MATH 435 - Some Answers
• Assignment 1, Question 6.
Define the inner Lebesgue measure m∗ (E) of a subset E of R by
m∗ (E) = sup{m(K) : K is compact, K ⊆ E}.
Assuming that E has finite outer measure, prove that E is measurable if and only
if m∗ (E) = m∗ (E).
If E is measurable and > 0, then (by standard results), there exists a closed set
F and an open set U such that F ⊆ E ⊆ U and m(U \ E) < 4 and m(E \ F ) < 4 .
You need a compact set in place of F . Let Fn = F ∩ [−n, n]. Now m(F ) = lim Fn
and so there exists a positive integer k such that m(Fk ) > m(F ) − 4 . The set Fk
is compact and by additivity and monotonicity of m, we get m(U ) − m(Fk ) < .
Now m(Fk ) ≤ m∗ (E) ≤ m∗ (E) ≤ m(U ). It follows that m∗ (E) − m∗ (E) < for
every positive . This proves the equality m∗ (E) = m∗ (E).
For the converse assume that m∗ (E) = m∗ (E) and > 0. Then (be definition of
sup and inf), there exists a compact set K and an open set U such that K ⊆ E ⊆ U
and m(U ) − m(K) < . So m∗ (U \ E).
• Assignment 2, Question 3 .
Recall the construction of a partition {Pq }q∈Q∩[0,1) of the interval [0, 1) where Pq =
P0 +̂q (addition mod 1).
(a) Prove that if E is a (Lebesgue ) measurable subset of P0 , them m(E) = 0.
Let {q} be an enumeration of the rationals in [0, 1). If E is measurable subset of
P0 , consider Ej = E +̂qj ⊆ P j. The sets Ej are disjoint and they have the same
measure. Their union
P U a subset
P of [0, 1], so has measure ≤ 1. By countable
additivity m(U ) = j M (Ej ) = j m(E). If m(E) > 0, then the sum is ∞ which
is impossible.
(b) If A is a subset of [0, 1) and m∗ (A) > 0, prove that there is a nonmeasurable
subset of A.
Let Aj = A ∩ Pj . Then Aj are disjoint. If they are all measurable, then by part (a),
each has measure 0. But then by countable subadditivity, m∗ (A) = m∗ (∪j Aj ) ≤
P
∗
j m (Aj ) = 0. A contradiction. So at least one Aj is nonmeasurable.
• Assignment 2, Question 5.
∞
For a sequence of sets {An } define lim inf An = ∪∞
k=1 ∩j=k Aj . prove that m(lim inf An ) ≤
lim inf m(An ).
∞
Let Bk = ∩∞
j=k Aj and B = lim inf An = ∪k=1 Bj . Then Bk is an increasing sequence
of measurable set and so µ(B) = lim µ(Bk ). Furthermore Bk ⊆ Ak , so lim µ(Bk ) =
lim inf µ(Bk ) ≤ lim inf µ(Bk ). Thus µ(lim inf An ) = µ(B) ≤ lim inf µ(An ).
Alternatively, let fn = χAn . Then, after a short argument, it can be shown that
lim inf fn = χlim inf An . Verify! Now Fatou’s lemma implies the assertion.
• The preamble to assignment 3 states:
• Be very explicit about usage of convergence theorems.
• Assignment 3, Question 5 Let {rn } be an enumeration of the rational numbers in
the interval (0,
and let fn (x) = (x − rn )−2/3 if rn < x < 1 and = 0 otherwise.
P1)
∞
Next let f = n=1 2−n fn .
R1
(a) Prove that 0 fn = 3(1 − rn )1/3 . Note: Calculus formulae are good enough
for the integral over [rn + , 1] as the function is continuous there. Justify
“passing to the limit”.
increasing
Let fnm = fn χ[rn ,rn +1/m] . The sequence {fnm }∞
m=1 is monotone
R
R with
limit fn . By the Monotone Convergence Theorem fn = limm→∞ fnm .
Now, for m large enough (so that rn +1/m < 1), the integral of fnm equals the
integral of the continuous function (x − rn )−2/3 on the interval [rn + 1/m, 1]
which equals 3(1 − rn )1/3 − 3m−1/3 . Take the limit.
R1
R1
Beware: You cannot say that 0 x1 sin x1 dx = limn→∞ 1/n x1 sin x1 dx. The
limit exists (and = π/2) but the function in not integrable over (0, 1).
(d) Redefine f by making it 0 on E0 , where E0 = {x : f (x) = ∞}. Call the
redefined function g. Prove that for every subinterval [a, b] of [0, 1] and every
positive real M , the set {x ∈ [a, b] | g(x) > M } has positive measure, i.e. g is
“essentially” unbounded on every interval.
One of the functions hn = 2−n fn will give you this, the others will just add
icing to the cake. The interval [a, b] includes a rational rn . Now hn (x) will be
> M if rn < x < rn + M −3/2 2−3n/2 . If the right end of this interval is > b,
replace it by b. In either case there is an interval I of positive length on which
hn is > M . Now I \ E0 has positive measure and on which g(x) ≥ hn (x) > M.