221 Analysis 2, 2008–09
Suggested solutions to exercise sheet 4
Let (X, M, m) be a measure space.
1. Fix a measurable set B ∈ M. Show that the function mB : M → [0, ∞]
given by mB (A) = m(A ∩ B) for A ∈ M, is a measure.
Solution. We have mB (∅) = m(∅ ∩B)
S = m(∅) = 0. If (Ai )i≥1 is a sequence
of pairwise disjoint sets in M and AS
= ∞
i=1 Ai , then (A
Si∞∩B)i≥1 is a sequence
∞
of pairwise disjoint sets in M and i=1 (Ai ∩ B) = ( i=1 Ai ) ∩ B = A ∩ B.
So
∞
X
mB (Ai ) =
i=1
∞
X
m(Ai ∩ B) = m
i=1
S∞
i=1 (Ai
∩ B) = m(A ∩ B) = mB (A).
So mB is countably additive. So mB is a measure.
2. (a) Let f : X → [0, ∞] be a nonnegative function and suppose that An ⊆ X
with An ↑ X. Show that if fn = f χAn then fn ↑ f .
(b) Show that if f : R → R is integrable (with respect to Lebesgue measure)
then
Z
Z n
f (x) dx = lim
f (x) dx.
n→∞
−n
[First consider what happens if f (x) ≥ 0 for all x ∈ R and use the
Monotone Convergence Theorem].
(c) Give an example
Z nof a function f : R → R which is not integrable, but
for which lim
f (x) dx = 0.
n→∞
−n
S∞
Solution. (a)
S∞We have A1 ⊆ A2 ⊆ A3 ⊆ . . . and n=1 An = X. Let x ∈ X.
We have x ∈ n=1 An , so x ∈ An for some n ≥ 1. Let n0 ≥ 1 be the smallest
such n. The sequence (fn (x))n≥1 = (0, 0, 0, . . . , 0, f (x), f (x), f (x), . . . ) is
{z
}
|
n0 −1
then increasing (since f (x) ≥ 0) with limit f (x).
So for any x ∈ X, fn (x) ↑ f (x). So fn ↑ f .
(b) Let An = [−n, n]. Suppose first that f : R → [0, ∞) and let fn = f χAn .
Since An ↑ R we have fn ↑ f by (a), so by the Monotone Convergence
Theorem,
Z n
Z
Z
Z
f (x) dx = fn (x) dx → f (x) dx = f (x) dx as n → ∞.
−n
1
if f : R → RR is any integrable function,
previous paragraph,
RNow
R n − then byR the
n
∞
∞
+
+
f (x) dx → −∞ f (x) dx and −n f (x) dx → −∞ f − (x) dx as n → ∞.
−n
So
Z n
Z n
Z n
+
f (x) dx =
f (x) dx −
f − (x) dx
−n
Z−n
Z −n
Z
+
−
→ f (x) dx − f (x) dx = f (x) dx as n → ∞.
R
(c)
Let
f
(x)
=
x.
Then
|f (x)| dx = ∞, so f is not integrable. However,
Rn
Rn
f
(x)
dx
=
x
dx
=
0
for any n.
−n
−n
For the next two questions, work with the measure space (R, M, m) where M
denotes the Lebesgue measurable subsets of R and m is Lebesgue measure.
3. (a) Let f : R → [0, ∞), f (x) = x χ[0,1) (x). Sketch the graph of f .
(b) Find a monotone increasing sequence (ϕn )n≥1 of simple nonnegative
measurable functions ϕn : R → [0, ∞) with ϕn ↑ f .
[Hint: look at the proof of Theorem 26].
R
(c) RFind ϕn dm and use the Monotone Convergence Theorem to find
f dm. [Don’t use the Fundamental Theorem of Calculus!]
(d) RLet I = R \ Q. Show that f = f χI almost everywhere, and compute
f dm.
I
Solution. (a) [omitted]
(b) The procedure in Theorem 26 is to approximate f (x) by rounding down
to the nearest integer multiple of 2−n . Since f (x) = x for x ∈ [0, 1), we have
(
k/2n
if x ∈ [0, 1) and k ∈ Z with k/2n ≤ x < (k + 1)/2n
ϕn (x) =
0
if x 6∈ [0, 1)
So
ϕn (x) =
n −1
2X
k=0
k
χ[k/2n ,(k+1)/2n ) .
