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1.7. Outer measures, pre-measures, product measures 159 on R, show that there exists a monotone function F : R → R such that μ = μF . Radon measures are studied in more detail in §1.10 of An epsilon of room, Vol. I. Exercise 1.7.12 (Near uniqueness). If F, F : R → R are monotone non- decreasing functions, show that μF = μF if and only if there exists a con- stant C ∈ R such that F+(x) = F+(x) + C and F−(x) = F−(x) + C for all x ∈ R. Note that this implies that the value of F at its points of discon- tinuity are irrelevant for the purposes of determining the Lebesgue-Stieltjes measure μF in particular, μF = μF+ = μF− . In the special case when F+(−∞) = 0 and F−(+∞) = 1, then μF is a probability measure, and F+(x) = μF ((−∞,x]) is known as the cumulative distribution function of μF . Now we give some examples of Lebesgue-Stieltjes measure. Exercise 1.7.13 (Lebesgue-Stieltjes measure, absolutely continuous case). (i) If F : R → R is the identity function F (x) = x, show that μF is equal to Lebesgue measure m. (ii) If F : R → R is monotone non-decreasing and absolutely continu- ous (which, in particular, implies that F exists and is absolutely integrable, show that μF = mF in the sense of Exercise 1.4.48, thus μF (E) = E F (x) dx for any Borel measurable E, and R f(x) dμF (x) = R f(x)F (x) dx for any unsigned Borel measurable f : R → [0, +∞]. In view of the above exercise, the integral R f dμF is often abbreviated R f dF , and referred to as the Lebesgue-Stieltjes integral of f with respect to F . In particular, observe the identity [a,b] dF = F+(b) − F−(a) for any monotone non-decreasing F : R → R and any −∞ b a +∞, which can be viewed as yet another formulation of the fundamental theorem of calculus. Exercise 1.7.14 (Lebesgue-Stieltjes measure, pure point case).