The support of a complex function \(f\) on a topological space \(X\) is the closure of the set \[\{x:f(x)\ne0\}\] The collection of all continuous complex functions on \(X\) whose support is compact is denoted by \(C_c(X)\), which is a vector space. The notation \[K\prec f\] means that \(K\) is a compact subset of \(X\), that \(f\in C_c(X),\;\;0\leq f(x)\leq 1\) for all \(x\in X\) and that \(f(x)=1\) for all \(x\in K\). The notation \[f\prec V\] means that \(V\) is open, that \(f\in C_c(X), \; 0\leq f(x)\leq 1\), and that the support of \(f\) lies in \(V\).
There exists open set between open set and its compact subset. Suppose \(X\) is a Hausdorff space, \(K\subset X\), \(K\) is compact, and \(p\in K^c\). Then there are open sets \(U\) and \(W\) such that \(p\in U, K\subset W\) and \(U\cap W=\varnothing\).
If \(q\in K\), there exist disjoint open sets \(U_q\) and \(V_q\), such that \(p\in U_q\) and \(q\in V_q\). Since \(K\) is compact, there are points \(q_1,q_2,\cdots,q_n\in K\) such that \[K\subset V_{q_1}\cup\cdots\cup V_{q_n}\] Then \[U=U_{q_1}\cap\cdots\cap U_{q_n}\] and \[W=V_{q_1}\cup\cdots\cup V_{q_n}\] satisfy that \(p\in U, K\subset W\) and \(U\cap W=\varnothing\). Since \(p\) can be every point in \(K^c\), then \(K^c\) must be open and \(K\) must be closed. Compact subsets of Hausdorff spaces are closed.The intersection of a collection of sets consists of the elements that are in every one of the sets. If the intersection of a collection of compact sets is empty, then the intersection of some finite subcollections of these sets are empty. If \(\{K_a\}\) is a collection of compact subsets of a Hausdorff space and if \(\bigcap_aK_a=\varnothing\). Fix a member \(K_1\) of \(\{K_a\}\), since no point of \(K_1\) belongs to every other \(K_a\), \(\{K_a^c\}\) is an open cover of \(K_1\). Hence \(K_1\subset K_{a_1}^c\cup\cdots\cup K_{a_n}^c\) for some finite collection \(\{K_{a_i}^c\}\). This implies that \[K_1\cap K_{a_1}\cap\cdots\cap K_{a_n}=\varnothing\] then some finite sub-collection of \(\{K_a\}\) also has empty intersection.
There is open set with compact closure between open set and its compact subset Suppose \(U\) is open in a locally compact Hausdorff space \(X\), \(K\subset U\), and \(K\) is compact. Then there is an open set \(V\) with compact closure such that \[K\subset V\subset \overline{V}\subset U\]
Since every point of \(K\) has a neighborhood with compact closure, and since \(K\) is covered by the union of finitely many of these neighborhoods, \(K\) lies in an open set \(G\) with compact closure. If \(U=X\), take \(V=G\). Otherwise, since There exists open set between open set and its compact subset, to each \(p\in U^c\) there corresponds an open set \(W_p\) such that \(K\subset W_p\) and \(p\notin\overline{W}_p\). Hence \(\{U^c\cap\overline{G}\cap\overline{W}_p\}\), where \(p\) ranges over \(U^c\), is a collection of compact sets with empty intersection. Since If the intersection of a collection of compact sets is empty, then the intersection of some finite subcollections of these sets are empty, then there are points \(p_1,\cdots,p_n\in U^c\) such that \[U^c\cap\overline{G}\cap\overline{W}_{p_1}\cap\cdots\cap\overline{W}_{p_n}=\varnothing\] The set \[V=G\cap W_{p_1}\cap\cdots\cap W_{p_n}\] with compact closure \(\overline{V}\) satisfy that \[K\subset V\subset \overline{V}\subset U\] since \[\overline{V}\subset \overline{G}\cap \overline{W}_{p_1}\cap\cdots\cap \overline{W}_{p_n}\]Let \(f\) be a real function on a topological space. If \[\{x:f(x)>a\}\] is open for every real \(a\), \(f\) is said to be lower semicontinuous. If \[\{x:f(x)<a\}\] is open for every real \(a\), \(f\) is said to be upper semicontinuous. The supremum of any collection of lower semicontinuous functions is lower semicontinuous. The infimum of any collection of upper semicontinuous functions is upper semicontinuous.
There exists mapping \(f\) those support is compact subset and subordinate to an open set and the compact set lies in the open set. Urysohn’s Lemma Suppose \(X\) is a locally compact Hausdorff space, \(V\) is open in \(X\), \(K\) is compact and \(K\subset V\), then there exists an \(f\in C_c(X)\), such that \[K\prec f\prec V\]
Put \(r_1=0,r_2=1\), and let \(r_3,r_4,r_5,\cdots\) be an enumeration of the rationals in \((0,1)\). Since There is open set with compact closure between open set and its compact subset, we can find open sets \(V_0\) and then \(V_1\) such that \(\overline{V}_0\) is compact and \[K\subset V_1\subset\overline{V}_1\subset V_0\subset\overline{V}_0\subset V\] Suppose \(n\ge 2\) and \(V_{r_1},\cdots,V_{r_n}\) have been chosen in such a manner that \(r_i<r_j\) implies \(\overline{V}_{r_j}\subset V_{r_i}\). Then one of the numbers \(r_1,\cdots,r_n\), say \(r_i\), will be the largest one which is smaller than \(r_{n+1}\), and another, say \(r_j\), will be the smallest one larger than \(r_{n+1}\). Since There is open set with compact closure between open set and its compact subset, we can find \(V_{r_{n+1}}\) so that \[\overline{V}_{r_j}\subset V_{r_{n+1}}\subset\overline{V}_{r_{n+1}}\subset V_{r_i}\] Continuing, we obtain a collection \(\{V_r\}\) of open sets, one for every rational \(r\in[0,1]\), with the following properties: \[K\subset V_1,\;\; \overline{V}_{0}\subset V\] each \(\overline{V}_{r}\) is compact, and \[s>r\quad implies \quad \overline{V}_{s}\subset V_r\] Define \[f_r(x)=\begin{cases} r & \text{if } x\in V_r\\ 0 & \text{otherwise} \end{cases}\] \[g_s(x)=\begin{cases} 1 & \text{if } x\in \overline{V}_s\\ s & \text{otherwise} \end{cases}\] and \[f=\sup_r f_r\quad g=\inf_s g_s\] Then by definition, \(f\) is lower semicontinuous and \(g\) is upper semicontinuous. It is clear that \(0\leq f(x)\leq 1\), that \(f(x)=1\) if \(x\in K\) and that \(f\) has its support in \(\overline{V}_0\). The inequality \(f_r(x)>g_s(x)\) is possible only if \(r>s,\quad x\in V_r,\quad x\notin\overline{V}_s\). But \(r>s\) implies \(V_r\subset V_s\). Hence \(f_r(x)\leq g_s(x)\) for all \(r\) and \(s\), so \(f(x)\leq g(x)\).
Suppose \(f(x)<g(x)\) for some \(x\). Then there are rationals \(r\) and \(s\) such that \(f(x)<r<s<g(x)\). Since \(f(x)<r\), we have \(x\notin V_r\); since \(g(x)>s\), we have \(x\in\overline{V}_s\). This is a contradiction. Hence \(f=g\).There exist partition of unity mappings subordinate to the open cover sets of compact set. Suppose \(V_1,\cdots,V_n\) are open subsets of a locally compact Hausdorff space \(X\), \(K\) is compact,and \[K\subset V_1\cup\cdots\cup V_n\] Then there exist functions \(h_i\prec V_i\quad (i=1,2,\cdots,n)\) such that \[h_1(x)+\cdots+h_n(x)=1\quad(x\in K)\] then the collection \(\{h_1,\cdots,h_n\}\) is called a partition of unity on \(K\), subordinate to the cover \(\{V_1,\cdots,V_n\}\).
