158 1. Measure theory for each compact sub-interval K of E. In a similar spirit, one can show that μ0(En) = inf U⊃En μ0(En) where U ranges over all open intervals containing En (Exercise!). Using the ε/2n trick, it thus suffices to show that μ0(K) n=1 μ0(Un) whenever Un is an open interval containing En. But by the Heine-Borel theorem, one can cover K by a finite number N n=1 Un of the Un, hence by finite subadditivity μ0(K) N n=1 μ0(Un) and the claim follows. As μ0 is now verified to be a pre-measure, we may use the Hahn- Kolmogorov extension theorem to extend it to a countably additive measure μ on a σ-algebra B that contains B0. In particular, B contains all the ele- mentary sets and hence (by Exercise 1.4.14) contains the Borel σ-algebra. Restricting μ to the Borel σ-algebra we obtain the existence claim. Finally, we establish uniqueness. If μ is another Borel measure with the stated properties, then μ (K) = |K|F for every compact interval K, and hence by (1.35) and upward monotone convergence, one has μ (I) = |I|F for every interval (including the unbounded ones). This implies that μ agrees with μ0 on B0, and thus (by Exercise 1.7.7, noting that μ0 is σ-finite) agrees with μ on Borel measurable sets. Exercise 1.7.10. Verify the claims marked “Exercise!” in the above proof. The measure μF given by the above theorem is known as the Lebesgue- Stieltjes measure μF of F . (In some texts, this measure is only defined when F is right-continuous, or equivalently if F = F+.) Exercise 1.7.11. Define a Radon measure on R to be a Borel measure μ obeying the following additional properties: (i) (Local finiteness) μ(K) for every compact K. (ii) (Inner regularity) One has μ(E) = supK⊂E,K compact μ(K) for every Borel set E. (iii) (Outer regularity) One has μ(E) = infU⊃E,U open μ(U) for every Borel set E. Show that for every monotone function F : R R, the Lebesgue-Stieltjes measure μF is a Radon measure on R conversely, if μ is a Radon measure
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