拓扑含空全交并，
开由开得即连续。
西代含空全补并，
开由可测即可测。
开由博雷即博雷，
可测博雷由可测。
Topology contains empty, whole, intersection and union sets,
continuous function maps onto open sets from open sets.
\(\sigma-slgebra\) contains empty, whole, complement and union sets,
measurable function maps onto open sets from measurable set.
Borel mapping maps onto open sets from Borel sets,
measurable function maps onto Borel sets from subset of \(\sigma-slgebra\) (measurable sets).
The empty set is \(\varnothing\). A collection \(\tau\) of subsets of a set \(X\) is said to be a topology in \(X\) if \(\tau\) has the following properties: (i) \(\varnothing\in\tau, X\in\tau\). (ii) If \(V_i\in\tau, i=1,2,\cdots,n\), then \(V_1\cap V_2\cap \cdots\cap V_n\in\tau\). (iii) If \(\{V_a\}\) is an arbitrary collection of members of \(\tau\), then \(\bigcup_aV_a\in\tau\). If \(\tau\) is a topology in \(X\), then \(X\) is called a topological space, and the members of \(\tau\) are called the open sets in \(X\). If \(X\) and \(Y\) are topological spaces and if \(f\) is a mapping of \(X\) into \(Y\), then \(f\) is said to be continuous provided that \(f^{-1}(V)\) is an open set in \(X\) for every open set \(V\) in \(Y\).
A collection \(\mathfrak M\) of subsets of a set \(X\) is said to be a a \(\sigma-algebra\) in \(X\) if \(\mathfrak M\) has the following properties: (i) \(X\in\mathfrak M\) (ii) If \(A\in \mathfrak M\), then \(A^c\in\mathfrak M\), where \(A^c\) is the complement of \(A\) relative to \(X\). (iii) If \(A=\bigcup_{n=1}^{\infty}A_n\) and if \(A_n\in\mathfrak M, n=1,2,\cdots,\) then \(A\in\mathfrak M\). If \(\mathfrak M\) is a \(\sigma\)-algebra in \(X\), then \(X\) is called a measurable space and the members of \(\mathfrak M\) are called the measurable sets in \(X\). If \(X\) is a measurable space, \(Y\) is a topological space and \(f\) is a mapping of \(X\) into \(Y\), then \(f\) is said to be measurable provided that \(f^{-1}(V)\) is a measurable set in \(X\) for every open set \(V\) in \(Y\).
A metric space is a set \(X\) in which a distance function (or matric) \(\rho\) is defined, with the following properties: (i) \(0\leq\rho(x,y)<\infty\) for all \(x, y\in X\) (ii) \(\rho(x,y)=0\) if and only if \(x=y\) (iii) \(\rho(x,y)=\rho(y,x)\) for all \(x, y\in X\) (iiii) \(\rho(x,y)\leq\rho(x,z)+\rho(z,y)\) for all \(x,y,z\in X\).
A mapping \(f\) of a metric space \(X\) into a metric space \(Y\) is continuous on \(X\) if and only if \(f^{-1}(V)\) is open in \(X\) for every open set \(V\) in \(Y\). Suppose \(f\) is continuous on \(X\) and \(V\) is an open set in \(Y\), we have to show that every point of \(f^{-1}(V)\) is an interior point of \(f^{-1}(V)\). Suppose \(p\in X\) and \(f(p)\in V\), since \(V\) is open, there exists \(\varepsilon>0\) that \(y\in V\) if \(d_Y(f(p),y)<\varepsilon\) and since \(f\) is continuous at \(p\), there exists \(\delta>0\) such that \(d_Y(f(x),f(p))<\varepsilon\) if \(d_X(x,p)<\delta\). Thus \(x\in f^{-1}(V)\) if \(d_X(x,p)<\delta\). Conversely, suppose \(f^{-1}(V)\) is open in \(X\) for every open set \(V\) in \(Y\). Fix \(p\in X\) and \(\varepsilon>0\), let \(V\) be the set of all \(y\in Y\) such that \(d_Y(y,f(p))<\varepsilon\), then \(V\) is open. Hence \(f^{-1}(V)\) is open, there exists \(\delta>0\) such that \(x\in f^{-1}(V)\) if \(d_X(p,x)<\delta\). But if \(x\in f^{-1}(V)\), then \(f(x)\in V\), so that \(d_Y(f(x),f(p))<\varepsilon\), then \(f\) is continuous on \(X\).