Since There is open set with compact closure between open set and its compact subset, each \(x\in K\) has a neighborhood \(W_x\) with compact closure \(\overline{W}_x\subset V_i\) for some \(i\) (depending on \(x\)). There are points \(x_1,\cdots,x_m\) such that \(W_{x_1}\cup\cdots\cup W_{x_m}\supset K\). If \(1\leq i\leq n\), let \(H_i\) be the union of those \(\overline{W}_{x_j}\) which lie in \(V_i\). By Urysohn’s lemma, there are functions \(g_i\) such that \(H_i\prec g_i\prec V_i\). Define \[\begin{align} h_1&=g_1\\ h_2&=(1-g_1)g_2\\ \cdots&\cdots\cdots\cdots\\ h_n&=(1-g_1)(1-g_2)\cdots(1-g_{n-1})g_n \end{align}\] Then \(h_i\prec V_i\). then \[h_1+h_2=g_1+(1-g_1)g_2=1+(g_1-1)-(g_1-1)g_2\\ =1+(g_1-1)(1-g_2)\\ =1-(1-g_1)(1-g_2)\] then \[h_1+h_2+\cdots+h_n=1-(1-g_1)(1-g_2)\cdots(1-g_n)\] Since \(K\subset H_1\cup\cdots\cup H_n\), at least one \(g_i(x)=1\) at each point \(x\in K\); hence \[h_1(x)+\cdots+h_n(x)=1\quad(x\in K)\]For vectors \(x\) and \(y\) and complex numbers \(a\) and \(b\), a linear transformation of a vector space \(V\) into another vector space \(V_1\) is a mapping \(\Lambda\) of a vector space \(V\) into another vector space \(V_2\) such that \[\Lambda(ax+by)=a\Lambda x+b\Lambda y\quad \forall x,y\in V\] In the special case in which \(a\Lambda x+b\Lambda y\) are scalars, \(\Lambda\) is called a linear functional. A linear functional is thus a complex function on \(V\) which satisfies \[\Lambda(ax+by)=a\Lambda x+b\Lambda y\] A positive linear functional means if \(\Lambda\in C_c(X),\quad f(X)\subset[0,\infty)\) then \(\Lambda f\in[0,\infty)\).
Riesz Representation Theorem Let \(X\) be a locally compact Hausdorff space, and let \(\Lambda\) be a positive linear functional on \(C_c(X)\). Then there exists a \(\sigma-algebra\) \(\mathfrak M\) in \(X\) which contains all Borel sets in \(X\), and there exists a unique positive measure \(\mu\) on \(\mathfrak M\) which represents \(\Lambda\) in the sense that
- \(\Lambda f=\int_Xfd\mu\) for every \(f\in C_c(X)\) and which has the following additional properties:
- \(\mu(K)<\infty\) for every compact set \(K\subset X\).
- For every \(E\in \mathfrak M\), we have \[\mu(E)=\inf\;\{\mu(V):E\subset V, \; V \text{ open}\}\]
- The relation \[\mu(E)=\sup\;\{\mu(K):K\subset E,\; K\text{ compact}\}\] holds for every open set \(E\) and for every \(E\in \mathfrak M\) with \(\mu(E)<\infty\)
- If \(E\in \mathfrak M,\quad A\subset E,\quad \mu(E)=0\), then \(A\in\mathfrak M\).
Let us begin by proving the uniqueness of \(\mu\). If \(\mu\) satisfies (c) and (d), it is clear that \(\mu\) is determined on \(\mathfrak M\) by its values on compact sets. Hence it suffices to prove that \(\mu_1(K)=\mu_2(K)\) for all \(K\), whenever \(\mu_1\) and \(\mu_2\) are measures for which the theorem holds. Fix \(K\) and \(\epsilon>0\), by (b) and (c), there exists a \(V\supset K\) with \(\mu_2(V)<\mu_2(K)+\epsilon\); by Urysohn’s lemma, there exists an \(f\) so that \(K\prec f\prec V\) hence \[\mu_1(K)=\int_X\chi_Kd\mu_1\leq\int_X f d\mu_1=\Lambda f=\int_Xfd\mu_2\\ \leq\int_X\chi_Vd\mu_2=\mu_2(V)<\mu_2(K)+\epsilon\] Thus \(\mu_1(K)\leq\mu_2(K)\). If we interchange the roles of \(\mu_1\) and \(\mu_2\), the opposite inequality is obtained, so \(\mu\) is unique.
Construction of \(\mu\) and \(\mathfrak M\). For every open set \(V\) in \(X\), define \[\mu(V)=\sup \;\{\Lambda f:f\prec V\}\] If \(V_1\subset U_2\), it is clear that \(\mu(V_1)\leq\mu(V_2)\). Hence \[\mu(E)=\inf\;\{\mu(V):E\subset V,\quad V\text{ open}\}\] if \(E\) is an open set, and it is consistent with \(\mu(V)=\sup \;\{\Lambda f:f\prec V\}\) to define \(\mu(E)\) by \(\mu(E)=\inf\;\{\mu(V):E\subset V,\quad V\text{ open}\}\) for every \(E\subset X\). Although we have defined \(\mu(E)\) for every \(E\subset X\), the countable additivity of \(\mu\) will be proved only on a certain \(\sigma-algebra\) \(\mathfrak M\) in \(X\).
Let \(\mathfrak M_F\) be the class of all \(E\subset X\) which satisfy two conditions: \(\mu(E)<\infty\), and \[\mu(E)=\sup\;\{\mu(K):K\subset E, \; K \text{ compact}\}\] Let \(\mathfrak M_F\) be the class of all \(E\subset X\) such that \(E\cap K\in \mathfrak M_F\) for every compact \(K\).
Proof that \(\mu\) and \(\mathfrak M\) have the required properties
\(\mu\) is monotone, i.e., that \(\mu(A)\leq\mu(B)\) if \(A\subset B\) and that \(\mu(E)=0\) implies \(E\in \mathfrak M_F\) and \(E\in \mathfrak M\). Thus (e) holds: If \(E\in \mathfrak M,\quad A\subset E,\quad \mu(E)=0\), then \(A\in\mathfrak M\) and so does (c) by definition: For every \(E\in \mathfrak M\), we have \[\mu(E)=\inf\;\{\mu(V):E\subset V, \; V \text{ open}\}\] The positivity of \(\Lambda\) implies that \(\Lambda\) is monotone: \(f\leq g\) implies \(\Lambda f\leq \Lambda g\). Since \[\Lambda g=\Lambda f+\Lambda(g-f)\] and \(g-f\ge 0\).
