Download M2PM1 Test 2 - Imperial College London

Imperial College London
M2PM1 Analysis 2
Progress Test 2
8 December 2011
1. [9 marks]
i. State the ε-criterion for Riemann integration.
ii. Let f : [0, 3] → R defined as follows: f (x) = x if x ∈ [0, 1], f (x) = x + 1 if x ∈ (1, 2] and
f (x) = x + 2 if x ∈ (2, 3]. Prove that f is Riemann integrable over [0, 3] and find its integral.
Z n+1
1
dx.
iii. Compute lim
n→∞ n
1 + x2011
iv. Let f be a function that is differentiable on some open interval containing a, and such
that f 00 (a) exists. Prove that
f (a + h) + f (a − h) − 2f (a)
→ f 00 (a) as h → 0.
h2
2. [8 marks]
i. Define what it means for a collection Σ of subsets of R to be an algebra and what it
means for Σ to be a σ-algebra.
ii. What is the Borel σ-algebra on R ?
iii. Prove that the set of rational numbers Q belongs to the Borel σ-algebra on R.
iv. Define what it means for a set to be compact.
v. Prove that the set [0, 1) is not compact.
3. [3 marks]
i. Give an example of a Borel set that is both closed and open. (no proof required).
ii. Give an example of a Riemann integrable function f : [−2, 2] → R for which
Z 2
Z 0
f (x)dx −
f (x)dx = 1.
−2
0
(no proof required).
iii. Give an example of a function f : R2 → R that is not continuous at (1, 1). (no proof
required).
1
Answers
1.
i. [1 mark] Let f : [a, b] → R be a bounded function. Suppose that for any ε > 0 there is
a partition Δ of [a, b] such that S(f, Δ) − s(f, Δ) ≤ ε. Then f is Riemann integrable over
[a, b].
ii. [4 marks] Let Δn be the partition of [0, 3] into 3n subintervals of equal length. Then
P
P2n+1 i−1
P3n
3n(3n−1)
1
1
i−1
i−1
+ 3 − n2 and
s(f, Δn ) = n1 n+1
i=1 n + n
i=n+2 ( n + 1) + n
i=2n+2 ( n + 2) =
2n2
P
P
P
3n
3n(3n+1)
i
1
i
+ 3. Then 0 ≤
S(f, Δn ) = n1 ni=1 ni + n1 2n
i=n+1 ( n + 1) + n
i=2n+1 ( n + 2) =
2n2
5
J(f ) − j(f ) ≤ S(f,
R Δ) − s(f, Δ) = n and the integrability of f follows as, necessarily
j(f ) = J(f ). Also f = j(f ) = supn→∞ s(f, Δn ) = 7.5.
1
iii. [2 marks] Using one of the results in the lectures, we have 1+(n+1)
2011
R
n+1
1
1
1
= (inf x∈[n,n+1] 1+x2011 )(n+1−n) ≤ n 1+x2011 dx ≤ (supx∈[n,n+1] 1+x2011 )(n+1−n) = 1+n12011 .
R n+1
Hence limn→∞ n 1+x12011 = 0 (Sandwich Theorem).
iv. [2 marks] By L’Hôpital’s rule, the given expression has the same limit as [f 0 (a +
h) − f 0 (a − h)]/2h, if this last limit exists. But then lim h→0 [f 0 (a + h) − f 0 (a − h)]/2h =
limh→0 12 [(f 0 (a + h) − f 0 (a))/h + (f 0 (a) − f 0 (a − h))/h = f 00 (a), since f 0 is differentiable at
a (f 00 is twice differentiable at a) and therefore both ratios on the right hand side of the
previous identity converge to f 00 (a).
2.
i. [2 marks] A collection Σ of subsets of R is an algebra if it satisfies the following three
axioms: a. R ∈ Σ. b. If A, B ∈ Σ, then A\B ∈ Σ. c. If A1 , A2 , ..., An ∈ Σ, then ∪Ai ∈ Σ.
A collection Σ of subsets of R is a σ-algebra if it is an algebra and has the property that
∪Ai ∈ Σ for any countable collection of sets Ai , i ≥ 1.
ii. [1 mark] The Borel σ-algebra on R is the σ-algebra generated by all open intervals
(a, b) ∈ R.
iii. [2 marks] For any q ∈ Q, the set (−∞, q) ∪ (q, +∞) = ∪k>1 (q − k, q) ∪ (q, q + k) is a
Borel set since it is a countable union of open intervals. Hence also its complement {q} is a
Borel set. Then Q = ∪q∈Q {q} is a Borel set as it is a countable union of Borel sets.
iv. [1 mark] A set K ⊂ R is called compact if it has the property that any open covering
of K contains a finite subcover.
v. [2 marks] The collection of sets (An )n , where An = (−1, 1 − n1 ), forms an open covering
of [0, 1). Assume that (An )n , contains a finite subcover. Then there exists a finite set
of indices {n1 , ..., nl } such that (Ani )i=1,...,l forms a cover for [0, 1). However ∪li=1 Anl =
(−1, 1 − min{n11,...,nl } ) 6 ⊃[0, 1). Hence [0, 1) cannot be compact.
3.
i. [1 mark] Take, for example, the set R. Give an example of a Borel set that is neither
closed not open (no proof required).
ii. [1 mark] Take, for example, the function defined as follows: f (x) = 1/2 if x ∈ [−2, 0],
f (x) = 0 if x ∈ (0, 2].
iii. [1 mark] Take, for example, the function f : R2 → R defined as
1 f or (x, y) 6= (1, 1)
.
f (x, y) =
0 f or (x, y) = (1, 1)
2