Question

does anyone know how to do this question5b in both direction？

5. Let 01, 02, ... be a strictly increasing sequence in (a,b), and let p > 0 be such that ... Pr = 1. Define a : [a,b] R as follows: a(x) = 0 if a srca, a(x) = p«ifa, <<< an+1, 1 and a(x) = 1 if supan Sasb. (a) For a sc<dsb, describe Aa = o(d) - a(c) in terms of an and Ph. (Thus, convince yourself that a is monotone increasing and that it is obtained by assigning "probability", or "mass". Pr to an and letting (x) equal the "total mass" of the interval (a,x).) (b) Show that a bounded fraction f : [a, b] → Ris Riemann-Stieltjes integrable with respect to a if and only if fla) = limf(x) exists and f(n) = f(a) for all n, and that in this case Isda - Şow).

Answer 1

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n a) We have to consider the following scenarios: c< a <sup an <d Case i) ; in this case, we have Aa = a(d) – a(c) = 1-0 =1 Σ Pk k=1 Σ ΡΕ n k:c<axsd c< a < d <sup an Case ii) ; in this case, (definition of supremum implies) there is an n such that an < d < an+1. Thus, we have Aa = a(d) - a(c) n ΣΡk – 0 k=1 n = Σοκ k=1 Σ ΡΕ n k:c<a<d aj <c<d<sup an Case iii) ; in this case, (definition of supremum implies) there is an n such that an < d < an+1, and n' such that An' <c< An' +1. Thus, we have Aa = a(d) - a(C)

ΣΡ -ΣΡΑ k=1 k=1 η k=n' +1 Σ ΡΕ Σ ΡΕ k:c<α<d Case iv) c<d< 01; in this case, we have Δα = α(d) – α(c) = 0 - 0 Σ ΡΕ k:c<α<d The last equality is due to the fact that the empty sum of positive terms is zero. Therefore, in all cases, we have Δα = α(d) – α(c) Σ ΡΕ k:c<α<d b) Notice that the formula in part a), Δα = α(d) – α(c) Σ Ρk k:c<α,<d implies that a is continuous at every point except at I = 01, 02,"; moreover, the discontinuity of a at these points I = 01, 02, ". is the left-discontinuity, that is (writing a(t) = a(2) - a(0), see case ii) above) lim α(α) = αακ) Τα

lim a(z) = a(ak) - pk rak < alak) Suppose that f : [a, b] + R is a bounded function. Because of left-disontinuity of a, we conclude that f is Riemann-integrable if and only if it is left continuous at all I = 01, 02,"; equivalently, the following holds: lim f(0) = f(an) Tan To find the integral, consider any mesh (or partitition or subdivision) of [a, b], given by P= {a = 20 <11 < ... <In=b} Then the lower integral is L(P, f,a) = (a(čk) – a(Ik-1)) inf f() re[TR-1, Ix]" k=1 n PI inf f(0) 1:38-1<a,SIK TE[Tk-1,5k) Now, as the mesh size goes to zero, each subinterval (2k-1, Ik] will contain at most one ar; thus, in the limit, we will have lim L(P, f,a) lim inf f(2) IA-IA-170- DE[Ix-1,6%] k=1 k=1 n ΣΡΑ = Pxf(ak) k=1