Step I
If \(E_1,E_2,\cdots\), are arbitrary subsets of \(X\), then \[\mu\Biggl(\overset{\infty}{\underset{i=1}{\bigcup}}E_i\Biggr)\leq\sum_{i=1}^{\infty}\mu(E_i)\] If \(V_1\) and \(V_2\) are open. Choose \(g\prec V_1\cup V_2\). Since There exist partition of unity mappings subordinate to the open cover sets of compact set, there are functions \(h_1\) and \(h_2\) such that \(h_i\prec V_i\) and \(h_1(x)+h_2(x)=1, \quad \forall x\in \text{supp }(g)\). Hence \(h_ig\prec V_i, \quad g=h_1g+h_2g\) and so \[\Lambda g=\Lambda(h_1g+h_2g)=\Lambda(h_1g)+\Lambda(h_2g)\leq \mu(V_1)+\mu(V_2)\\ (\forall g\prec V_1\cup V_2)\] then \[\mu(V_1\cup V_2)\leq\mu(V_1)+\mu(V_2)\] If \(\mu(E_i)=\infty\) for some \(i\), then \[\mu\Biggl(\overset{\infty}{\underset{i=1}{\bigcup}}E_i\Biggr)\leq\sum_{i=1}^{\infty}\mu(E_i)\] is true. Suppose therefore that \(\mu(E_i)<\infty\) for every \(i\). Choose \(\epsilon>0\), since \[\mu(E)=\inf\;\{\mu(V):E\subset V,\quad V\text{ open}\}\] there are open sets \(V_i\supset E_i\) such that \[\mu(V_i)<\mu(E_i)+2^{-i}\epsilon\quad(i=1,2,3,\cdots)\] Put \(V=\bigcup_{1}^{\infty}V_i\) and choose \(f\prec V\). Since \(f\) has compact support, we see that \(f\prec V_1\cup\cdots\cup V_n\) for some \(n\). Applying induction to \[\mu(V_1\cup V_2)\leq\mu(V_1)+\mu(V_2)\] we obtain \[\Lambda f\leq\mu(V_1\cup\cdots\cup V_n)\leq\mu(V_1)+\cdots+\mu(V_n)\leq\sum_{i=1}^{\infty}\mu(E_i)+\epsilon\] Since this holds for every \(f\prec V\) and since \(\bigcup E_i\subset V\), it follows that \[\mu\Biggl(\overset{\infty}{\underset{i=1}{\bigcup}}E_i\Biggr)\leq\mu(V)\leq\sum_{i=1}^{\infty}\mu(E_i)+\epsilon\] Step II
If \(K\) is compact, then \(K\in\mathfrak M_F\) and \[\mu(K)=\inf\;\{\Lambda f: K\prec f\}\] This implies assertion (b) \(\mu(K)<\infty\) for every compact set \(K\subset X\).
If \(K\prec f\) and \(0<a<1\), let \(V_a=\{x:f(x)>a\}\). Then \(K\subset V_a\) and \(ag\leq f\) whenever \(g\prec V_a\). Hence \[\mu(K)\leq\mu(V_a)=\sup\;\{\Lambda g:g\prec V_a\}\leq a^{-1}\Lambda f\] Let \(a\to 1\), to conclude that \[\mu(K)\leq \Lambda f\] Thus \(\mu(K)<\infty\). Since \(K\) evidently satisfies \[\mu(E)=\sup\;\{\mu(K):K\subset E, \; K \text{ compact}\},\quad K\in \mathfrak M_F\] If \(\epsilon>0\), there exists \(V\supset K\) with \(\mu(V)<\mu(K)+\epsilon\), by Urysohn’s lemma, \(K\prec f\prec V\) for some \(f\). Thus \[\Lambda f\leq\mu(V)<\mu(K)+\epsilon\] then \[\mu(K)=\inf\;\{\Lambda f: K\prec f\}\]
Step III
Every open set satisfies \[\mu(E)=\sup\;\{\mu(K):K\subset E, \; K \text{ compact}\}\] Hence \(\mathfrak M_F\) contains every open set \(V\) with \(\mu(V)<\infty\). Let \(a\) be a real number such that \(a<\mu(V)\). There exists an \(f\prec V\) with \(a<\Lambda f\). If \(W\) is any open set which contains the support \(K\) of \(f\), then \(f\prec W\), hence \(\Lambda f\leq \mu(W)\). Thus \(\Lambda f\leq \mu(K)\). This exhibits a compact \(K\subset V\) with \(a<\mu(K)\), so that \[\mu(E)=\sup\;\{\mu(K):K\subset E, \; K \text{ compact}\}\] holds for \(V\).
Step IV
Suppose \(E=\bigcup_{i=1}^{\infty}E_i\), where \(E_1,E_2,E_3,\cdots\) are’pairwise disjoint members of \(\mathfrak M_F\). Then \[\mu(E)=\sum_{i=1}^{\infty}\mu(E_i)\] If \(\mu(E)<\infty\), then \(E\in\mathfrak M_F\). We first show that \[\mu(K_1\cup K_2)=\mu(K_1)+\mu(K_2)\] if \(K_1\) and \(K_2\) are disjoint compact sets. Choose \(\epsilon>0\), by Urysohn’s lemma, there exists \(f\in C_c(X)\) such that \(f(x)=1\) on \(K_1\), \(f(x)=0\) on \(K_2\), and \(0\leq f\leq1\). By Step II there exists \(g\) such that \[K_1\cup K_2\prec g,\quad \Lambda g<\mu(K_1\cup K_2)+\epsilon\] Note that \(K_1\prec fg\) and \(K_2\prec(1-f)g\). Since \(\Lambda\) is linear, it follows from \(\mu(K)\leq\Lambda f\) that \[\mu(K_1)+\mu(K_2)\leq\Lambda(fg)+\Lambda(g-fg)=\Lambda g<\mu(K_1\cup K_2)+\epsilon\] since Step I \[\mu\Biggl(\overset{\infty}{\underset{i=1}{\bigcup}}E_i\Biggr)\leq\mu(V)\leq\sum_{i=1}^{\infty}\mu(E_i)+\epsilon\] then \[\mu(K_1\cup K_2)=\mu(K_1)+\mu(K_2)\]
If \(\mu(E)=\infty\), \[\mu(E)=\sum_{i=1}^{\infty}\mu(E_i)\] follows from Step I. Assume therefore that \(\mu(E)<\infty\) and choose \(\epsilon>0\). Since \(E_i\in \mathfrak M_F\), there are compact sets \(H_i\subset E_i\) with \[\mu(H_i)>\mu(E_i)-2^{-i}\epsilon\quad(i=1,2,3,\cdots)\] Putting \(K_n=H_1\cup\cdots\cup H_n\) and using induction on \[\mu(K_1\cup K_2)=\mu(K_1)+\mu(K_2)\] then \[\mu(E)\ge\mu(K_n)=\sum_{i=1}^{n}\mu(H_i)>\sum_{i=1}^{n}\mu(E_i)-\epsilon,\quad \forall n,\forall \epsilon>0\] from Step I, \[\mu(E)=\sum_{i=1}^{\infty}\mu(E_i)\]
But if \(\mu(E)<\infty\) and \(\epsilon>0\), \[\mu(E)=\sum_{i=1}^{\infty}\mu(E_i)\] shows \[\mu(E)\leq\sum_{i=1}^{N}\mu(E_i)+\epsilon\] for some \(N\). By \[\mu(E)\ge\mu(K_n)=\sum_{i=1}^{n}\mu(H_i)>\sum_{i=1}^{n}\mu(E_i)-\epsilon,\quad \forall n,\forall \epsilon>0\] it follows that \(\mu(E)\leq\mu(K_N)+2\epsilon\) and this shows that \(E\) satisfies \[\mu(E)=\sup\;\{\mu(K):K\subset E, \; K \text{ compact}\}\] hence \(E\in\mathfrak M_F\).
Step V
If \(A\in\mathfrak M_F,\quad \epsilon>0\), there is a compact \(K\) and an open \(V\) such that \(K\subset E\subset V\) and \(\mu(V-K)<\epsilon\).