2n
(c) We have
Z
n −1
Z 2X
k
χ[k/2n ,(k+1)/2n ) dm
ϕn dm =
2n
k=0
=
=
n −1
2X
k=0
n −1
2X
k=0
k
m[ kn , k+1
n )
2n 2 2
k 1
·
2n 2n
n
2 −1
1 X
1
= 2n
k = 2n · 21 (2n − 1) · 2n
2 k=0
2
=
1
2n − 1
= (1 − 2−n ).
n
2·2
2
2
Since ϕn ↑ f , by the Monotone Convergence Theorem we have
Z
Z
1
1
f dm = lim
ϕn dm = lim (1 − 2−n ) = .
n→∞
n→∞ 2
2
(d) If x ∈ I then f (x) = f χI (x). So
{x ∈ X : f (x) 6= f χI (x)} ⊆ R \ I = Q.
Since Q is countable it has Lebesgue measure zero; so {x ∈ X R: f (x) 6=
fR χI (x)} alsoRhas measure zero. So f = f χI almost everywhere. So f dm =
f χI dm = I f dm.
4. (a) Let c ∈ R and let A be a Lebesgue
measurable
subset of R. Show that
R
R
χA (x − c) = χA+c (x) and χA+c (x) dx = χA (x) dx.
(b) Show that if ϕ : R → [0, ∞) is a simple nonnegative
measurable
function
R
R
and ϕc : R → [0, ∞), ϕc (x) = ϕ(x − c) then ϕc dm = ϕ dm.
(c) Use the Monotone Convergence Theorem to show that if f : R → [0, ∞]
is a nonnegative measurable
function
and fc : R → [0, ∞] is given by
R
R
fc (x) = f (x − c) then fc dm = f dm.
(d) Show that if f : R → R is an integrable functionR and fc : RR → R is given
by fc (x) = f (x − c) then fc is integrable, and fc dm = f dm.
[You may use the fact that fc is measurable without proof.]
Solution. (a) Recall that
Lebesgue measure is translation
invariant, so
R
R
m(A + c) = m(A). But χA+c dm = m(A + c) and χA dm = m(A). Also,
χA+c (x) = 1 ⇐⇒ x ∈ A + c ⇐⇒ x − c ∈ A ⇐⇒ χA (x − c) = 1, and
otherwise χA+c (x) = 0 = χA (x − c). So χA+c = χA .
P
P
P
(b) Let ϕ(x) = ni=1 ai χAi . Then ϕc (x) = ni=1 ai χAi (x−c) = ni=1 ai χAi +c (x).
So
Z
Z
Z
Z
n
n
X
(a) X
ϕc dm =
ai χAi +c dm =
ai χAi dm = ϕ dm.
i=1
i=1
(c) Let ϕn be a monotone increasing sequence of simple nonnegative measurable functions with ϕn ↑ f (such a sequence exists by Theorem 26). Then
ϕn,c (x) = ϕn (x − c) defines a monotone increasing Rsequence of Rsimple nonnegative measurable functions with ϕn,c ↑ fc and ϕn,c dm = ϕn dm for
each n ≥ 1 by (b). So by the Monotone Convergence Theorem,
Z
Z
Z
Z
(b)
M CT
M CT
fc dm = lim
ϕn,c dm = lim
ϕn dm =
f dm.
n→∞
n→∞
R
R
(d) If f is integrable then |f | dm < ∞, so by (c), we have |fc | dm < ∞.
So fc is integrable, and since (fc )+ and (fc )− are nonnegative measurable
functions, we can apply (c) to deduce:
Z
Z
Z
Z
Z
Z
(c)
+
−
+
−
fc dm = (fc ) dm − (fc ) dm = f dm − f dm = f dm.
3
5. Let f, g : X → [−∞, ∞] be two measurable functions.
(a) If q ∈ Q, show that g −1 [−∞, q) and f −1 (q, ∞] ∩ g −1 [−∞, q) are both
measurable sets.
(b) Show that for any x ∈ X with f (x) > g(x), there is a rational number q
with g(x) < q < f (x). Deduce that
[
{x ∈ X : f (x) > g(x)} =
f −1 (q, ∞] ∩ g −1 [−∞, q).
q∈Q
Why does this show that this set is measurable?
