(-\text{min}(f,0))=\text{max}(f,0)+\text{min}(f,0)\] and apply the preceding construction to \(f^+\) and \(f^-\).
Suppose the range of simple function \(s\) consists of the distinct numbers \(c_1,\cdots,c_n\) Let \(E_i=\{x|s(x)=c_i\},\quad(i=1,\cdots,n)\) and \[K_{E_i}(x)=\begin{cases} 1, & (x\in E) \\[2ex] 0, & (x\notin E) \end{cases}\] Suppose \[s(x)=\sum_{i=1}^{n}c_iK_{E_i}(x)\quad(x\in X, c_i>0)\] is measurable and suppose \(E\in\mathfrak M\). We define \[I_E(s)=\sum_{i=1}^{n}c_i\mu(E\cap E_i)\] If \(f\) is measurable and nonnegative, we define \[\int_Efd\mu=\text{sup }I_E(s)\] where the sup is taken over all measurable simple functions \(s\) such that \(0\leq s\leq f\). \(\int_Efd\mu\) is called the Lebesgue integral of \(f\), with respect to the measure \(\mu\), over the set \(E\).
Let \(f\) be measurable and if at least one of the two integrals \[\int_Ef^+d\mu=\int_E\text{max}(f,0)d\mu,\quad\int_Ef^-d\mu=\int_E-\text{min}(f,0)d\mu\] is finite, we define \[\int_Efd\mu=\int_Ef^+d\mu-\int_Ef^-d\mu\] If both integrals are finite, then \(\int_Efd\mu\) is finite and we say that \(f\) is Lebesgue integrable on \(E\) with respect to \(\mu\), we write \(f=\mathscr L(\mu)\) on \(E\).\(f\) is integrable on \(E\) only if its integral over \(E\) is finite.
Suppose \(f\) is measurable and nonnegative on \(X\), for \(A\in \mathfrak M\), define \[\phi(A)=\int_Afd\mu\] then \(\phi\) is countably additive on \(\mathfrak M\). To prove this, we have to show that if \(A_n\in\mathfrak M(n=1,2,3,\cdots), A_i\cap A_j=0\) for \(i\ne j\), and \(A=\bigcup_1^{\infty}A_n\), \[\phi(A)=\sum_{n=1}^{\infty}\phi(A_n)\] If \(f\) is a characteristic function, then the countable additivity of \(\phi\) is precisely the same as the countable additivity of \(\mu\), since \[\int_AK_Ed\mu=\mu(A\cap E)\] If \(f\) is simple, the \(f\) is of the form \[s(x)=\sum_{i=1}^{n}c_iK_{E_i}(x)\quad(x\in X, c_i>0)\] and countably additive. In the general case, we have, for every measurable simple function \(s\) such that \(0\leq s\leq f\), \[\int_Asd\mu=\sum_{n=1}^{\infty}\int_{A_n}sd\mu\leq\sum_{n=1}^{\infty}\phi(A_n)\] therefore by \[\int_Efd\mu=\text{sup }I_E(s)\] we have \[\phi(A)\leq\sum_{n=1}^{\infty}\phi(A_n)\] Suppose \(\phi(A_n)<+\infty\) for every \(n\). Given \(\varepsilon>0\) we can choose a measurable function \(s\) such that \(0\leq s \leq f\) and such that \[\int_{A_1}sd\mu\ge\int_{A_1}fd\mu-\varepsilon,\quad \int_{A_2}sd\mu\ge\int_{A_2}fd\mu-\varepsilon\] Hence \[\phi(A_1\cup A_2)\ge\int_{A_1\cup A_2}sd\mu=\int_{A_1}sd\mu+\int_{A_2}sd\mu\ge\phi(A_1)+\phi(A_2)-2\varepsilon\] so that \[\phi(A_1\cup A_2)\ge \phi(A_1)+\phi(A_2)\] It follows that we have for every \(n\), \[\phi(A_1\cup\cdots\cup A_n)\ge \phi(A_1)+\cdots+\phi(A_n)\] Since \(A\supset A_1\cup\cdots\cup A_n\), then \[\phi(A)\ge \sum_{n=1}^{\infty}\phi(A_n)\] then \[\phi(A)=\sum_{n=1}^{\infty}\phi(A_n)\]
Let us write \(f\sim g\) on \(E\) if the set \[\{x|f(x)\ne g(x)\}\cap E\] has measure zero. Then \(f\sim g\) implies \(g\sim f\); and \(f\sim g, g\sim h\) implies \(f\sim h\). If \(f\sim g\) on \(E\) we have \[\int_Afd\mu=\int_Agd\mu\] provided the integrals exist, for every measurable subset \(A\) of \(E\).