The local definition of continuity: A mapping \(f\) of \(X\) into \(Y\) is said to be continuous at the point \(x_0\in E\) if to every neighborhood (open set) \(V\) of \(f(x_0)\), there corresponds a neighborhood \(W\) of \(x_0\) such that \(f(W)\subset V\). The global definition of continuity: Let \(X\) and \(Y\) be topological spaces. A mapping \(f\) of \(X\) into \(Y\) is continuous if and only if \(f\) is continuous at every point of \(X\). If \(f\) is continuous and \(x_0\in X\), then \(f^{-1}(V)\) is a neighborhood of \(x_0\), for every neighborhood \(V\) of \(f(x_0)\). Since \(f(f^{-1}(V))\subset V\), it follows that \(f\) is continuous at \(x_0\). If \(f\) is continuous at every point of \(X\) and if \(V\) is open in \(Y\), every point \(x\in f^{-1}(V)\) has a neighborhood \(W_x\) such that \(f(W_x)\subset V\). Therefore \(W_x\subset f^{-1}(V)\). It follows that \(f^{-1}(V)\) is the union of the open sets \(W_x\), so \(f^{-1}(V)\) is itself open. Thus \(f\) is continuous.
Let \(X,Y,Z\) be topological spaces, if \(f:X\to Y\) and \(g:Y\to Z\) are continuous, if \(h=g\circ f\) then \(h:X\to Z\) is continuous. If \(V\) is open in \(Z\), then \(g^{-1}(V)\) is open in \(Y\), and \[h^{-1}(V)=f^{-1}(g^{-1}(V))\] if \(f\) is continuous, it follows that \(h^{-1}(V)\) is open, then \(h:X\to Z\) is continuous.
Let \(Y,Z\) be topological spaces and \(g:Y\to Z\) is continuous, if \(V\) is open in \(Z\), then \(g^{-1}(V)\) is open in \(Y\), and \[h^{-1}(V)=f^{-1}(g^{-1}(V))\] if \(X\) is a measurable space and \(f:X\to Y\) is measurable, it follows that \(h^{-1}(V)\) is measurable, and if \(h=g\circ f\), then \(h:X\to Z\) is measurable.
Let \(u\) and \(v\) be real measurable functions on a measurable space \(X\), let \(\Phi\) be a continuous mapping of the plane into a topological space \(Y\), and define \[h(x)=\Phi(u(x),v(x)),\quad(x\in X)\] Then \(h:X\to Y\) is measurable. Put \(f(x)=(u(x),v(x))\) Then \(h=\Phi\circ f\) and \(f\) is measurable and maps \(X\) into the plane. If \(R\) is any open rectangle in the plane, with sides parallel to the axes, then \(R\) is the cartesian product of two segments \(I_1\) and \(I_2\) and since \(u\) and \(v\) are measurable, \[f^{-1}(R)=u^{-1}(I_1)\cap v^{-1}(I_2)\] is measurable set. Every open set \(V\) in the plane is a countable union of such rectangles \(R_i\) and since \[f^{-1}(V)=f^{-1}\Biggl(\overset{\infty}{\underset{i=1}{\bigcup}}R_i\Biggr)=\overset{\infty}{\underset{i=1}{\bigcup}}f^{-1}(R_i)\] then \(f^{-1}(V)\) is measurable. Then from the conclusion of 7), \(h:X\to Y\) is measurable.
Let X be a measurable space. If \(f=u+iv\), where \(u\) and \(v\) are real measurable functions on \(X\), then \(f\) is a complex measurable function on \(X\). If \(f=u+iv\) is a complex measurable function on \(X\), then from the conclusion of 7), with \(g(z)=\text{Re}(z)\), \(g(z)=\text{Im}(z)\), \(g(z)=|z|\), we know that \(u\) and \(v\) and \(|f|\) are real measurable functions on \(X\).
If \(f\) and \(g\) are complex measurable functions on \(X\), then \[f+g=(\text{Re}(f)+\text{Re}(g))+(\text{Im}(f)+\text{Im}(g))i\] and \[fg=\Bigl(\text{Re}(f)\text{Re}(g)-\text{Im}(f)\text{Im}(g)\Bigr)+\Bigl(\text{Re}(f)\text{Im}(g)+\text{Im}(f)\text{Re}(g)\Bigr)i\] are complex measurable functions.
