McGill University Department of Mathematics and Statistics
MATH 243 Analysis 2, Winter 2023
Assignment 5: Solutions
1 ifx∈{n1:n∈N}
Prove that f is Riemann integrable on [0,1] and compute
0 otherwise
Hint: You may use either the squeeze theorem or the ε-criterion for Riemann integrability.
Either way, this is a challenging question!
1. Solution: Using the squeeze theorem
Let 0 < ε < 1 be arbitrary. Define α,ω : [0,1] → R as follows:
0 if 0 ≤ x ≤ 2ε α(x):= f(x) if2ε
We still need to determine the value of
• Alternatively, one can define α ≡ 0 on [0,1] with only minor modifications to the proof above.
• Integrability of α and ω (as well as the values of their integrals) can also be determined with additivity: α and ω are both constant on [0, 2ε ] and are thus Riemann integrable. On [2ε,1], we have that α = ω and this function is constantly zero, except at finitely many points, and is thus Riemann integrable with integral 0. Thus
1 ε/2 1
α= α+ α=0+0=0
1 ε/2 1 ε ε
ω= ω+ ω=2+0=2 0 0 ε/2
2. Solution: Using the ε-criterion for Riemann integrability
Let0<ε<1,andletx1:=2ε.Notethat[x1,1]∩Sisfinite,sincen1 ≥2ε ⇔n≤2ε.Thismeans that f is constantly zero on [x1, 1], except at the finitely many points in [x1, 1]∩S. Consequently, f is Riemann integrable on [x1,1], and it follows from the ε-criterion for Riemann integrability that there exists a partition Q = {x1,x2,...,xk = 1} of [x1,1] such that U(f,Q)−L(f,Q) < 2ε. Observe that L(f, Q) = 0, since each of the partition intervals of Q contains at least one irrational number and thus at least one point not in S. This implies that U(f,Q) < 2ε.
Now consider the partition P := {0} ∪ Q = {x0 = 0,x1,...,xk = 1}. Then L(f,P) = 0 (again since each of the partition intervals of P contains at least one irrational number and thus at least one point not in S) and
U ( f , P ) = 1 · ( x 1 − x 0 ) + U ( f , Q ) = x 1 + U ( f , Q ) = 2ε + U ( f , Q ) < 2ε + 2ε = ε
Hence U(f,P) − L(f,P) = U(f,P) < ε. It now follows from the ε-criterion for Riemann
integrability that f is Riemann integrable on [0, 1]. We still need to find the value of the integral:
0=L(f,P)≤L(f)=
f =U(f)≤U(f,P)<ε
f < ε, from which it follows that f = 0.
Hence ∀ε > 0 : 0 ≤
3. Solution: Using the ε-criterion for Riemann integrability
LetS:={n1 :n∈N};letε>0bearbitrary.ChooseN∈NsuchthatN3 <ε⇔N>3ε.Let x1 := N1 and divide the interval [x1,1] into N2 −1 subintervals of equal width
1 − N1 N − 1 1
∆x = N2 − 1 = N(N2 − 1) = N(N + 1)
We obtain the partition P := (x0 = 0,x1 = N1 ,x2,…,xN2 = 1). Note that the first partition interval [0, x1 = 1 ] contains infinitely many points of S (which are 1 , 1 , . . . ), whereas the
union of the remaining N2 − 1 intervals [x1, x2], [x2, x3],. . . ,[xN2−1, xN2 ] contains exactly N
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points of S (which are 1, 12 , . . . , N1 ). Each of these N points is contained in at most 2 partition intervals.
Mi := sup{f(x) : x ∈ [xi−1,xi]} can thus be equal to 1 for at most 2N of these N2 −1 intervals; in all other cases the supremum is 0. We thus have
N2 N2 U(f,Pn)=Mi(xi −xi−1)=1· 1 +Mi(xi −xi−1)
i=1 N i=2 =∆x
≤ 1 +2N∆x= 1 +2N· 1 = 1 + 2 < 1 + 2 = 3 <ε N N N(N+1) N N+1 N N N
For L(f, Pn) we immediately have that L(f, Pn) = 0, since each of the partition intervals contains at least one irrational number and thus at least one point not in S. We thus have
U (f, P ) − L(f, P ) = U (f, P ) < ε
It now follows from the ε-criterion for Riemann integrability that f is Riemann integrable. We still need to find the value of the integral:
Hence ∀ε > 0 : 0 ≤
f < ε, from which it follows that f = 0.
0=L(f,P)≤L(f)=
f =U(f)≤U(f,P)<ε
2. Consider the function f : [0, 1] → R,
1 if 1
Wewillusethesqueezetheorem. Letε>0. Letα≡0on[a,b]andletω:=g. Then 0 = α(x) ≤ f(x) ≤ ω(x) = g(x) for all x ∈ [a,b]. α and ω are also both Riemann integrable.
Lastly, (ω − α) = g = 0 < ε. All conditions of the squeeze theorem are thus satisfied and
we conclude that f is Riemann integrable on [a, b]. Furthermore, since α ≤ f ≤ ω on [a, b] we
3. Let f,g : [a,b] → R be functions such that 0 ≤ f ≤ g on [a,b] and b
g = 0. Show that f is
Riemann integrable on [a, b] and that Solution:
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α= 0=0≤ f≤ ω= g=0 aaaaa
4. Consider the function f : [0, 1] → R,
1 if 1