Our definitions show that there exist \(K\subset E\) and \(V\supset E\) so that \[\mu(V)-\frac{\epsilon}{2}<\mu(E)<\mu(K)+\frac{\epsilon}{2}\] Since \(V-K\) is open, \(V-K\in \mathfrak M_F\). Step IV implies that \[\mu(K)+\mu(V-K)=\mu(V)<\mu(K)+\epsilon\]
Step VI
If \(A, \; B\in\mathfrak M_F\), then \(A-B,\;A\cup B,\;A\cap B\) belong to \(\mathfrak M_F\).
If \(\epsilon>0\). Step V shows that there are sets \(K_1\) and \(V_i\) such that \(K_1\subset A\subset V_1,\quad K_2\subset B\subset V_2\) and \(\mu(V_i-K_i)<\epsilon\) for \(i=1,2\). Since \[A-B\subset V_1-K_2\subset(V_1-K_1)\cup(K_1-V_2)\cup(V_2-K_2)\] Step I shows that \(\mu(A-B)\leq\epsilon+\mu(K_1-V_2)+\epsilon\) Since \(K_1-V_2\) is a compact subset of \(A-B\), then \((A-B)\) satisfies \[\mu(E)=\sup\;\{\mu(K):K\subset E, \; K \text{ compact}\}\] so that \(A-B\in \mathfrak M_F\) Since \(A\cup B=(A-B)\cup B\), an application of Step IV shows that \(A\cup B\in \mathfrak M_F\). Since \(A\cap B=A-(A-B)\) we also have \(A\cap B\in \mathfrak M_F\).
Step VII
\(\mathfrak M\) is a \(\sigma-algebra\) in \(X\) which contains all Borel sets.
Let \(K\) be an arbitrary compact set in \(X\). If \(A\in \mathfrak M\), then \(A^c\cap K=K-(A\cap K)\), so that \(A^c\cap K\) is a difference of two members of \(\mathfrak M_F\). Hence \(A^c\cap K\in \mathfrak M_F\) and we conclude: \(A\in \mathfrak M\) implies \(A^c\in \mathfrak M\). Next, suppose \(A\in\bigcup_{1}^{\infty} A_i\), where each \(A_i\in \mathfrak M\). Put \(B_1=A_1\cap K\) and \[B_n=(A_n\cap K)-(B_1\cup\cdots\cup B_{n-1})\quad(n=2,3,4\cdots)\] Then \(\{B_n\}\) is a disjoint sequence of members of \(\mathfrak M_F\), by Step VI, and \(A\cap K=\bigcup_{1}^{\infty} B_n\). It It follows from Step IV that \(A\cap K\in \mathfrak M_F\). Hence \(A\in\mathfrak M\). Finally, if \(C\) is closed, then \(C\cap K\) is compact, hence \(C\cap K\in \mathfrak M_F\), so \(C\in \mathfrak M\). In particular, \(X\in \mathfrak M\). Then \(\mathfrak M\) is a \(\sigma-algebra\) in \(X\) which contains all closed subsets of \(X\). Hence \(\mathfrak M\) contains all Borel sets in \(X\).
Step VIII
\(\mathfrak M_F\) consists of precisely those sets \(E\in\mathfrak M\) for which \(\mu(E)<\infty\).
This implies assertion (d) The relation \[\mu(E)=\sup\;\{\mu(K):K\subset E,\; K\text{ compact}\}\] holds for every open set \(E\) and for every \(E\in \mathfrak M\) with \(\mu(E)<\infty\).
If \(E\in \mathfrak M_F\), Step II and VI imply that \(E\cap K\in \mathfrak M_F\) for every compact \(K\), hence \(E\in\mathfrak M\). Conversely, suppose \(E\in\mathfrak M\) and \(\mu(E)<\infty\) and choose \(\epsilon>0\). There is an open set \(V\supset E\) with \(\mu(V)<\infty\), by Step III and V, there is a compact \(K\subset V\) with \(\mu(V-k)<\epsilon\). Since \(E\cap K\in \mathfrak M_F\), there is a compact set \(H\subset E\cap K\) with \[\mu(E\cap K)<\mu(H)+\epsilon\] Since \(E\subset(E\cap K)\cup(V-K)\), it follows that \[\mu(E)\leq\mu(E\cap K)+\mu(V-K)<\mu(H)+2\epsilon\] which implies that \(E\in \mathfrak M_F\).
Step IX
\(\mu\) is a measure on \(\mathfrak M\).
The countable additivity of \(\mu\) on \(\mathfrak M\) follows immediately from Steps IV and VIII.
Step X
For every \(f\in C_c(X), \Lambda f=\int_X fd\mu\)
This proves (a) \(\Lambda f=\int_Xfd\mu\) for every \(f\in C_c(X)\). It is enough to prove this for real \(f\). Also it is enough- to prove the inequality \[\Lambda f\leq\int_Xfd\mu,\quad \forall \text { real }f\in C_c(X)\] Once the inequality is established, the linearity of \(\Lambda\) shows that \[-\Lambda f=\Lambda(-f)\leq\int_X(-f)d\mu=-\int_Xfd\mu\] which together with \[\Lambda f\leq\int_Xfd\mu\] shows that equality holds \[\Lambda f=\int_Xfd\mu\] Let \(K\) be the support of a real \(f\in C_c(X)\), let \([a,b]\) be an interval which contains the range of \(f\). choose \(\epsilon>0\), and choose \(y_i\), for \(i=0,1,\cdots,n\) so that \(y_i-y_{i-1}<\epsilon\) and \[y_0<a<y_1<\cdots<y_n=b\] Put \[E_i=\{x:y_{i-1}<f(x)\leq y_i\}\cap K\quad(i=1,\cdots,n)\] Since \(f\) is continuous, \(f\) is Borel measurable, and the sets \(E_i\) are therefore disjoint Borel sets whose union is \(K\). There are open sets \(V_i\supset E_i\) such that \[\mu(V_i)<\mu(E_i)+\frac{\epsilon}{n}\quad(i=1,\cdots,n)\] and such that \(f(x)<y_i+\epsilon\) for all \(x\in V_i\). By theorem There exist partition of unity mappings subordinate to the open cover sets of compact set, there are functions \(h_i\prec V_i\) such that \(\sum h_i=1\) on \(K\). Hence \(f=\sum h_i f\) and Step II shows that \[\mu(K)\leq\Lambda(\sum h_i)=\sum\Lambda h_i\] Since \(h_if\leq(y_i+\epsilon)h_i\) and since \(y_i-\epsilon<f(x)\) on \(E_i\), we have \[\begin{align} \Lambda f&=\sum_{i=1}^{n}\Lambda(h_if)\leq\sum_{i=1}^{n}(y_i+\epsilon)\Lambda h_i\\ &=\sum_{i=1}^{n}(|a|+y_i+\epsilon)\Lambda h_i-|a|\sum_{i=1}^{n}\Lambda h_i\\ &\leq\sum_{i=1}^{n}(|a|+y_i+\epsilon)\Biggl[\mu(E_i)+\frac{\epsilon}{n}\Biggr]-|a|\mu(K)\\ &=\sum_{i=1}^{n}(|a|+y_i+\epsilon)\mu(E_i)+\sum_{i=1}^{n}(|a|+y_i+\epsilon)\frac{\epsilon}{n}-|a|\mu(K)\\ &=\sum_{i=1}^{n}(y_i-\epsilon)\mu(E_i)+2\epsilon\mu(K)+\frac{\epsilon}{n}\sum_{i=1}^{n}(|a|+y_i+\epsilon)\\ &\leq\int_Xfd\mu+\epsilon\Biggl[2\mu(K)+|a|+b+\epsilon\Biggr] \end{align}\] then \[\Lambda f\leq\int_Xfd\mu,\quad \forall \text { real }f\in C_c(X)\]
A measure \(\mu\) on the \(\sigma-algebra\) of all Borel sets in a locally compact Hausdorff space \(X\) is called a Borel measure on \(X\). If \(\mu\) is positive, for every Borel set \(E\subset X,\quad E\in \mathfrak M\), we have \[\mu(E)=\inf\;\{\mu(V):E\subset V, \; V \text{ open}\}\] then \(E\) is outer regular. If \[\mu(E)=\sup\;\{\mu(K):K\subset E,\; K\text{ compact}\}\] holds for every open set \(E\) and for every \(E\in \mathfrak M\) with \(\mu(E)<\infty\), then \(E\) is inner regular. If every Borel set in \(X\) is both outer and inner regular, \(\mu\) is called regular.