If \(E\) is a measurable set in \(X\) and if \[\chi_E(x)=\begin{cases} 1 & \text{if } x\in E\\ 0 & \text{if } x\notin E\\ \end{cases}\] then \(\chi_E(x)\) is a measurable function. We call \(\chi_E(x)\) the characteristic function of the set \(E\).
If \(\mathscr F\) is any collection of subsets of \(X\), there exists a smallest \(\sigma-algebra\) \(\mathfrak M^*\) in \(X\) such that \(\mathscr F\subset\mathfrak M^*\). This \(\mathfrak M^*\) is called the \(\sigma-algebra\) generated by \(\mathscr F\). Let \(\Omega\) be the family of all \(\sigma-algebra\) \(\mathfrak M\) in \(X\) which contains \(\mathscr F\). Since the collection of all subsets of \(X\) is such a \(\sigma-algebra\), then \(\Omega\) is not empty. Let \(\mathfrak M^*\) be the intersection of all \(\mathfrak M\in\Omega\), it is clear that \(\mathscr F\in\mathfrak M^*\) and that \(\mathfrak M^*\) lies in every \(\sigma-algebra\) in \(X\) which contains \(\mathscr F\). Then we have to show that \(\mathfrak M^*\) is itself a \(\sigma-algebra\). If \(A_n\in\mathfrak M^*, n=1,2,3,\cdots,\) and if \(\mathfrak M\in\Omega\), then \(A_n\in\mathfrak M\), since \(\mathfrak M\) is a \(\sigma-algebra\), so \(\bigcup A_n\in\mathfrak M\). Since \(\bigcup A_n\in\mathfrak M\) for every \(\mathfrak M\in\Omega\), then \(\bigcup A_n\in\mathfrak M^*\). Since, \(A\in\mathfrak M^*\) and \(A^c\in\mathfrak M^*\), where \(A^c\) is the complement of \(A\) relative to \(\mathscr F\). The 3 properties of \(\sigma-algebra\) are verified and \(\mathfrak M^*\) is a \(\sigma-algebra\).
From the last theorem, let \(X\) be a topological space. Then there exists a smallest \(\sigma-algebra\) \(\mathscr B\) in \(X\) such that every open set in \(X\) belongs to \(\mathscr B\). The members of \(\mathscr B\) are called the Borel sets of \(X\). A Borel set is any set in a topological space that can be formed from open sets (or, equivalently, from closed sets) through the operations of countable union, countable intersection, and relative complement. All closed sets, all countable unions of closed sets (called \(F_{\sigma}\)’s) and all countable intersections of open sets (called \(G_{\delta}\)’s) are Borel sets. For a topological space \(X\), the collection of all Borel sets on \(X\) forms a \(\sigma-algebra\), known as the Borel \(\sigma-algebra\). The Borel \(\sigma-algebra\) on \(X\) is the smallest \(\sigma-algebra\) containing all open sets (or, equivalently, all closed sets). Every closed set in \(R^1\) can be generated by intersection of open sets \[[a,b]=\overset{\infty}{\underset{n=1}{\bigcap}}\Bigl(a-\frac{1}{n},b+\frac{1}{n}\Bigr)\] Every open set in \(R^1\) can be generated by union of closed sets \[(a,b)=\overset{\infty}{\underset{n=1}{\bigcup}}\Bigl[a+\frac{1}{n},b-\frac{1}{n}\Bigr]\]
Since \(\mathscr B\) is a \(\sigma-algebra\), we consider the measurable space \((X,\mathscr B)\). If \(f:X\to Y\) is a continuous mapping of \(X\), where \(Y\) is any topological space, then \(f^{-1}(V)\in\mathscr B\) for every open set \(V\) in \(Y\). In other words, every continuous mapping of \(X\) is Borel measurable mapping.
