Midterm Solutions
Math 461, Spring 2016
Problem 1.
If S and T are sets, the symmetric difference of S and T is the set
S 4 T = (S ∩ T c ) ∪ (S c ∩ T ).
If {En } is a sequence of measurable sets in [0, 1], we say that {En } converges to a measurable
set E if
lim m(En 4 E) = 0.
n→∞
Part (a)
Proposition. Let E1 , E2 , . . . and E be measurable sets in [0, 1] and suppose that χEn → χE
pointwise almost everywhere on [0, 1]. Then {En } converges to E.
Proof. Since χEn → χE pointwise almost everywhere, we know that |χEn −χE | → 0 pointwise
almost everywhere. But |χEn − χE | ≤ 1 for all n, so by the bounded convergence theorem
Z
lim m(En 4 E) = lim
n→∞
n→∞
χEn 4E dm
X
|χEn − χE | dm =
= lim
n→∞
Z
Z
Z
X
lim |χEn − χE | dm =
X n→∞
0 dm = 0.
X
Part (b)
One simple example is the following sequence {En } of closed intervals:
[0, 1],
0, 21 , 12 , 1 ,
1 1 2 2 0, 3 , 3 , 3 , 3 , 1 ,
1
0, 4 , . . .
Since m(En ) → 0 as n → ∞, this sequence converges to the empty set. However, for each
number x ∈ [0, 1], there exist infinitely many terms of this sequence that contain x, as well
as infinitely many that do not contain x, and therefore {χEn (x)} does not converge.
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Part (c)
Proposition. Let {En } be a sequence of measurable sets in [0, 1], and suppose that
∞
X
m(En 4 En+1 ) < ∞.
n=1
Then the sequence {χEn } converges pointwise almost everywhere.
Proof. For each N ∈ N, let
[
ZN =
(En 4 En+1 ).
n≥N
and let Z =
T
N ∈N
ZN . Since
m(ZN ) ≤
X
m(En 4 En+1 ),
n≥N
we know that m(ZN ) → 0 as N → ∞, and hence m(Z) = 0. But if x ∈ Z c , then x ∈ ZNc for
some N ∈ N, and hence x ∈
/ En 4 En+1 for all n ≥ N . It follows that χn (x) = χn+1 (x) for
all n ≥ N , so the sequence {χn (x)} converges. Thus {χEn } converges pointwise on Z c .
Part (d)
Proposition. Let {En } be a sequence of measurable sets in [0, 1], and suppose that for
every > 0, there exists an N ∈ N so that
i, j ≥ N
⇒
m(Ei 4 Ej ) < .
Then {En } converges to some measurable set E.
Proof. Let N1 < N2 < · · · be an increasing sequence so that
i, j ≥ Nk
⇒
m(Ei 4 Ej ) <
1
2k
for each k, and let Fk = ENk . Then m(Fk 4 Fk+1 ) < 1/2k for each k, so by part (c) the
sequence {χFk } converges pointwise almost everywhere. The limit is a measurable function
with range {0, 1}, so χFk → χE pointwise almost everywhere for some measurable set E. By
part (a), it follows that {Fk } converges to E.
To prove that {En } converges to E, observe that for n ≥ Nk we have
m(En 4 E) ≤ m(En 4 Fk ) + m(Fk 4 E) ≤
1
+ m(Fk 4 E).
2k
But the quantity on the right goes to 0 as k → ∞, so it follows that m(En 4 E) → 0
as n → ∞.
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Problem 2.
If f : R → R is a bounded measurable function and g : R → R is an L1 function, the
convolution of f and g is the function f ∗ g : R → R defined by
Z
(f ∗ g)(x) =
f (x − t) g(t) dm(t).
R
(Note that this integral always exists, so f ∗ g is a well-defined function.)
Part (a)
Proposition. Let fn : R → R be a uniformly bounded sequence of measurable functions
converging pointwise to a function f : R → R, and let g : R → R be an L1 function. Then
fn ∗ g converges pointwise to f ∗ g.
Proof. Let x ∈ R, and let M > 0 so that |fn | ≤ M for all M . Then |fn (x − t)g(t)| ≤ |M g(t)|
for all n ∈ N and all t ∈ R. But |M g| is an L1 function, so by the dominated convergence
theorem
Z
lim (fn ∗ g)(x) = lim
fn (x − t) g(t) dm(t)
n→∞
n→∞
R
Z
Z
lim fn (x − t) g(t) dm(t) =
=
f (x − t) g(t) dm(t) = (f ∗ g)(x).
R n→∞
R
Part (b)
Proposition. If f : R → R is a bounded continuous function and g : R → R is an L1
function, then f ∗ g is continuous.
Proof. Let x ∈ R, and let {xn } be a sequence in R that converges to x. We will prove that
(f ∗ g)(xn ) converges to (f ∗ g)(x). Since f is continuous, we know that
lim f (xn − t) = f (x − t)
n→∞
for all t ∈ R. Since f is bounded, there exists a constant M > 0 so that |f | ≤ M . Then
|f (xn − t) g(t)| ≤ |M g(t)| for all n ∈ N and all t ∈ R. But |M g| is an L1 function, so by the
dominated convergence theorem
Z
lim (f ∗ g)(xn ) = lim
n→∞
n→∞
f (xn − t) g(t) dm(t)
R
Z
=
Z
lim f (xn − t) g(t) dm(t) =
f (x − t) g(t) dm(t) = (f ∗ g)(x).
R n→∞
R
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