If \(f\in\mathscr L(\mu)\) on \(E\), then \(|f|\in\mathscr L(\mu)\) and \[\Biggl|\int_Efd\mu\Biggl|\leq\int_E|f|d\mu\]. For \(A\in \mathfrak M,\quad \phi(A)=\int_Afd\mu\) then \(\phi\) is countably additive on \(\mathfrak M\) and \[\phi(A)=\sum_{n=1}^{\infty}\phi(A_n)\] Write \(E=E_1+E_2\), where \(f(x)\ge0\) on \(A\) and \(f(x)<0\) on \(B\), then \[\int_E|f|d\mu=\int_{E_1}|f|d\mu+\int_{E_2}|f|d\mu=\int_{E_1}f^+d\mu+\int_{E_2}f^-d\mu<+\infty\] so that \(|f|\in\mathscr L(\mu)\). Since \(f\leq|f|\) and \(-f\leq|f|\) we see that \[\int_Efd\mu\leq\int_E|f|d\mu,\quad -\int_Efd\mu\leq\int_E|f|d\mu\] then \[\Biggl|\int_Efd\mu\Biggl|\leq\int_E|f|d\mu\]
Suppose \(E\in \mathfrak M\), let \(\{f_n\}\) be a sequence of measurable functions such that \[0\leq f_1(x)\leq f_2(x)\leq\cdots\leq \infty\quad(x\in E)\] Let \(f\) be defined by \[f_n(x)\to f(x)\quad(x\in E)\] as \(n\to\infty\) Then \[\int_Ef_nd\mu\to\int_Efd\mu\quad(n\to\infty)\] This theorem is called Lebesgue’s monotone convergence theorem. Since \[0\leq f_1(x)\leq f_2(x)\leq\cdots\leq \infty\quad(x\in E)\] it is clear that as \(n\to\infty\), \[\int_Ef_nd\mu\to a\] for some \(a\); and since \(\int f_n\leq\int f\), we have \[a\leq\int_E fd\mu\] Choose \(c\) such that \(0<c<1\) and let \(s\) be a simple measurable function such that \(0\leq s\leq f\). Put \[E_n=\{x|f_n(x)\ge cs(x)\}\quad(n=1,2,3,\cdots)\] then each \(E_n\) is measurable, and \(E_1\subset E_2\subset E_3\subset \cdots\) and since for \(x\in E\), if \(f(x)>0\), then \(x\in E_1\), if \(f(x)>0\), then \(cs(x)<f(x)\), hence \(x\in E_n\) for some \(n\), then \[E=\overset{\infty}{\underset{n=1}{\bigcup}}E_n\] Since \(f_n(x)\to f(x)\quad(x\in E),\text{ as } n\to \infty\) For every \(n\), \[\int_Ef_nd\mu\ge\int_{E_n}f_nd\mu\ge c\int_{E_n}sd\mu\] Since the integral is a countably additive set function, and since \[E=\overset{\infty}{\underset{n=1}{\bigcup}}E_n\] as \(n\to \infty\) then \[a\ge \int_{E_n}f_nd\mu\ge c\int_{E_n}sd\mu =c\int_{E}sd\mu\] let \(c\to1\) we see that \[a\ge \int_{E}sd\mu\] since \[\int_Efd\mu=\text{sup }\int_Esd\mu\] then \[a\ge\int_Efd\mu\] then \[\int_Ef_nd\mu\to\int_Efd\mu\quad(n\to\infty)\]
If \(f_n:X\to[0,\infty]\) is measurable, for \(n=1,2,3,\cdots\) and \[f(x)=\sum_{n=1}^{\infty}f_n(x)\quad(x\in X)\] there are sequences of simple measurable functions \(\{s_i'\}\), \(\{s_i''\}\) such that \(s_i'\to f_1\) and \(s_i''\to f_2\). If \(s_i=s_i'+s_i''\), then \(s_i\to f_1+f_2\) and the monotone convergence theorem shows that \[\int_X(f_1+f_2)d\mu=\int_Xf_1d\mu+\int_Xf_2d\mu\] Next, put \(g_N=f_1+\cdots+f_N\). The sequence \(\{g_N\}\) converges monotonically to \(f\), and \[\int_Xg_Nd\mu=\sum_{n=1}^{N}\int_Xf_nd\mu\] Applying the monotone convergence theorem once more, we obtain \[\int_Xfd\mu=\sum_{n=1}^{\infty}\int_Xf_nd\mu\]
