MAT-4-EAP Essentials in Analysis and Probability M09

1. (a) State what is meant by an algebra and a σ-algebra of subsets of a set Ω.In each of the following two cases state, with reasons, whether or not A is an algebra of subsets

of Ω:

(i) A is the collection of all subsets of Ω = 1, 2, 3, · · · , 10 having an even number of elements.

(ii) A is the collection of all subsets E of Ω = R with the property that −x ∈ E whenever x ∈ E.[10 marks]

(b) Let A be the collection of all subsets A of R such that either A or R\A is countable. Show thatA is a σ-algebra.

[7]

(c) What is meant by a Borel subset of R?Let (fn) be a sequence of continuous real-valued functions of R, and let F be the set of all x ∈ R

such that the sequence (fn(x)) is bounded.Show that F is a Borel set. [8]

2. (a) Define the notions of additivity and countable additivity of a set function on an algebra ofsubsets of a set Ω.

Let A be the algebra of all subsets of N which are either finite or have finite complement, anddefine µ on A by µ(E) = 0 if E is finite and µ(E) = 1 if Ec is finite. Show that µ is additive butnot countable additive.

[8]

(b) Describe the construction of the Cantor middle third set E and show that E has Lebesguemeasure 0.

[7](c) Consider an infinite sequence of independent tosses of a fair coin. Let Xn be the number ofheads in the first n tosses. Show that, if 0 < α < 1

2, and 0 < s ≤ 1, then P(Xn ≤ αn) ≤ φ(s)n

where φ(s) = 12s−α(1 + s).

By making a suitable choice of s deduce that P(Xn ≤ αn) ≤ un for n = 1, 2, · · · , whereu = 2αα(1− α)1−α−1. Show also that u < 1.

[10]

[OVER]

3 (a) Define the notion of simple function on a measure space (Ω,F , µ) and explain briefly (withoutproofs) how simple functions are used to define the integral

∫fdµ of a nonnegative measurable

function f .[10]

(b) State the Monotone Convergence Theorem.

Show that (1 − x2)−1/2 =∑∞

n=0 4−n(2nn

)x2n for 0 ≤ x < 1 and by integrating over [0, 1] show

thatπ

2=

∞∑n=0

4−n

2n + 1

(2n

n

)[9]

(c) Give an example of a sequence of nonnegative Borel functions (fn) on [0, 1] such that fn converges

pointwise on [0, 1] to a limit f but∫ 1

0fn(x)dx does not converge to

∫ 1

0f(x)dx.

[6]

4. (a) State the Borel–Cantelli theorem on the probability of infinitely many occurrences of asequence of events.

[5]Let (Xn) be a sequence of independent random variables each having a geometric distribution

so that P(Xn = k) = pqk−1 for k = 1, 2, · · · . Let β = (log 1q)−1. Show that, with probability 1,

Xn > β log n for infinitely many n.Show also that if α > β then, with probability 1, Xn > α log n for only finitely many n.

[10]

(b) State the Dominated Convergence Theorem, and apply it to the series 1− x + x2 − x3 + · · · toshow that ∫ 1

0

dx

1 + x= 1− 1

2+

1

3− 1

4+ · · ·

Deduce the sum of the series on the right.[10]

[End of Paper]

Solutions

1. (a) A collection A of subsets of Ω is an algebra if (i) Ω ∈ A, (ii) A ∈ A implies Ac ∈ A and (iii)A, B ∈ A implies A ∪B ∈ A. For a σ-algebra (iii) is replaced by A1, A2, · · · ∈ A implies ∪An ∈ A.(i) Not an algebra since A = 1, 2 and B = 2, 3 belong to A but A ∪B = 1, 2, 3 does not.(ii) This is an algebra. Suppose E and F are in A. Then if x ∈ E ∪ F we have x ∈ E so −x ∈ E,likewise −x ∈ F , so −x ∈ E ∪ F , and hence E ∪ F ∈ A. And if y ∈ Ec then y /∈ E so −y /∈ E,since −y ∈ E would imply y = −(−y) ∈ E which is false, hence −y ∈ Ec and thus Ec ∈ A.

(b) As Rc = ∅ is countable, R ∈ A.Now suppose A ∈ A. Then if A is countable R\Ac = A is countable, whereas if R\A is countable

then Ac = R\A is countable, so in either case Ac ∈ A.Next suppose A1, A2, · · · ∈ A and let A = ∪kAk. If all Ak are countable then A is the union of a

sequence of countable sets and hence countable. Otherwise there exists k such that Ack is countable

and then Ac ⊆ Ack is also countable. In either case A ∈ A.

(c) A Borel set is a set belonging to the σ-algebra B generated by all intervals of the form [a, b)with a, b ∈ R.

Note that any open set in R is a countable union of intervals of this form, and hence is a Borelset, and it follows that any closed set is also Borel.

Now F is the set of all x ∈ R such that there exists k ∈ N such that for all n ∈ N we have|fn(x)| ≤ k. Thus F = ∪∞

k=1 ∩∞n=1 Ekn where Ekn = x : |fn(x)| ≤ k. Now Ekn is closed since fn is

continuous, hence Ekm ∈ B, and as B is a σ-algebra it follows that F ∈ B.

2. (a) An additive set function is a mapping µ : A → [0,∞] such that µ(∅) = 0 and µ(E ∪ F ) =µ(E) + µ(F ) whenever E and F are disjoint sets in A. µ is countably additive if in additionµ(E) =

∑∞n=1 µ(En) whenever E1, E2, · · · are disjoint sets in A with union E ∈ A.