Suppose \(\mathfrak M\) is a \(\sigma-algebra\) in \(X\), and \(Y\) is a topological space. Let \(f\) maps \(X\) into \(Y\). If \(\Omega\) is the collection of all sets \(E\subset Y\) such that \(f^{-1}(E)\in\mathfrak M\), then \[f^{-1}(Y)=X\in\mathfrak M\] if \(f^{-1}(A)\in\mathfrak M\) then \[f^{-1}(A^c)=f^{-1}(Y-A)=X-f^{-1}(A)\in\mathfrak M\] and \[f^{-1}(A_1\cup A_2\cup\cdots)=f^{-1}(A_1)\cup f^{-1}(A_2)\cup\cdots\in\mathfrak M\] then \(\Omega\) is a \(\sigma-algebra\) in \(Y\).
Suppose \(\mathfrak M\) is a \(\sigma-algebra\) in \(X\), and \(Y\) is a topological space. Let \(f\) maps \(X\) into \(Y\). If \(f\) is measurable and \(E\) is a Borel set in \(Y\) and \(\Omega\) is the collection of all sets \(E\subset Y\) such that \(f^{-1}(E)\in\mathfrak M\), the measurability of \(f\) implies that \(\Omega\) contains all open sets in \(Y\), and since \(\Omega\) is a \(\sigma-algebra\), \(\Omega\) contains all Borel sets in \(Y\), then \(f^{-1}(E)\mathfrak M\).
Suppose \(\mathfrak M\) is a \(\sigma-algebra\) in \(X\), and \(Y\) is a topological space. Let \(f\) maps \(X\) into \(Y\). If \(Y=[-\infty,\infty]\) and \(f^{-1}((a,\infty])\in\mathfrak M\) for every real \(a\), let \(\Omega\) be the collection of all sets \(E\subset Y\) such that \(f^{-1}(E)\in\mathfrak M\). Choose a real number \(a\), and choose \(a_n<a\) and \(a_n\to a\) as \(n\to\infty\). Since \((a_n,\infty]\in\Omega\) for each \(n\), and \[[-\infty,a)=\overset{\infty}{\underset{n=1}{\bigcup}}[-\infty,a_n]=\overset{\infty}{\underset{n=1}{\bigcup}}(a_n,\infty]^c\] and since \(\Omega\) is a \(\sigma-algebra\) in \(Y\), we see that \([-\infty,a)\in\Omega\). The same is then true of \[(a,b)=[-\infty,b)\cap (a,\infty]\] Since every open set in \([-\infty,\infty]\) is a countable union of segments of the above types, \(\Omega\) contains every open set. Thus \(f\) is measurable.
Suppose \(\mathfrak M\) is a \(\sigma-algebra\) in \(X\), and \(Y\) is a topological space. Let \(f\) maps \(X\) into \(Y\). If \(f\) is measurable, if \(Z\) is a topological space, if \(g:Y\to Z\) is a Borel mapping, and if \(h=g\circ f\), let \(V\subset Z\) be open, then \(g^{-1}(V)\) is a Borel set of \(Y\), and since \[h^{-1}(V)=f^{-1}(g^{-1}(V))\in\mathfrak M\] then \(h:X\to Z\) is measurable.
Let \(\{a_n\}\) be a sequence in \([-\infty,\infty]\) and put \[b_k=\sup\,\{a_k,a_{k+1},a_{k+2},\cdots\}\quad(k=1,2,3,\cdots)\] and \[\beta=\inf\,\{b_1,b_2,b_3,\cdots\}\] we call \(\beta\) the upper limit of \(\{a_n\}\) and write \[\beta=\inf_{k\ge1}\Bigl\{\sup_{i\ge k} a_i\Bigr\}=\lim_{n\to\infty}\sup\, a_n\] Then \(b_1\ge b_2\ge b_3\ge\cdots\), so that \(b_k\to\beta\) as \(k\to\infty\), and there is a subsequence \(\{a_{n_i}\}\) of \(\{a_{n}\}\) such that \(\{a_{n_i}\}\to\beta\) as \(i\to\infty\) and \(\beta\) is the largest number with this property. If we put \[b_k=\inf\,\{a_k,a_{k+1},a_{k+2},\cdots\}\quad(k=1,2,3,\cdots)\] and \[\beta=\sup\,\{b_1,b_2,b_3,\cdots\}\] we call \(\beta\) the lower limit of \(\{a_n\}\) and write \[\beta=\sup_{k\ge1}\Bigl\{\inf_{i\ge k} a_i\Bigr\}=\lim_{n\to\infty}\inf\, a_n\] Then \[\lim_{n\to\infty}\inf\, a_n=-\lim_{n\to\infty}\sup\, (-a_n)\] if \(\{a_n\}\) converges, then \[\lim_{n\to\infty}\sup\, a_n=\lim_{n\to\infty}\inf\, a_n=\lim_{n\to\infty}\, a_n\]
Suppose \(f_n\) is a sequence of extended-real functions on a set \(X\), then \(\underset{n}{\sup}\,f_n\) and \(\displaystyle\lim_{n\to\infty}\sup\,f_n\) are the functions defined on \(X\) by \[\Bigl(\underset{n}{\sup}\,f_n\Bigr)(x)=\underset{n}{\sup}(f_n(x))\] \[\Bigl(\lim_{n\to\infty}\sup\,f_n\Bigr)(x)=\lim_{n\to\infty}\sup\,(f_n(x))\] If \[f(x)=\lim_{n\to\infty} f_n(x)\] the limit being assumed to exist at every \(x\in X\), then \(f\) is the pointwise limit of the sequence \(\{f_n\}\).