Fatou’s theorem Suppose \(E\in\mathfrak M\) If \(\{f_n\}\) is a sequence of nonnegative measurable function. For \(n=1,2,3,\cdots\) and \(x\in E\), \[g_k(x)=\inf_{i\ge k} f_i(x)\quad(k=1,2,3,\cdots; x\in E)\] Then \(g_k\leq f_k\) and \(g_k\) is measurable on \(E\), and \[0\leq g_1(x)\leq g_2(x)\leq\cdots\] then \[\int_Eg_kd\mu\leq \int_Ef_kd\mu\quad(k=1,2,3,\cdots)\] and by definition \[g_k(x)\to\lim\inf f_n(x)\quad(k\to\infty)\] then the monotone convergence theorem shows therefore that as \(k\to\infty\), \[\int_Eg_kd\mu\to\int_E\Biggl(\lim_{n\to\infty}\text{inf }f_n\Biggr)d\mu\] and \[\int_Ef_kd\mu\to\lim_{n\to\infty}\inf\int_E f_nd\mu\] then \[\int_E\Biggl(\lim_{n\to\infty}\inf f_n\Biggr)d\mu\leq\lim_{n\to\infty}\inf\int_E f_nd\mu\quad(x\in E)\]
Suppose \(f:X\to[0,\infty]\) is measurable, and \[\varphi(E)=\int_Efd\mu\quad(E\in\mathfrak M)\] Let \(E_1,E_2,E_3,\cdots\) be disjoint members of \(\mathfrak M\) whose union is \(E\). Then \[\chi_Ef=\sum_{j=1}^{\infty}\chi_{E_j}f\] and \[\varphi(E)=\int_X\chi_Efd\mu\] \[\varphi(E_j)=\int_X\chi_{E_j}fd\mu\] Then \[\varphi(E)=\sum_{j=1}^{\infty}\varphi(E_j)\] since \(\varphi(\varnothing)=0\) then \(\varphi\) is a measure on \(\mathfrak M\).
And for every measurable \(g\) on \(X\) with range in \([0,\infty]\), then whenever \(g=\chi_E\) for some \(E\in\mathfrak M\) \[\int_Xgd\varphi=\int_Xgfd\mu\] or \[d\varphi=fd\mu\] holds for every simple measurable function \(g\), and the general case follows from the monotone convergence theorem.We define \(L^1(\mu)\) to be the collection of all complex measurable functions \(f\) on \(X\) for which \[\int_X|f|d\mu=\int_X(f^++f^-)d\mu<\infty\] since \(f\) is measurable, then \(|f|\) is measurable. The members of \(L^1(\mu)\) are called Lebesgue integrable functions (with respect to \(\mu\)). If \(f=u+iv\), where \(u\) and \(v\) are real measurable functions on \(X\), and if \(f\in L^1(\mu)\), we define \[\int_Xfd\mu=\int_Xf^+d\mu-\int_Xf^-d\mu=\int_Xu^+d\mu-\int_Xu^-d\mu+i\int_Xv^+d\mu-i\int_Xv^-d\mu\] for every measurable set \(E\). The four functions \(u^+,u^-,v^+,v^-\) are measurable, real, and nonnegative, hence the four integrals exist with the definition \[\int_Efd\mu=\sup\int_E\sum_{i=1}^{n}a_i\chi_{A_i}d\mu\] where \(a_1,a_2,\cdots,a_n\) are the distinct values of a simple function and \(0\leq\sum_{i=1}^{n}a_i\chi_{A_i}\leq f\). Furthermore, we have \(u^+\leq|u|=u^++u^-<|f|=f^++f^-<\infty\), so that each of these four integrals is finite.