A set \(E\) in a topological space is called \(\sigma-compact\) if \(E\) is a countable union of compact sets. A set \(E\) in a measure space (with measure \(\mu\)) is said to have \(\sigma-finite\; measure\) if \(E\) is a countable union of sets \(E_i\) with \(\mu(E_i)<\infty\). If \(E\in \mathfrak M\) and \(E\) has \(\sigma\)-finite measure, then \(E\) is inner regular.
Suppose \(X\) is a locally compact, \(\sigma\)-compact Hausdorff space. A \(\sigma-algebra\) \(\mathfrak M\) in \(X\) which contains all Borel sets in \(X\), and a unique positive measure \(\mu\) on \(\mathfrak M\) which represents \(\Lambda\), a positive linear functional on \(C_c(X)\). then
- If \(E\in\mathfrak M\) and \(\epsilon>0\), there is a closed set \(F\) and an open set \(V\) such that \(F\subset E\subset V\) and \(\mu(V-F)<\epsilon\).
- \(\mu\) is a regular Borel measure on \(X\).
- If \(E\in\mathfrak M\), there are sets \(A\) and \(B\) such that \(A\) is an \(F_{\sigma}\) (All closed sets, all countable unions of closed sets), \(B\) is a \(G_{\delta}\) (All countable intersections of open sets), \(A\subset E\subset B\) and \(\mu(B-A)=0\)
Let \(X=K_1\cup K_2\cup K_3\cup\cdots\), where each \(K_n\) is compact. If \(E\in\mathfrak M\) and \(\epsilon>0\), then \(\mu(K_n\cap E)<\infty\) and there are open sets \(V_n\supset (K_n\cap E)\) such that \[\mu(V_n-(K_n\cap E))<\frac{\epsilon}{2^{n+1}}\quad(n=1,2,3,\cdots)\] If \(V=\bigcup V_n\) then \(V-E\subset\bigcup(V_n-(K_n\cap E))\), so that \[\mu(V-E)<\frac{\epsilon}{2}\] Apply this to \(E^c\) in place of \(E\): There is an open set \(W\supset E^c\) such that \(\mu(W-E^c)<\frac{\epsilon}{2}\). Since \(W^c\subset E\) and \(E-W^c=W-E^c\), then \(W^c\subset E\subset V\) and \(\mu(V-W^c)<\mu(V-E)<\frac{\epsilon}{2}\). This proves (a).
Every closed set \(F\subset X\) is \(\sigma\)-compact, because \(F=\bigcup(F\cap K_n)\). Hence (a) implies that every set \(E\in\mathfrak M\) is inner regular. This proves (b).
If we apply (a) with \(\epsilon=1/j\quad(j=1,2,3,\cdots)\) we obtain closed sets \(F_j\) and open sets \(V_j\) such that \(F_j\subset E\subset V_j\) and \(\mu(V_j-F_j)<1/j\). Put \(A=\bigcup F_j,\quad B=\bigcap V_j\). Then \(A\subset E\subset B\), \(A\) is an \(F_{\sigma}\) (All closed sets, all countable unions of closed sets), \(B\) is a \(G_{\delta}\) (All countable intersections of open sets), and \(\mu(B-A)=0\) since \(B-A\subset V_j-F_j,\quad j=1,2,3,\cdots\). This proves (c).
Let \(X\) be a locally compact Hausdorffspace in which every open set is \(\sigma-compact\). Let \(\lambda\) be any positive Borel measure on \(X\) such that \(\lambda(K)<\infty\) for every compact set \(K\), then \(\lambda\) is regular.
Put \(\Lambda f=\int_Xfd\lambda,\quad f\in C_c(X)\). Since \(\lambda(K)<\infty\) for every compact \(K\), \(\Lambda\) is a positive linear functional on \(C_c(X)\), and there is a regular measure \(\mu\), \[\int_Xfd\lambda=\int_Xfd\mu\quad(f\in C_c(X))\] We will show that \(\lambda=\mu\). Let \(V\) be open in \(X\). Then \(V=\bigcup K_i,\quad(i=1,2,3,\cdots)\), where \(K_i\) is compact. By Urysohn’s lemma we can choose \(f_i\) so that \(K_i\prec f_i\prec V\). Let \(g_n=\max(f_1,\cdots,f_n)\). Then \(g_n\in C_c(X)\) and \(g_n(x)\) increases to \(\chi_V(x)\) at every point \(x\in X\). Hence \[\int_Xfd\lambda=\int_Xfd\mu\] and the monotone convergence theorem imply \[\lambda(V)=\lim_{n\to\infty}\int_Xg_nd\lambda=\lim_{n\to\infty}\int_Xg_nd\mu=\mu(V)\] Then \(\lambda=\mu\).
Now let \(E\) be a Borel set in \(X\) and choose \(\epsilon>0\). Then there is a closed set \(F\) and an open set \(V\) such that \(F\subset E\subset V\) and \(\mu(V-F)<\epsilon\). Hence \(\mu(V)\leq \mu(F)+<\epsilon\leq\mu(E)+<\epsilon\). Since \(V-F\) is open and \(\lambda(V)=\mu(V)\), then \(\lambda(V-F)<\epsilon\), hence \(\lambda(V)\leq\lambda(E)+\epsilon\) then \[\lambda(E)\leq\lambda(V)=\mu(V)\leq\mu(E)+\epsilon\] \[\mu(E)\leq\mu(V)=\lambda(V)\leq\lambda(E)+\epsilon\] so \[|\lambda(E)-\mu(E)|<\epsilon\] for every \(\epsilon>0\) then \(\lambda(E)=\mu(E)\)Euclidean \(k\)-dimensional space \(R^k\) is the set of all points \(\mathbf x=\{a_1,\cdots,a_k\}\) whose coordinates \(a_i\) are real numbers. If \(\mathbf a\in R^k,\quad \delta>0\), we call the set \[Q(\mathbf a;\delta)=\{\mathbf x:a_i\leq x_i<a_i+\delta,\quad1\leq i\leq k\}\] \(\delta-box\) with corner at \(\mathbf a=(a_1,\cdots,a_k)\), each \(\delta-box\) only contains one corner. A set of the form \[W=\{\mathbf x:a_i\leq x_i\leq b_i,\quad 1\leq i\leq k\}\] is called a \(k-cell\). its volume is defined to be \[\text{vol }(W)=\prod_{i=1}^{k}(b_i-a_i)\]
There exists a positive complete measure \(m\) defined on a \(\sigma-algebra\) \(\mathfrak M\) in \(R^k\), with the following properties:
- \(m(W)=\text{vol }(W)\) for every \(k-cell\) \(W\).