If \(f_n:X\to[-\infty,\infty]\) is measurable, for \(n=1,2,3,\cdots\), and \[g=\sup_{n\ge1}\,f_n\] \[h=\lim_{n\to\infty}\sup f_n\] since \[g^{-1}((a,\infty])=\overset{\infty}{\underset{n=1}{\bigcup}}f_n^{-1}((a,\infty])\] then \(g\) is measurable. If \[g=\inf_{n\ge1}\,f_n\] then \[g^{-1}([-\infty,a))=\overset{\infty}{\underset{n=1}{\bigcap}}f_n^{-1}([-\infty,a))\] then \(g\) is measurable. If \[h=\lim_{n\to\infty}\sup f_n=\inf_{k\ge1}\Bigl\{\sup_{i\ge k} f_i\Bigr\}\] then \(h\) is measurable.
The functions \[f^+=\max\,\{f,0\}\] \[f^-=-\min\,\{f,0\}\] are called positive and negative parts of \(f\). We have \[|f|=f^++f^-\] and \[f=f^+-f^-\] which is a representation of \(f\) as a difference of two nonnegative functions.
A complex function \(s\) on a measurable space \(X\) whose range consists of only finitely many points is called a simple function.. If \(a_1,\cdots,a_n\) are the distinct values of a simple function \(s\), and if we set \[A_i=\{x:s(x)=a_i\}\] and \[\chi_{A_i}=\begin{cases} 1 & \text{if } x\in A_i\\ 0 & \text{if } x\notin A_i\\ \end{cases}\] is the characteristic function. Then \[s=\sum_{i=1}^{n}a_i\chi_{A_i}\] Then \(s\) is measurable if and only if each of the sets \(A_i\) is measurable.
Let \(f:X\to[0,\infty]\) be measurable. There exist simple measurable functions \(s_n\) on \(X\) such that \[0\leq s_1\leq s_2\leq\cdots\leq f\] \[\underset{n\to\infty}{s_n}(x)\to f(x)\quad\forall x\in X\] Put \(\delta_n=2^{-n}\), to each positive integer \(n\) and each real number \(t\) corresponds a unique integer \(k_n(t)\) that satisfies \[k_n(t)\delta_n\leq t<(k_n(t)+1)\delta_n\] Define \[\varphi_n(t)=\begin{cases} k_n(t)\delta_n & \text{if } 0\leq t<n\\ n & \text{if } n\leq t\leq\infty\\ \end{cases}\] Each \(\varphi_n\) is then a Borel function on \([0,\infty]\), \[t-\delta_n<\varphi_n(t)\leq t,\quad 0\leq t\leq n,\quad 0\leq\varphi_1\leq\varphi_2\leq\cdots\leq t\] and \[\underset{n\to\infty}{\varphi_n}(t)\to t,\quad\forall t\in[0,\infty]\] It follows that the functions \[s_n=\varphi_n\circ f\] satisfy \[0\leq s_1\leq s_2\leq\cdots\leq f\] \[\underset{n\to\infty}{s_n}(x)\to f(x)\quad\forall x\in X\] and they are measurable.
