4. In lecture, we've mentioned how the lower and upper Darboux sums 'bound' possibleRiemann sums. In this problem, let's make that statement more rigorous.(a) Let f: [a, b] → R be a bounded function, and let P = {xo,...,n} be a partitionof [a, b]. We say a set of points 7 := {₁,..., Cn} is a tagging of P if x₁-1 ≤ Cj ≤ xifor all i= 1,..., n.Given any partition P and tagging 7 of P, show that71L(P, f) ≤ f(c) Ax¡ ≤ U(P, ƒ)i=1(b) Suppose f: [a, b] → R is Riemann integrable. Show that for all e > 0, there existsa partition P such that for any tagging 7,T[1-Monce-ΣΔε; <εi=1(c) Given a bounded function f [a, b] → R and n € N, define the right uniformRiemann sum Rn(f, [a, b]) asRn(f, [a, b])Ef(x):=72Σf(xi) Axi=1where Ar= (b-a)/n and x₁ = a + (b-a)(i/n).Let f: RR be the Dirichlet function, which satisfiesxEQ{10x & QShow that the sequences {R₂(f, [0, 1])}_₁, {R₂(ƒ, [1, 1+√2])}_₁, and {R₂(f, [0, 1+√2])} converge, butlim (R₂ (ƒ, [0, 1]) + R₂(ƒ, [1,1 + √2])) ‡ lim R₁(f, [0,1 + √2])84x84x(Remark: You may use the fact that if r € Q and x & Q, then r + x & Q, andrx Q if r #0.):(d) Prove or disprove (i.e. by counterexample): A bounded function f [a, b] → Ris Riemann integrable if and only if the sequence of right uniform Riemann sums{R₂(f, [a, b])} converges as n → ∞o.

Since you have posted a question with multiple subparts, we will solve the first three subparts foryou. To get remaining subparts solved please repost the complete question and mention the subparts (a) Given f:a,b→ℝ be a bounded function. Let P=x0,x1,⋯,xn be a partition of a,b. Also τ:=c1,c2,……,cn is called tagging of P if xi-1≤ci≤xi for all i=1,2,…,n. For any tagging τ of P, we need to show that LP,f≤∑i=1nfci∆xi≤UP,f. We know that, for any partition P=x0,x1,⋯,xn, the lower LP,f and upper Riemann sum of a function f on a,b can be calculated as: LP,f=∑i=1nmi∆xiUP,f=∑i=1nMi∆xi Where mi=inffx: xi-1≤x≤xiMi=supfx: xi-1≤x≤xiSince mi=inffx: xi-1≤x≤xi and Mi=supfx: xi-1≤x≤xi for all i=1,2,…,n, therefore: mi≤fx≤Mi for all xi-1≤x≤xi and for all i=1,2,…,n. Since τ:=c1,c2,……,cn is a tagging of P, therefore mi≤fci≤Mi for i=1,2,…,n. Therefore: mi∆xi≤fci∆xi≤Mi∆xi for all i=1,2,…,n. Therefore by taking summation, we have: ∑i=1nmi∆xi≤∑i=1nfci∆xi≤∑i=1nMi∆xiLP,f≤∑i=1nfci∆xi≤UP,f Hence LP,f≤∑i=1nfci∆xi≤UP,f.(b) Given f:a,b→ℝ be a Riemann integrable function. We know that a function f is said to be Riemann integrable, if for every ε>0, there exists a partition P of a,b such that UP,f-LP,f<ε and in this case we have ∫abf-LP,f=∫abf-UP,f<ε. Since f:a,b→ℝ is Riemann integrable function, therefore for every ε>0, there exists a partition P of a,b such that UP,f-LP,f<ε2 and ∫abf-LP,f=∫abf-UP,f<ε2. From part a, we have LP,f≤∑i=1nfci∆xi≤UP,f for any tagging τ of P, therefore: ∑i=1nfci∆xi-LP,f<ε2 and UP,f-∑i=1nfci∆xi<ε2. Now consider: ∫abf-∑i=1nfci∆xi=∫abf-LP,f+LP,f-∑i=1nfci∆xi≤∫abf-LP,f+LP,f-fci∆xi<ε2+ε2=ε Hence ∫abf-∑i=1nfci∆xi<ε.(c) The given function is fx=1x∈ℚ0x∉ℚ For a bounded function f, the right uniform Riemann sum is defined as: Rnf,a,b:=∑i=1nfxi∆x, where ∆x=b-anand xi=a+b-ain. (i) Now consider Rnf,0,1, therefore ∆x=1-0n and xi=0+1-0in. Therefore Rnf,0,1 can be calculated as: Rnf,0,1=∑i=1nfxi∆x=∑i=1nfin1n=∑i=1n11n ∵in∈ℚ=1 Hence limn→∞Rnf,0,1=limn→∞1=1.(ii) Now consider Rnf,1,1+2, therefore ∆x=1+2-1n=2n and xi=1+1+2-1in=1+2in. Therefore Rnf,1,1+2 can be calculated as: Rnf,1,1+2=∑i=1nfxi∆x=∑i=1nf1+2in2n=∑i=1n02n ∵1+2in∉ℚ=0 Hence limn→∞Rnf,1,1+2=limn→∞0=0. (iii) Now consider Rnf,0,1+2, therefore ∆x=1+2-0n=1+2n and xi=0+1+2-0in=1+2in. Therefore Rnf,0,1+2 can be calculated as: Rnf,0,1+2=∑i=1nfxi∆x=∑i=1nf1+2in1+2n=∑i=1n01+2n ∵1+2in∉ℚ=0 Hence limn→∞Rnf,0,1+2=limn→∞0=0. By (i), (ii) and (iii), we observe that limn→∞Rnf,0,1+limn→∞Rnf,1,1+2≠limn→∞Rnf,0,1+2.