- \(\mathfrak M\) contains all Borel sets in \(R^k\); \(E\in\mathfrak M\) if and only if there are sets \(A\) and \(B\subset R^k\) such that \(A\subset E\subset B\), \(A\) is an \(F_{\sigma}\) (All closed sets, all countable unions of closed sets), \(B\) is a \(G_{\delta}\) (All countable intersections of open sets), and \(m(B-A)=0\) and \(m\) is regular.
- \(m\) is translation-invariant, \[m(E+x)=m(E)\] for every \(E\in\mathfrak M\) and every \(x\in R^k\).
- If \(\mu\) is any positive translation-invariant Borel measure on \(R^k\) such that \(\mu(K)<\infty\) for every compact set \(K\), then there is a constant \(c\) such that \(\mu(E)=cm(E)\) for all Borel sets \(E\subset R^k\).
- To every linear transformation \(T\) of \(R^k\) into \(R^k\) corresponds a real number \(\Delta(T)\) such that \[m(T(E))=\Delta(T)m(E)\] for every \(E\in \mathfrak M\). In particular, \(m(T(E))=m(E)\) when \(T\) is a rotation.
The members of \(\mathfrak M\) are the Lebesgue measurable sets in \(R^k\); \(m\) is the Lebesgue measure on \(R^k\).
For \(n=1,2,3,\cdots\), let \(P_n\) be the set of \(x\in R^k\) whose coordinates are integer multiples of \(2^{-n}\) and let \(\Omega_n\) be the collection of all \(2^{-n}\) boxes with corners at points of \(P_n\). If \(f\) is any complex function on \(R^k\), with compact support, define \[\Lambda_nf=2^{-nk}\sum_{x\in P_n}f(x)\quad(n=1,2,3,\cdots)\] Now suppose \(f\in C_c(R^k)\), \(f\) is real, \(W\) is an open \(k-cell\) which contains the \(\text{supp }f\) and \(\epsilon>0\). \(\Biggl[\) Let \(f\) be a mapping of a metric space \(X\) into a metric space \(Y\), we say \(f\) is uniformly continuous on \(X\) if every \(\epsilon>0\) there exists \(\delta>0\) such that \(d_Y(f(p),f(q))<\epsilon\) for all \(p\) and \(q\) in \(X\) for which \(d_X(p,q)<\delta\). \(\Biggr]\) \(\Biggl[\) Let \(f\) be a continuous mapping of a compact metric space \(X\) into a metric space \(Y\), then \(f\) is uniformly continuous on \(X\). Given \(\epsilon>0\), since \(f\) is continuous, we can associate each point \(p\in X\) a positive number \(\phi(p)\) such that \(\underset{q\in X}{d_X(p,q)}<\phi(p)\) implies that \(d_Y(f(p), f(q))<\frac{\epsilon}{2}\). Let \(J(p)\) be the set of all \(q\in X\) for which \[d_X(p,q)<\frac{1}{2}\phi(p)\], since \(p\in J(p)\), the collection of all sets \(J(p)\) is an open cover of \(X\); and since \(X\) is compact, there is a finite set of points \(p_1,\cdots,p_n\) in \(X\) that \(X\subset J(p_1)\cup\cdots\cup J(p_n)\). Let \(\delta=\frac{1}{2}\text{min}[\phi(p_1),\cdots,\phi(p_n)]\), then \(\delta>0\). Now let \(p,q\) be the points of \(X\), such that \(d_X(p,q)<\delta\), since \(X\subset J(p_1)\cup\cdots\cup J(p_n)\), there is an integer \(m, 1\le m\le n\) such that \(p\in J(p_m)\), hence \(d_X(p,p_m)<\frac{1}{2}\phi(p_m)\) and we also have \[d_X(q,p_m)\le d_X(q,p)+d_X(p,p_m)<\delta+\frac{1}{2}\phi(p_m)\le \phi(p_m)\] therefore \[d_Y(f(p),f(q))\le d_Y(f(p),f(p_m))+d_Y(f(q),f(p_m))<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon\] which means \(f\) is uniformly continuous on \(X\). \(\Biggr]\) The uniform continuity of \(f\) shows that there is an integer \(N\) and that there are functions \(g\) and \(h\) with support in \(W\), such that \(g\) and \(h\) are constant on each box belonging to \(\Omega_N\), \(g\leq f\leq h\) and \(h-g<\epsilon\). If \(n>N\) and \(Q\in\Omega_N\), then \(\text{Vol }(Q)=2^{-Nk}\) and the set \(P_n\) has \(2^{(n-N)k}\) points in \(Q\). Then \[\Lambda_Ng=\Lambda_ng\leq\Lambda_nf\leq\Lambda_nh=\Lambda_Nh\] Thus the upper and lower limits of \(\{\Lambda_nf\}\) differ by at most \(\epsilon\text{ vol }(W)\), and since \(\epsilon\) was arbitrary, we have proved the existence of \[\Lambda f=\lim_{n\to\infty}\Lambda_nf\quad(f\in C_c(R^k))\] It is immediate that \(\Lambda\) is a positive linear functional on \(C_c(R^k)\). (In fact, \(\Lambda f\) is precisely the Riemann integral of \(f\) over \(R^k\)).
We define \(m\) and \(\mathfrak M\) to be the measure and \(\sigma-algebra\) associated with this \(\Lambda\) as in Riesz Representation Theorem. Since Riesz Representation Theorem gives us a complete measure and since \(R^k\) is \(\sigma-compact\), then (b) \(\mathfrak M\) contains all Borel sets in \(R^k\); \(E\in\mathfrak M\) if and only if there are sets \(A\) and \(B\subset R^k\) such that \(A\subset E\subset B\), \(A\) is an \(F_{\sigma}\) (All closed sets, all countable unions of closed sets), \(B\) is a \(G_{\delta}\) (All countable intersections of open sets), and \(m(B-A)=0\) and \(m\) is regular.
To prove (a), let \(W\) be the open cell \[W=\{\mathbf x:a_i<x_i<b_i,\quad 1\leq i\leq k\}\] Let \(E_r\) be the union of those boxes belonging to \(\Omega_r\) whose closures lie in \(W\), choose \(f_r\) so that \(\overline{E}_r\prec f_r\prec W\) and put \(g_r=\max\{f_1,\cdots,f_r\}\). The construction of \(\Lambda\) shows that \[\text{vol }(E_r)\leq\Lambda f_r\leq \Lambda g_r\leq \text{vol }(W)\] As \(r\to\infty\), \(\text{vol }(E_r)\to\text{vol }(W)\) and \[\Lambda g_r=\int g_rdm\to m(W)\] by the monotone convergence theorem, since \(g_r(\mathbf x)\to \chi_W(\mathbf x)\) for all \(\mathbf x\in R^k\). Thus \(m(W)=\text{vol }(W)\) for every open cell \(W\), and since every \(k-cell\) is the intersection of a decreasing sequence of open \(k-cells\), we obtain (a) \(m(W)=\text{vol }(W)\) for every \(k-cell\) \(W\).