A positive measure is a function \(\mu\), defined on a \(\sigma-algebra\) \(\mathfrak M\), whose range is in \([0,\infty]\) and which is countably additive. This means that if \(\{A_i\}\) is a disjoint countable collection of members of \(\mathfrak M\), then \[\mu\Biggl(\overset{\infty}{\underset{i=1}{\bigcup}}A_i\Biggr)=\sum_{i=1}^{\infty}\mu(A_i)\] we assume that \(\mu(A)<\infty\) for at least one \(A\in\mathfrak M\). A measure space is a measurable space which has a positive measure defined on the \(\sigma-algebra\) of its measurable sets, which can be written as an “ordered triples” \((X, \mathfrak M, \mu)\), where \(X\) is a set, \(\mathfrak M\) is a \(\sigma-algebra\) and \(\mu\) is a measure defined on \(\mathfrak M\). Similarly, measurable spaces are “ordered pairs” \((X, \mathfrak M)\). Since the set \(X\) is the largest member of \(\mathfrak M\), so if we know \(\mathfrak M\) we also know \(X\). Similarly, every measure has a \(\sigma-algebra\) for its domain, by definition, so if we know a measure \(\mu\) we also know the \(\sigma-algebra\) \(\mathfrak M\) on which \(\mu\) is defined and we know the set \(X\) in which \(\mathfrak M\) is a \(\sigma-algebra\). So we can use the expressions like “Let \(\mu\) be a measure (on X or on \(\mathfrak M\))”. A complex measure is a complex-valued countably additive function defined on a \(\sigma-algebra\). Similarly, a topological space is an ordered pair \((X,\tau)\), where \(\tau\) is a topology in the set \(X\), which is a topological space, and the significant data are contained in \(\tau\), not in \(X\).
Let \(\mu\) be a positive measure on a \(\sigma-algebra\) \(\mathfrak M\), take \(A\in\mathfrak M\) and \(\mu(A)<\infty\), and take \(A_1=A\) and \(A_2=A_3=\cdots=\varnothing\) then \[\mu(A_1)=\mu(A)=\mu(A_1+\varnothing+\cdots+\varnothing)=\mu(A_1)+\mu(\varnothing)+\cdots+\mu(\varnothing)\] then \[\mu(\varnothing)=0\]
Let \(\mu\) be a positive measure on a \(\sigma-algebra\) \(\mathfrak M\), then \[\mu(A_1\cup\cdots\cup A_n)=\mu(A_1)+\cdots+\mu(A_n)\] if \(A_1,\cdots,A_n\) are pairwise disjoint members \(\mathfrak M\).
Let \(\mu\) be a positive measure on a \(\sigma-algebra\) \(\mathfrak M\), if \(A\in\mathfrak M,\; B\in\mathfrak M,\; A\subset B\), then since \(B=A\cup(B-A)\) and \(A\cap(B-A)=\varnothing\) then \(\mu(B)=\mu(A)+\mu(B-A)\ge\mu(A)\)
Let \(\mu\) be a positive measure on a \(\sigma-algebra\) \(\mathfrak M\), if \(A=\bigcup_{n=1}^{\infty}A_n, \; A_n\in\mathfrak M,\; A_1\subset A_2\subset A_3\subset\cdots\), put \(B_1=A_1,\; B_n=A_n-A_{n-1}\quad n=2,3,4,\cdots\) then \(B_n\in\mathfrak M,\;B_i\cap B_j=\varnothing,\;i\ne j\) \(A_n=B_1\cup\cdots\cup B_n\) and \(A=\bigcup_{i=1}^{\infty}B_i\) Hence \[\mu(A_n)=\sum_{i=1}^{n}\mu(B_i)\] and \[\mu(A)=\sum_{i=1}^{\infty}\mu(B_i)\] since \[\sum_{i=1}^{n}\mu(B_i)\to\sum_{i=1}^{\infty}\mu(B_i)\] as \(n\to\infty\) then \[\underset{n\to\infty}{\mu(A_n)}\to \mu(A)\]