Let A, B ∈ A be disjoint. Then Ac and Bc cannot both be finite. If A and B are both finitethen A ∪ B is also finite and µ(A ∪ B) = 0 = µ(A) + µ(B). If A and Bc are finite then (A ∪ B)c

is finite, so µ(A ∪ B) = 1 = 0 + 1 = µ(A) + µ(B). So in all cases µ(A ∪ B) = µ(A) + µ(B). Alsoµ(∅) = 0 as ∅ is countable, and so µ is additive.

Now let An = n for n ∈ N. Then ∪nAn = N and µ(N) = 1 but µ(An) = 0 for each n so∑µ(An) = 0 and so µ is not countably additive.

(b) Let E0 = [0, 1]. Then construct a set E1 by removing the ‘middle third’ of E0, i.e. by removing(1

3, 2

3). We get E1 = [0, 1

3] ∪ [2

3, 1], a union of two intervals of length 1

3. Then remove the middle

third of each of these two intervals, and obtain E2 = [0, 19] ∪ [2

9, 1

3] ∪ [2

3, 7

9] ∪ [8

9, 1] which is a union

of 4 intervals of length 19. Continuing in this way, we obtain after n steps a set En which is a union

of 2n closed intervals of length 3−n. Note that En+1 ⊆ En. Let E = ∩∞n=0En.

For each n, λ(En) = (23)n so λ(E) ≤ (2

3)n → 0 so λ(E) = 0.

(c) Xn has a binomial B(n, 12) distribution so P(Xn ≤ αn) = 2−n

∑k≤αn

(nk

)≤ 2−ns−αn

∑k≤αn

(nk

)sk ≤

2−ns−αn∑n

k=0

(nk

)sk = 2−ns−αn(1 + s)n = φ(s)n.

We have φ′(s) = 12−αs−α−1 + (1 − α)s−α = 1

2sα−1−α + (1 − α)s. Thus φ′(s0) = 0 for

s0 = α1−α

< 1 and φ′(s) is positive for s > s0 and negative for 0 < s < s0 so φ has a minimum at s0.

Now φ(s0) = 12( α

1−α)−α 1

1−α= u and so P(Xn ≤ αn) ≤ un. And φ(1) = 1 and so u = φ(s0) < 1.

3. (a) A simple function is a measurable function which takes only finitely many values. Asimple function f can be expressed as f =

∑ni=1 λiχEi

where Ei ∈ F . Then one defines∫

fdµ =∑ni=1 λiµ(Ei) and shows that this is independent of the choice of representation. One also shows

that f ≤ g implies∫

fdµ ≤∫

gdµ for simple f and g. Then if f is nonnegative and measurable,one constructs an increasing sequence (fn) of simple functions such that fn → f pointwise. Then

the sequence (∫

fndµ) is increasing, so tends to a limit, which is defined to be∫

fdµ. To show itis well-defined one checks that any other increasing sequence of simple functions converging to fgives the same result.

(b) The MCT states that if (fn) is an increasing sequence of nonnegative measurable functions andf = limn fn then

∫fndµ →

∫fdµ.

For 0 ≤ x < 1, we have (1−x2)−1/2 = 1+(−12)(−x2)+

(− 12)(− 3

2)

2!(−x2)2+· · · =

∑∞n=0

1.3···(2n−1)n!2n x2n =∑∞

n=0

(2nn

)4−nx2n. Since all the terms are nonnegative the MCT applied to the partial sums gives∫ 1

0(1− x2)−1/2dx =

∑∞n=0

(2nn

)4−n

∫ 1

0x2ndx =

∑∞n=0

4−n

2n+1

(2nn

). The integral is [sin−1 x]10 = π

2.

(c) Let fn(x) = n if x ∈ (0, 1n) but 0 otherwise. Then fn(x) → 0 for all x ∈ [0, 1] (since fn(0) = 0

for all n, and if x > 0 then fn(x) = 0 for n > 1x). So fn → f pointwise where f(x) = 0 but

∫ 1

0fn = 1

for all n while∫ 1

0f = 0.

4. (a) Borel-Cantelli: if E1, E2, · · · is a sequence of events and F = ∩∞m=1 ∪∞

n=m En then (i) if∑P(En) < ∞ then P(F ) = 0 and (ii) if

∑P(En) = ∞ and E1, E2, · · · are independent then

P(F ) = 1.We have P(Xn > k) = qk for k ∈ N so P(Xn > β log n) = P(Xn > bβ log nc) = qbβ log nc ≥

qβ log n = n−1 and∑

n−1 diverges so the first result follows from part (ii) of the Borel-Cantellitheorem.

If α > β, fix σ with α > σ > β. Then for n large enough bα log nc > σ log n so P(Xn >α log n) = qbα log nc < qσ log n = n−σ/β. As σ/β > 1, the series

∑n−σ/β converges so the second

result follows from part (i) of the Borel-Cantelli theorem.

(b) The DCT states that if (fn) is a sequence of measurable functions on Ω such that fn → fa.e., and if there is a nonnegative function φ ∈ L1(µ) such that for each n, |fn| ≤ φ a.e., then∫

fndµ →∫

fdµ.

Let fn(x) = 1 − x + x2 − · · · + (−1)nxn = 1−(−x)n+1

1+x. Then fn(x) → 1

1+xpointwise on [0, 1)

and |fn(x)| ≤ 2 on [0, 1) so the DCT, applied to the sequence (fn) and Lebesgue measure on [0, 1],

and with φ(x) = 2, gives∫ 1

0fn →

∫ 1

0dx

1+x. Since

∫ 1

0fn = 1 − 1

2+ 1

3− · · · + (−1)n

n+1it follows that∫ 1

0dx

1+x=

∑∞n=0

(−1)n

n+1.

Since∫ 1

0dx

1+x= [log x]10 = log 2 it follows that

∑∞n=0

(−1)n

n+1= log 2.