Lebesgue’s dominated convergence theorem Suppose \(E\in\mathfrak M\) Let \(\{f_n\}\) be a sequence of measurable functions such that \[f_n(x)\to f(x)\quad(x\in E, n\to\infty)\] If there exists a function \(g\in L^1(\mu)\) on \(E\), such that \[|f_n(x)|\leq g(x)\quad(n=1,2,3,\cdots,x\in E)\] \(\{f_n\}\) is said to be dominated by \(g\), then \[\lim_{n\to\infty}\int_Ef_nd\mu=\int_Efd\mu\] Since \(|f_n(x)|\leq g(x)\), we have \(f^+\leq g\) and \(f^-\leq g\), then \(f_n\in L^1(\mu)\) and \(f\in L^1(\mu)\) on \(E\). Since \(f_n+g\ge 0\), Fatou’s theorem shows that \[\int_E(f+g)d\mu\leq\lim_{n\to\infty}\text{inf }\int_E (f_n+g)d\mu\] or \[\int_Efd\mu\leq\lim_{n\to\infty}\text{inf }\int_E f_nd\mu\] Since \(g-f_n\ge0\) then \[\int_E(g-f)d\mu\leq\lim_{n\to\infty}\text{inf }\int_E (g-f_n)d\mu\] so that \[-\int_Efd\mu\leq\lim_{n\to\infty}\text{inf }\Biggl[-\int_Ef_nd\mu\Biggr]\] which is the same as \[\int_Efd\mu\ge\lim_{n\to\infty}\text{sup }\int_Ef_nd\mu\] then \[\lim_{n\to\infty}\int_Ef_nd\mu=\int_Efd\mu\]
If \(\mu\) is a measure on a \(\sigma-algebra\) \(\mathfrak M\) and if \(E\in \mathfrak M\), the statement" \(P\) holds almost everywhere on \(E\)" (abbreviated to “\(P\) holds \(a.e.\) on \(E\)”) means that there exists an \(N\in \mathfrak M\) such that \(\mu(N) = 0, N\subset E\), and \(P\) holds at every point of \(E - N\). If \(f\) and \(g\) are measurable function~ and if \[\mu\Bigl(\{x:f(x)\ne g(x)\}\Bigr)=0\] we say that \(f=g\; a.e.\;[\mu]\) on \(X\) and we write \(f\sim g\). If \(f\sim g\) for every \(E\in\mathfrak M\), \[\int_Efd\mu=\int_Egd\mu\], where \(E\) is the union of the disjoint sets \(E-N\) and \(E\cap N\). On \(E-N\), \(f=g\) and \(\mu(E\cap N)=0\). Thus sets of measure \(0\) are negligible in integration.
Let \((X,\mathfrak M,\mu)\) be a measure space, let \(\mathfrak M^*\) be the collection of all \(E\subset X\) for which there exist sets \(A\) and \(B\in\mathfrak M\) such that \(A\subset E\subset B\) and \(\mu(B-A)=0\), and define \(\mu(E)=\mu(A)\) in this situation.
First, suppose \(A\subset E\subset B\), \(A_1\subset E\subset B_1\), and \(\mu(B-A)=\mu(B_1-A_1)=0\). Since \[A-A_1\subset E-A_1\subset B_1-A_1\] we have \(\mu(A-A_1)=0\), hence \(\mu(A)=\mu(A\cap A_1)\) and similarly \(\mu(A_1)=\mu(A_1\cap A)\), then \(\mu(A_1)=\mu(A)\), then \(\mu\) is well defined for every \(E\in\mathfrak M^*\).
Next, (i) Since \(X\in\mathfrak M\) and \(\mathfrak M\subset\mathfrak M^*\) then \(X\in \mathfrak M^*\). (ii) If \(A\subset E\subset B\), then \(B^c\subset E^c\subset A^c\), since \[A^c-B^c=A^c\cap B=B-A\] thus \(E\in\mathfrak M^*\) implies \(E^c\in\mathfrak M^*\). (iii) If \(A_i\subset E_i\subset B_i,\; E=\bigcup E_i,\; A=\bigcup A_i,\; B=\bigcup B_i\), then \(A\subset E\subset B\) and \[B-A=\overset{\infty}{\underset{1}{\bigcup}}(B_i-A)\subset\overset{\infty}{\underset{1}{\bigcup}}(B_i-A_i)\] Since countable unions of sets of measure zero have measure zero, then \(E\in\mathfrak M^*\) if \(E_i\in\mathfrak M^*,\; i=1,2,3,\cdots\). So \(\mathfrak M^*\) has the three defining properties of a \(\sigma-algebra\), Then \(\mathfrak M^*\) is a \(\sigma-algebra\).