Let \(\mu\) be a positive measure on a \(\sigma-algebra\) \(\mathfrak M\), if \(A=\bigcap_{n=1}^{\infty}A_n,\;A_n\in\mathfrak M,\;A_1\supset A_2\supset A_3\supset\cdots\) and \(\mu(A_1)<\infty\) then put \(C_n=A_1-A_n\) then \(C_1\subset C_2\subset C_3\subset\cdots\) and \(\mu(C_n)=\mu(A_1)-\mu(A_n)\) and \(A_1-A=\bigcup C_n\) and so \[\mu(A_1)-\mu(A)=\mu(A_1-A)=\lim_{n\to\infty}\mu(C_n)=\mu(A_1)-\lim_{n\to\infty}\mu(A_n)\] then \[\underset{n\to\infty}{\mu(A_n)}\to \mu(A)\]
For any \(E\subset X\), where \(X\) is any set, if \(E\) is an infinite set, define \(\mu(E)=\infty\) if \(E\) is an infinite set, and if \(E\) is a finite set, define \(\mu(E)\) as the number of points in \(E\). This \(\mu\) is called the counting measure on \(X\). Fix \(x_0\in X\), for any \(E\subset X\) define \[\mu(E)=\begin{cases} 1 & \text{if } x_0\in E\\ 0 & \text{if } x_0\notin E\\ \end{cases}\] \(\mu\) is called the unit mass concentrated at \(x_0\). Let \(\mu\) be the counting measure on the set \(\{1,2,3,\cdots\}\), let \(A_n=\{n,n+1,n+2,\cdots\}\). Then \(\bigcap A_n=\varnothing\) but \(\mu(A_n)=\infty,\;n=1,2,3,\cdots\) This shows that the hypothesis \[\mu(A_1)<\infty\] in last theorem is necessary.
If \(s:X\to[0,\infty)\) is a measurable simple function, of the form \[s=\sum_{i=1}^{n}a_i\chi_{A_i}\] where \(a_1,\cdots,a_n\) are the distinct values of \(s\) and \[\chi_{A_i}=\begin{cases} 1 & \text{if } x\in A_i\\ 0 & \text{if } x\notin A_i\\ \end{cases}\] If \(E\in\mathfrak M\) we define \[\int_Esd\mu=\sum_{i=1}^{n}a_i\mu(A_i\cap E)\] if \(a_i=0\) for some \(i\) and that \(\mu(A_i\cap E)=\infty\), we define \(0\cdot\infty=0\).
If \(f:X\to[0,\infty)\) is measurable, and \(E\in\mathfrak M\), we define \[\int_Efd\mu=\sup\int_Esd\mu\] and if \(f\ge0\) \[\int_Efd\mu=\int_X\chi_Efd\mu\] the supremum being taken over all simple measurable functions \(s\) such that \(0\leq s\leq f\). \(\int_Efd\mu\) is called the Lebesgue integral of \(f\) over \(E\), with respect to the measure \(\mu\).Let \[s=\sum_{i=1}^{n}a_i\chi_{A_i}\] and \[t=\sum_{j=1}^{m}b_j\chi_{B_j}\] be nonnegative measurable simple functions on \(X\). For \(E\in\mathfrak M\), define \[\varphi(E)=\int_Esd\mu\] and if \(E_1,E_2,\cdots,\) are disjoint members of \(\mathfrak M\) whose union is \(E\), the countable additivity of \(\mu\) shows that \[\varphi(E)=\varphi(\sum_{r=1}^{\infty} E_r)=\sum_{i=1}^{n}a_i\mu(A_i\cap E)=\sum_{i=1}^{n}a_i\sum_{r=1}^{\infty}\mu(A_i\cap E_r)\\=\sum_{r=1}^{\infty}\sum_{i=1}^{n}a_i\mu(A_i\cap E_r)=\sum_{r=1}^{\infty}\varphi(E_r)\] which is countably additive. Also, \(\varphi(\varnothing)=0\), so that \(\varphi\) is not identically \(\infty\). Then \(\varphi\) is a measure on \(\mathfrak M\).
Let \(E_{ij}=A_i\cap B_j\) then \[\int_{E_{ij}}(s+t)d\mu=(a_i+b_j)\mu(E_{ij})\] and \[\int_{E_{ij}}sd\mu+\int_{E_{ij}}td\mu=a_i\mu(E_{ij})+b_j\mu(E_{ij})=(a_i+b_j)\mu(E_{ij})=\int_{E_{ij}}(s+t)d\mu\] and sum up all the \({E_{ij}}\), we get \[\int_{E}(s+t)d\mu=\int_{E}sd\mu+\int_{E}td\mu\]
References:
1. Rudin W. Real and complex analysis. Tata McGraw-hill education; 2006.