Mat 357 Assignment 5 solutions

Mat 357 Assignment 5 solutions
Jordan Bell
April 14, 2015
6.8 (i) Suppose that m∗ (A) < ∞. If you can completely digest this argument, you can probably do the case of m∗ (A) = ∞. (And probably use the
result for finite outer measure, rather than having to rewrite the argument from
scratch.)
Let A ⊂ R. Because outer measure is defined as an infimum, for each n
1
∗
there is
Tnan open set Un containing A such that m(Un ) ≤ m (A) + n . Let
Vn = i=1 Ui , each of which is an open
set containing A and which satisfies
T∞
m(Vn ) ≤ m(Un ) ≤ m∗ (A)+ n1 . Let H = n=1 Vn , which is a Gδ set that contains
A, and for each n, m(H) ≤ Vn ≤ m∗ (A) + n1 . Because this is true for all n it
follows that m(H) ≤ m∗ (A). But because A ⊂ H, m∗ (A) ≤ m∗ (H) = m(H),
so m(H) = m∗ (A).
If E is a measurable set such that A ⊂ E, m(H \ E) = m(H ∩ E c ) =
m(H) − m(H ∩ E) = m∗ (A) − m(H ∩ E). Because H and E both contain A,
m(H ∩ E) ≥ m∗ (A), hence
m(H \ E) = m∗ (A) − m(H ∩ E) ≤ m∗ (A) − m∗ (A) = 0,
hence m(H \ E) = 0.
If H1 is a hull of A and H2 is a hull of A, then, because H1 is a hull and
H2 is a measurable set containing A, m(H1 \ H2 ) = 0, and because H2 is a hull
and H1 is a measurable set containing A, m(H2 \ H1 ) = 0, so H1 and H2 differ
by a set of measure 0, showing A has a hull and that modulo sets of measure 0
it has a unique hull.
(ii) This is like (i), except we use m∗ rather than m and we use closed sets
instead of open sets and unions instead of intersections.
6.54 (a) If fn : [a, b] → R converge almost uniformly to f , then there is a
set S ⊂ [a, b] with m(S) = 0 such that fn converge uniformly to f on [a, b] \ S.
Thus, for any > 0 we have m(S) < , which shows that fn converges nearly
uniformly to f .
Let fn = χ[1/n,2/n] . fn converges pointwise to 0 (for any x, eventually x is
not in the window [1/n, 2/n]). Thus if fn converges almost uniformly, its limit
is 0. Suppose by contradiction that there is some S ⊂ [a, b] with m(S) = 0
such that for any η > 0 there is some n0 such that if n ≥ n0 and x ∈ [0, 2] \ S
then |fn (x)| ≤ η (namely, that fn converges uniformly to 0 on [0, 2] \ S). Take
η < 1. Because m(S) = 0, for each n the set [1/n, 2/n] \ S has positive measure
1
so is nonempty; let xn be an element of it. In particular, xn0 ∈ [0, 2] \ S and
|fn0 (xn0 )| = 1 > η, a contradiction.
On the other hand, let 1 > > 0 and let S = [0, /2], so m(S) < . For
n ≥ 2 we have [1/n, 2/n] ⊂ S, and so for x ∈ [0, 2] \ S we have |fn (x)| = 0,
which shows that fn converges uniformly to 0 on [0, 2] \ S. That is, fn converges
nearly uniformly to 0.
In summary, if a sequence converges almost uniformly to f then it converges
nearly uniformly to f , but there are sequences that converge nearly uniformly
and not almost uniformly.
(b) (i) It is apparent that for a fixed k, X(k, 1) ⊃ X(k, 2) ⊃ · · · and that for
a fixed l, X(1, l) ⊂ X(2, l) ⊂ · · · .
The set E = {x ∈ [a, b] : fn (x) does not converge} is measurable, and we
are given that m(E) = 0. For x ∈ [a, b], x ∈ E if and only if for there is some
l ≥ 1 such that for all k ≥ 1 such that x 6∈ X(k, l). That is,
E=
∞
∞ \
[
X(k, l)c .
l=1 k=1
Because m(E) = 0 and E is aTunion, each of the sets in the union has measure
∞
0. HenceSthe complement of k=1 X(k, l)c differs from [a, b] by a measure zero
∞
set, i.e. k=1 X(k, l) differs from [a, b] by a measure zero set.
(ii) Because X(1, l) ⊂ X(2, l) ⊂ · · · , we have X(1, l)c ⊃ X(2,
l)c ⊃ · · · , and
T
T∞
∞
c
c
so k=1 X(k, l) = liml→∞ X(k, l) . And since the measure of k=1 X(k, l)c is
0, the measure of X(k, l)c tends to 0 as k → ∞. Thus, for > 0 there is some kl
(depending on and l, where we only indicate the dependence on l) such that
m(X(kl , l)c ) < 2l .
T∞
(iii)SWe continue to use the same as above. Let X = l=1 X(kl , l). Then
∞
X c = l=1 X(kl , l)c , and so
m(X c ) ≤
∞
X
m(X(kl , l)c ) <
l=1
∞
X
= .
2l
l=1
Now we show to fn converges to f uniformly on X. Let η > 0. For kl ≥
and n ≥ kl and for x ∈ X, because x ∈ X(kl , l) we have
|fn (x) − f (x)| <
1
η
1
≤ η.
l
That is, fn converges to f uniformly on X.
56. (a) If x ∈ Q, for sufficiently large n we have cos(πn!x) = 1, hence
limk→∞ limn→∞ (cos(πn!x))k = 1 = χQ (x).
If x 6∈ Q, then for each n we have | cos(πn!x)| < 1, for which
lim lim (cos(πn!x))k = 0 = χQ (x).
n→∞ k→∞
2