(c) Deduce that the set {x ∈ X : f (x) 6= g(x)} is measurable.
Solution. (a) We have [−∞, q) = {−∞} ∪ A where A is the Borel set
A = (−∞, q). Since g is measurable, g −1 [−∞, q) is measurable.
Since f is measurable, f −1 (q, ∞] is measurable. The measurable sets form
a σ-algebra, so are closed under [countable] intersections. So f −1 (q, ∞] ∩
g −1 [−∞, q) is measurable.
(b) If f (x) > g(x) then the open interval (g(x), f (x)) ⊆ R is not empty. So
it contains a rational number q; so g(x) < q < f (x).
S
Let S = q∈Q f −1 (q, ∞] ∩ g −1 [−∞, q). If x ∈ X with f (x) > g(x) then
f (x) > q and g(x) < q for some q ∈ Q, so x ∈ f −1 (q, ∞] ∩ g −1[−∞, q); so
x ∈ S.
Conversely, if x ∈ X with x ∈ S then x ∈ f −1 (q, ∞] ∩ g −1 [−∞, q) for some
q ∈ Q. So f (x) > q and g(x) < q, so f (x) > q > g(x). So f (x) > g(x).
Since Q is countable, S is the countable union of the measurable sets f −1 (q, ∞]∩
g −1 [−∞, q) for q ∈ Q. The measurable sets form a σ-algebra, so are closed
under countable unions. So S is measurable.
(c) For x ∈ X we have f (x) 6= g(x) ⇐⇒ f (x) > g(x) or f (x) < g(x). So
{x ∈ X : f (x) 6= g(x)} = {x ∈ X : f (x) > g(x)} ∪ {x ∈ X : g(x) > f (x)}.
By part (b), both of the sets in this union are measurable, so {x ∈ X : f (x) 6=
g(x)} is measurable.
6. Let f, g : X → [−∞, ∞] be two measurable functions.
(a) Show that if α ∈ R and q ∈ Q then f −1 (q, ∞] ∩ g −1(α − q, ∞] is a
measurable set.
(b) Suppose that g(x) 6= −∞ for all x ∈ X. The function f + g is then well
defined. Let α ∈ R.
Show that if f (x) + g(x) > α then there is a rational number q with
α − g(x) < q < f (x). Deduce that
[
f −1 (q, ∞] ∩ g −1(α − q, ∞].
(f + g)−1(α, ∞] =
q∈Q
Why does this show that this set is measurable?
(c) Deduce that f + g is a measurable function.
4
Solution. (a) Since f and g are measurable, f −1 (q, ∞] and g −1 (α − q, ∞]
are both measurable sets. The measurable sets form a σ-algebra, so are closed
under [countable] intersections. So f −1 (q, ∞] ∩ g −1(α − q, ∞] is measurable.
(b) If f (x) + g(x) > α then f (x) > α − g(x), so there is a rational number q
in the non-empty open interval (α − g(x), f (x)); then α − g(x) < q < f (x).
S
Let S = q∈Q f −1 (q, ∞] ∩ g −1(α − q, ∞]. If x ∈ (f + g)−1(α, ∞] then f (x) +
g(x) > α, so there is a rational number q ∈ Q with α − g(x) < q < f (x). So
f (x) > q ⇐⇒ x ∈ f −1 (q, ∞] and g(x) > α − q ⇐⇒ x ∈ g −1 (α − q, ∞]. So
x ∈ f −1 (q, ∞] ∩ g −1 (α − q, ∞]; so x ∈ S.
Conversely, if x ∈ S then there is q ∈ Q with x ∈ f −1 (q, ∞] ∩ g −1 (α − q, ∞],
so f (x) > q and g(x) > α − q, so f (x) + g(x) > q + α − q = α, so x ∈
(f + g)−1 (α, ∞].
So S = (f + g)−1(α, ∞], and this is the countable union of the measurable
sets f −1 (q, ∞] ∩ g −1 (α − q, ∞] for q ∈ Q. So this set is measurable.
(c) In (b) we showed that (f + g)−1 (α, ∞] is measurable for all α ∈ R. So
f + g is measurable.
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