If \(\lambda\) is a positive Borel measure on \(R^k\) and \(\lambda(E)=m(E)\) for all boxes \(E\), then the same equality holds for all open sets \(E\), by property: \(\Biggl(\) For \(n=1,2,3,\cdots\), let \(P_n\) be the set of \(x\in R^k\) whose coordinates are integer multiples of \(2^{-n}\) and let \(\Omega_n\) be the collection of all \(2^{-n}\) boxes with corners at points of \(P_n\). Then every nonempty open set in \(R^k\) is a countable union of disjoint boxes belonging to \(\Omega_1\cup\Omega_2\cup\Omega_3\cup\cdots\). Since if \(V\) is open, every \(\mathbf x\in V\) lies in an open ball which lies in \(V\); let \[Q(\mathbf a;\delta)=\{\mathbf x:a_i\leq x_i<a_i+\delta,\quad1\leq i\leq k\}\] hence \(\mathbf x\in Q\subset V\) for some \(Q\) belonging to some \(\Omega_n\). In other words, \(V\) is the union of all boxes which lie in \(V\) and which belong to some \(\Omega_n\). From this collection of boxes, select those which belong to \(\Omega_1\), and remove those in \(\Omega_2,\Omega_3,\cdots\) which lie in any of the selected boxes. From the remaining collection, select those boxes of \(\Omega_2\) which lie in \(V\), and remove those in \(\Omega_3,\Omega_4, \cdots\) which lie in any of the selected boxes. Since if \(n\) fixed, each \(\mathbf x\in R^k\) lies in one and only one member of \(\Omega_n\) and if \(Q_1\in \Omega_n,\;Q_2\in\Omega_r,\;r<n\) then either \(Q_1\subset Q_2\) or \(Q_1\cap Q_2=\varnothing\), then every nonempty open set in \(R^k\) is a countable union of disjoint boxes belonging to \(\Omega_1\cup\Omega_2\cup\Omega_3\cup\cdots\) \(\Biggr)\) and therefore for all Borel sets \(E\), since \(\lambda\) and \(m\) are regular \(\Biggl[\) Let \(X\) be a locally compact Hausdorffspace in which every open set is \(\sigma-compact\). Let \(\lambda\) be any positive Borel measure on \(X\) such that \(\lambda(K)<\infty\) for every compact set \(K\), then \(\lambda\) is regular \(\Biggr]\).
To prove (c), fix \(\mathbf x\in R^k\) and define \(\lambda(E)=m(E+\mathbf x)\), then \(\lambda\) is a measure; by (a), \(\lambda(E)=m(E)\) for all boxes, hence \(m(E+\mathbf x)=m(E)\) for all Borel sets \(E\). The same equality holds for every \(E\in \mathfrak M\), because of (b). Then (c) \(m\) is translation-invariant, \[m(E+x)=m(E)\] for every \(E\in\mathfrak M\) and every \(x\in R^k\). Suppose next that \(\mu\) satisfies the hypotheses of (d). Let \(Q_0\) be a 1-box, put \(c=\mu(Q_0)\). Since \(Q_0\) is the union of \(2^{nk}\) disjoint \(2^{-n}\) boxes that are translates of each other, we have \[2^{nk}\mu(Q)=\mu(Q_0)=cm(Q_0)=c\cdot 2^{nk}m(Q)\] for every \(2^{-n}\)-box \(Q\). Since every nonempty open set in \(R^k\) is a countable union of disjoint boxes belonging to \(\Omega_1\cup\Omega_2\cup\Omega_3\cup\cdots\) then \(\mu(E)=cm(E)\) for all open sets \(E\subset R^k\). Then (d) If \(\mu\) is any positive translation-invariant Borel measure on \(R^k\) such that \(\mu(K)<\infty\) for every compact set \(K\), then there is a constant \(c\) such that \(\mu(E)=cm(E)\) for all Borel sets \(E\subset R^k\).
To prove (e), let \(T:R^k\to R^k\) be linear. If the range of \(T\) is a subspace \(Y\) of lower dimension, then \(m(Y)=0\) and the desired conclusion holds with \(\Delta(T)=0\). In the other case, \(T\) is a \(1-to-1\) map of \(R^k\) onto \(R^k\) whose inverse is also linear. Thus \(T\) is a homeomorphism of \(R^k\) onto \(R^k\), so that \(T(E)\) is a Borel set for every Borel set \(E\), and we can define a positive Borel measure \(\mu\) on \(R^k\) by \[\mu(E)=m(T(E))\] The linearity of \(T\), combined with the translation-invariance of \(m\), gives \[\mu(E+\mathbf x)=m(T(E+\mathbf x))=m(T(E)+T\mathbf x)=m(T(E))=\mu(E)\] Thus \(\mu\) is translation-invariant, and since (d) If \(\mu\) is any positive translation-invariant Borel measure on \(R^k\) such that \(\mu(K)<\infty\) for every compact set \(K\), then there is a constant \(c\) such that \(\mu(E)=cm(E)\) for all Borel sets \(E\subset R^k\), and since (b) \(\mathfrak M\) contains all Borel sets in \(R^k\); \(E\in\mathfrak M\) if and only if there are sets \(A\) and \(B\subset R^k\) such that \(A\subset E\subset B\), \(A\) is an \(F_{\sigma}\) (All closed sets, all countable unions of closed sets), \(B\) is a \(G_{\delta}\) (All countable intersections of open sets), and \(m(B-A)=0\) and \(m\) is regular, then (e) To every linear transformation \(T\) of \(R^k\) into \(R^k\) corresponds a real number \(\Delta(T)\) such that \[m(T(E))=\Delta(T)m(E)\] for every \(E\in \mathfrak M\). To find \(\Delta(T)\), we need to know \(m(T(E))/m(E)\) for one set \(E\) with \(0<m(E)<\infty\). If \(T\) is a rotation, let \(E\) be the unit ball of \(R^k\); then \(T(E)=E\) and \(\Delta(T)=1\), then \(m(T(E))=m(E)\) when \(T\) is a rotation.
Every set of positive measure has nonmeasurable subsets. If \(A\subset R^1\) and every subset of \(A\) is Lebesgue measurable, then \(m(A)=0\).
A group is a set equipped with a binary operation that combines any two elements to form a third element in such a way that four conditions called group axioms are satisfied, namely closure, associativity, identity and invertibility. A subgroup \(H\) of a group \(G\) can be used to decompose the underlying set of \(G\) into disjoint equal-size pieces called cosets. There are two types of cosets: left cosets and right cosets. Cosets (of either type) have the same number of elements (cardinality) as does \(H\). Furthermore, \(H\) itself is a coset, which is both a left coset and a right coset. The number of left cosets of \(H\) in \(G\) is equal to the number of right cosets of \(H\) in \(G\). The common value is called the index of \(H\) in \(G\) and is usually denoted by \([G : H]\). Such as \[\underset{G}{\underbrace{\left| \begin{array}{cc|c} \hline 0&4&H\\ 1&5&1+H\\ 2&6&2+H\\ 3&7&3+H\\ \hline \end{array} \right|}}\] Since \(R^1\) is a group relative to addition, let \(Q\) be the subgroup that consists of the rational numbers, and let \(E\) be a set that contains exactly one point from each coset of \(Q\) in \(R^1\). Then if \(r\in Q,\;s\in Q,\; r\ne s\) \[(E+r)\cap(E+s)=\varnothing\] and every \(x\in R^1\) lies in \(E+r\) for some \(r\in Q\).
