Measure Theory/Riesz' representation theorem

Theorem (Riesz' representation theorem) Let be a locally compact Hausdorff space and let be a positive linear functional on . Then, there exists a -field containing all Borel sets of and a unique measure such that for all for all compact If , and then If , and then The measure space is complete Proof Recall the Urysohn's lemma: If is a locally compact Hausdorff space and if is open and is compact with then there exists with satisfying and . This is written in short as We shall first prove that if such a measure exists, then it is unique. Suppose are measures that satisfy (1) through (5) It suffices to show that for every compact Let be compact and let be given. By (3), there exists open with such that Urysohn's lemma implies that there exists such that (1) implies that . But , that is . We can similarly show that . Thus, Suppose is open in , define If are open , then If is a subset of then define Define Let monotonicity of is obvious for all subsets of Let with It is obvious that which implies that . Hence, we have that is complete. Step 1 Suppose is a sequence of subsets of . Then, Proof Let be open subsets of . We wish to show that . Given , let be such that (so that ) and . This is possible because . Now by Urysohn's lemma we can find , such that and on with . Thus, and on . As is a linear functional for all , Thus, for every , i.e. If is a sequence of members of , there exist open such that given , . Define , is open. Let . Then but Thus, Step 2 If is compact, then and Proof It suffices to show that for every compact Let and , define Then is open and Then, by Urysohn's lemma, there exists such that , and hence, on Since https://4dspace.mtts.org.in/expository-article-download.php?ai=160https://laurent.claessens-donadello.eu/pdf/giulietta.pdf, search for the label THOooTWZWooHqGDAx., we have As , we have and hence, Let be By definition, there exists open such that By Urysohn's lemma, there exists such that , which implies that , that is . Hence, Step 3 Every open set satisfies If is open and , , then Proof Let be open. Let such that . It suffices to show that there exists compact such that . By definition of , there exists such that and Let . Obviously . Let be open such that , then and hence , further Thus, Step 4 Suppose is a sequence of pairwise disjoint sets in and let . Then, Proof If , by step 1, we are done. If is finite then, and hence, is countably additive on Suppose, are compact and disjoint then ; Claim: As is a locally compact Hausdorff space, there exist disjoint open sets , with , Hence, by Urysohn's lemma, there exists such that and Now, and , Thus, Assume . Given , there exists compact such that Let . Obviously, is compact. Thus, and hence, By step 1, we have . Thus, Step 5 If and then there exists compact and open with and Proof is open. As , there exist compact and open such that with Now, (by step 4) Thus, Step 6 is a field of subsets of Proof Let and let be given. There exist compact and open such that , with Write As is a closed subsetof , it is compact Then, thus, is finite and hence, Now write and Step 7 is a -field containing all Borel sets Proof Let be closed Then, is compact for every compact Therefore and hence (by definition) and hence, has all closed sets. In particular, Let . Then, and and hence, Now let where We know that for every compact Let , . , but and hence, Step 8 Proof Let . Then for every compact Now, let , . Given , there exists open such that , , that is, . Further, there exists compact such that implies that , that is, there exists compact such that Therefore, implies that As is arbitrary, we are done. Step 9 For , Proof Without loss of generality, we may assume that is real valued. It is obvious from the definition of that Let . Hence, as is continuous, is compact. and we can write for some . Let . Let be an -fine partition of Let . As is continuous, is compact, is measurable for every , and hence, Hence, we can find open sets such that for all We know that if compact with open then there exists wiht and on Hence, there exist functions such that on . Thus, for all By step 2, we have on each Thus, As is arbitrary, we have which completes the proof.