Finally, if the sets \(E_i\) are disjoint, the same is true of the sets \(A_i\), and we conclude that \[\mu(E)=\mu(A)=\sum_{1}^{\infty}\mu(A_i)=\sum_{1}^{\infty}\mu(E_i)\] so \(\mu\) is countably additive on \(\mathfrak M^*\) and \(\mu\) is a measure on \(\mathfrak M^*\). This extended measure \(\mu\) is called complete, since all subsets of sets of measure \(0\) are now measurable; the \(\sigma-algebra\) \(\mathfrak M^*\) is called the \(\mu-completion\) of \(\mathfrak M\).Suppose \(\{f_n\}\) is a sequence of complex measurable functions defined a.e. on \(X\) such that \[\sum_{n=1}^{\infty}\int_X|f_n|d\mu<\infty\]
Let \(S_n\) be the set on which \(f_n\) is defined, so that \(\mu(S_n^c)=0\). For \(x\in S=\bigcap S_n\), \(\mu(S^c)=0\) since \[\sum_{n=1}^{\infty}\int_X|f_n|d\mu=\int_X\sum_{n=1}^{\infty}|f_n|d\mu=\int_S\sum_{n=1}^{\infty}|f_n|d\mu<\infty\] If \(E=\{x\in S:\sum|f_n(x)|<\infty\}\) then \(\mu(E^c)=0\) Then the series \[f(x)=\sum_{n=1}^{\infty}f_n(x)\] converges absolutely for every \(x\in E\), converges for almost all \(x\in X\), and if \(f(x)=\sum_{n=1}^{\infty}f_n(x)\) for \(x\in E\), then \(|f(x)|\leq \sum|f_n(x)|\) on \(E\), since \[\int_S\sum_{n=1}^{\infty}|f_n|d\mu<\infty\] so that \(f\in L^1(\mu)\) on \(E\). If \(g_n=f_1+\cdots+f_n\), then \(|g_n|\leq\sum|f_n(x)|\), \(g_n(x)\to f(x)\) for all \(x\in E\) and Lebesgue’s Dominated Convergence Theorem shows \[\int_Xf(x)d\mu=\lim_{n\to\infty}\int_Xg_n(x)d\mu=\int_X\sum f_n(x)d\mu=\sum_{n=1}^{\infty}\int_Xf_n(x)d\mu\]If \(f\in L^1(\mu)\), then there is a complex number \(a\) with \(|a|=1\), which is the conjugate of \[\frac{\int_Xfd\mu}{|\int_Xfd\mu|}\] such that \(a\int_Xfd\mu=|\int_Xfd\mu|\). Let \(u\) be the real part of \(af\), then \(u\leq|af|=|f|\). Since \(\int_Xafd\mu\) is real, then \[\Biggl|\int_Xfd\mu\Biggr|=a\int_Xfd\mu=\int_Xafd\mu=\int_Xud\mu\leq\int_X|f|d\mu\]
Suppose \(\mu(X)<\infty\),\(f\in L^1(\mu)\), \(S\) is a closed set in the complex plane, and the averages \[A_E(f)=\frac{1}{\mu(E)}\int_Efd\mu\] lie in \(S\) for every \(E\in \mathfrak M\) with \(\mu(E)>0\). Let \(\Delta\) be a closed circular disc (with center at \(a\) and radius \(r>0\)) in \(S^c\). Since \(S^c\) is the union of countably many such discs. If \(\mu(E)>0, E=f^{-1}(\Delta)\), since \(A_E(f)\in S\), \[|A_E(f)-a|=\frac{1}{\mu(E)}\Biggl|\int_E(f-a)d\mu\Biggr|\leq\frac{1}{\mu(E)}\int_E|f-a|d\mu\leq r\] is impossible, hence \(\mu(E)=0, E=f^{-1}(\Delta)\).
References:
1. Rudin W. Real and complex analysis. Tata McGraw-hill education; 2006.