Suppose \(x\in(E+r)\cap(E+s)\), then \[x=y+r=z+s,\quad y\in E,z\in E,y\ne z\] But \(y-z=s-r\in Q\), so that \(y\) and \(z\) lie in the same coset of \(Q\), a contradiction. Then if \(r\in Q,\;s\in Q,\; r\ne s\) \[(E+r)\cap(E+s)=\varnothing\]
Let \(y\) be the point of \(E\) that lies in the same coset as \(x\), put \(r=x-y\). Fix \(t\in Q\) and put \(A_t=A\cap(E+t)\). By hypothesis \(A_t\) is measurable. Let \(K\subset A_t\) be compact, let \(H\) be the union of the translates \(K+r\), where \(r\) ranges over \(Q\cap[0,1]\). Then \(H\) is bounded, hence \(m(H)<\infty\). Since \(K\subset E+t\), then the sets \(K+r\) are pairwise disjoint. Thus \(m(H)=\sum_{r}m(K+r)\). But \(m(K+r)=m(K)\), then \(m(K)=0\). This holds for every compact \(K\subset A_t\). Hence \(m(A_t)=0\). Since \(A=\bigcup A_t\), where \(t\) ranges over \(Q\). Since \(Q\) is countable, we conclude that \(m(A)=0\).Lusin’s Theorem Let \(X\) be a locally compact Hausdorff space, \(\mu\) is a measure on \(X\). In particular, \(\mu\) could be Lebesgue measure on some \(R^k\). Suppose \(f\) is a complex measurable function on \(X\), \(\mu(A)<\infty,\;f(x)=0,\; x\notin A\). Then there exists a \(g\in C_c(X)\) such that \[\mu(\{x:f(x)\ne g(x)\})<\epsilon,\; \epsilon>0\] Furthermore, we may arrange it so that \[\underset{x\in X}{\sup}|g(x)|\leq\underset{x\in X}{\sup}|f(x)|\]
Assume first that \(0\leq f<1\) and that \(A\) is compact. Attach a sequence \(\{s_n\}\) to \(f\) such that \(0\leq s_1\leq s_2\leq \cdots\leq f\) \(\Biggl[\) 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. \(\Biggl]\) Put \(t_1=s_1\) and \(t_n=s_n-s_{n-1},\; n=2,3,4,\cdots\) Then \(2^nt_n\) is the characteristic function of a set \(T_n\subset A\) and \[f(x)=\sum_{n=1}^{\infty}t_n(x)\quad(x\in X)\] Fix an open set \(V\) such that \(A\subset V\) and \(\overline{V}\) is compact. There are compact sets \(K_n\) and open sets \(V_n\) such that \(K_n\subset T_n\subset V_n\subset V\) and \(\mu(V_n-K_n)<2^{-n}\epsilon\). By Urysohn’s lemma, there are functions \(h_n\) such that \(K_n\prec h_n\prec V_n\). Define \[g(x)=\sum_{n=1}^{\infty}2^{-n}h_n(x)\quad(x\in X)\] This series converges uniformly on \(X\), so \(g\) is continuous. The support of \(g\) lies in \(\overline{V}\). Since \(2^{-n}h_n(x)=t_n(x)\) except in \(V_n-K_n\), we have \(g(x)=f(x)\) except in \(\bigcup(V_n-K_n)\), and this latter set has measure less than \(\epsilon\). Thus if \(A\) is compact and \(0\leq f\leq 1\) \[\mu(\{x:f(x)\ne g(x)\})<\epsilon,\; \epsilon>0\] The compactness of \(A\) is easily removed, for if \(\mu(A)<\infty\) then \(A\) contains a compact set \(K\) with \(\mu(A-K)\) smaller than any preassigned positive number. Next, if \(f\) is a complex measurable function and if \(B_n=\{x:|f(x)|>n\}\), since \(\Biggl[\) 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 \[\underset{n\to\infty}{\mu(A_n)}\to \mu(A)\] \(\Bigg]\) then \(\bigcap B_n=\varnothing\), so \(\mu(B_n)\to 0\). Since \(f\) coincides with the bounded function \((1-\chi_{B_n})\cdot f\) except on \(B_n\), \[\mu(\{x:f(x)\ne g(x)\})<\epsilon,\; \epsilon>0\] follows in the general case.
Finally, let \(R=\sup\;\{|f(x)|:x\in X\}\) and define \(\varphi(z)=z\) if \(|z|\leq R\), and \(\varphi(z)=Rz/|z|\) if \(|z|>R\). Then \(\varphi\) is a continuous mapping of the complex plane onto the disc of radius \(R\). If \(g\) satisfies \[\mu(\{x:f(x)\ne g(x)\})<\epsilon,\; \epsilon>0\] and \(g_1=\varphi\circ g\), then \(g_1\) satisfies \[\mu(\{x:f(x)\ne g(x)\})<\epsilon,\; \epsilon>0\] and \[\underset{x\in X}{\sup}|g(x)|\leq\underset{x\in X}{\sup}|f(x)|\]Vitali-Caratheodory Theorem Suppose \(f\in L^1(\mu)\), \(f\) is real-valued, and \(\epsilon >0\). Then there exist functions \(u\) and \(v\) on \(X\) such that \(u\leq f\leq v\), \(u\) is upper semicontinuous \[\{x:u(x)<a\}\] is open for every real \(a\), and bounded above, \(v\) is lower semicontinuous \[\{x:v(x)>a\}\] is open for every real \(a\), and bounded below, and \[\int_X(v-u)d\mu<\epsilon\]
Assume first that \(f\ge0\) and that \(f\) is not identically \(0\). Since \(f\) is the pointwise limit of an increasing sequence of simple functions \(s_n\), \(f\) is the sum of the simple functions \(t_n=s_n-s_{n-1},\quad s_0=0\) and since \(t_n\) is a linear combination of characteristic functions, there are measurable sets \(E_i\) (not necessarily disjoint) and constants \(c_i>0\) such that \[f(x)=\sum_{i=1}^{\infty}c_i\chi_{E_i}(x)\quad(x\in X)\] Since \[\int_Xfd\mu=\sum_{i=1}^{\infty}c_i\mu(E_i)\] the series converges. There are compact sets \(K_i\) and open sets \(V_i\) such that \(K_i\subset E_i\subset V_i\) and \[c_i\mu(V_i-K_i)<2^{-i-1}\epsilon\quad(i=1,2,3,\cdots)\] Put \[v=\sum_{i=1}^{\infty}c_i\chi_{V_i}\quad u=\sum_{i=1}^{N}c_i\chi_{K_i}\] where \(N\) is chosen so that \[\sum_{N+1}^{\infty}c_i\mu(E_i)<\frac{\epsilon}{2}\] Then \(v\) is lower semicontinuous, \(u\) is upper semicontinuous, \(u\leq f\leq v\), and \[v-u=\sum_{i=1}^{N}c_i(\chi_{V_i}-\chi_{K_i})+\sum_{N+1}^{\infty}c_i\chi_{V_i}\\ \leq\sum_{i=1}^{\infty}c_i(\chi_{V_i}-\chi_{K_i})+\sum_{N+1}^{\infty}c_i\chi_{E_i}\] Then \[\int_X(v-u)d\mu\leq\int_X\Biggl(\sum_{i=1}^{\infty}c_i(\chi_{V_i}-\chi_{K_i})+\sum_{N+1}^{\infty}c_i\chi_{E_i}\Biggr)d\mu\\ <\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon\]
In the general case, write \(f=f^+-f^-\), attach \(u_1\) and \(v_1\) to \(f^+\), attach \(u_2\) and \(v_2\) to \(f^-\), and put \(u=u_1-v_2,\quad v=v_1-u_2\). Since \(u_1\) and \(-v_2\) are upper semicontinuous and since the sum of two upper semicontinuous functions is upper semicontinuous, \(v_1\) and \(-u_2\) are lower semicontinuous and the sum of two lower semicontinuous functions is lower semicontinuous. \(u\) and \(v\) have the desired properties.
BIBLIOGRAPHY
1. Rudin W. Real and complex analysis. Tata McGraw-hill education; 2006.