Real Analysis HW 7 Solutions
Problem 44: Let f be integrable over R and > 0. Establish the following three approximation properties.
R
(i) There is a simple function η on R which has finite support and R |f − η| < (ii) RThere is a step function s on R which vanishes outside a closed, bounded interval and
|f − s| < .
R
(iii) RThere is a continuous function g on R which vanishes outside a bounded set and
|f − g| < .
R
Solution:
(i) As shown in Problem 24, if f is non-negative we may find an increasing sequence of
non-negative simple functions {φn } with finite support such that φn → f pointwise. It
follows by monotone convergence that we can find a φ such that
Z
Z
|f − φ| =
f − φ < .
R
R
For generalR f we write f = f + − fR− , and find η1 and η2 simple and of finite support
such that R |f + − η1 | < /2 and R |f − − η2 | < /2. Since f + and f − have disjoint
support, we see that η1 and η2 must also have disjoint support, therefore η = η1 − η2
is also simple with finite support and it follows that
Z
Z
Z
+
|f − η| ≤
|f − η1 | + |f − − η2 | < .
R
R
R
(ii) By part (i), since we can approximate by simple functions, by the triangle inequality
it suffices to show that the characteristic function χE of a bounded measurable set
E can be approximated by step functions. Note that since E isSmeasurable, we can
∞
find a disjoint collection of open intervals {Ik }∞
k=1 such that O =
k=1 Ik , and m(O ∼
E) S
< /2. Since O must have finite measure we can find an N large enough such that
m( ∞
k=N +1 Ik ) < /2. Therefore
N
X
s=
χIn
k=1
1
is a step function and
Z
|χE − s| ≤
R
N Z
X
k=1
≤m
Z
∞
X
|χE∩Ik − χIk | +
R
N
[
!
Ik ∼ E
+m
k=1
χE∩Ik
R
k=N +1
∞
[
!
Ik ∩ E
k=N +1
∞
[
≤ m(O ∼ E) + m
!
Ik
< .
k=N +1
(iii) Using part (ii), once again we see by the triangle inequality that it suffices to show
that any characteristic function of a bounded interval χ[a,b] can be approximated by a
continuous function. Let g be the continuous function which is 1 on [a + /2, b − /2]
and linearly interpolated to 0 outside of [a, b], then
Z
|χ[a,b] − g| < m([a, a + /2) ∪ (b − /2, b]) = .
R
Problem 46: (Riemann-Lebesgue) Let f be integrable over (−∞, ∞). Show that
Z ∞
lim
f (x) cos nxdx = 0.
n→∞
−∞
Solution: Using the result from problem 44, we Rknow there exists a step functon s, vanishing
outside of a closed bounded interval such that R |f − s| < /2. Let s be the step function
which we write in cannonical form as
s=
K
X
sk χ(ak ,bk ) ,
k=1
where {(ak , bk )} are a disjoint collection of bounded open intervals and {sk } are distinct.
Note that it doesn’t matter that we don’t define s at the end points of the intervals since
they are a set of measure 0. We see that
Z
∞
−∞
Z
K
X
s(x) cos nxdx ≤
|sk | =
k=1
K
X
k=1
≤
bk
ak
cos nxdx
|sk |
| sin nbk − sin nak |
n
2K max{|si |}
.
n
2
Therefore if n > N ≡ 4K max{si }/, we conclude
Z ∞
Z
Z ∞
≤
f
(x)
cos
nxdx
|f
(x)
−
s(x)|dx
+
−∞
−∞
∞
−∞
s(x) cos nxdx
< /2 + /2 = .
Problem 47: Let f be integrable over (−∞, ∞).
(i) Show that for each t,
∞
Z
Z
∞
f (x + t) dx.
f (x) dx =
−∞
−∞
(ii) Let g be a bounded measurable function on R. Show that
Z ∞
g(x) · [f (x) − f (x + t)]dx = 0.
lim
t→0
−∞
Solution:
P
(i) Note that this result is true for simple functions. Let ϕ = nk=1 ci χEk be simple, then
since χE (x + t) = χE−t (x), we haven by the translation invariance of Lebesgue
Z
ϕ(x + t) dx =
R
n
X
ck m(Ek − t) =
n
X
Z
ck m(Ek ) =
ϕ(x) dx.
R
k=1
k=1
For the case of integrable f , we may restrict ourselves to f non-negative, since we
may always write any integrable f as f = f + − f − , where f + (x) = max{f (x), 0} and
f − (x) = max{−f (x), 0}. For such a non-negative f , we know that there is a sequence
of simple functions with finite support {ϕn }∞
n=1 such that ϕn ≤ f with ϕn converging
to f pointwise and monotonically. In particular ϕ(· + t) → f (· + t) monotonically for
every fixed t. It follows from the monotone convergence theorem that
Z
Z
Z
Z
f (x) dx.
f (x + t) dx = lim
ϕn (x + t) dx = lim
ϕn (x) dx =
R
n→∞
n→∞
R
R
R
(ii) Let M < ∞ be a constant such that |g| ≤ M . Since f is integrable, we know by
Problem 44
R that for every > 0 there exists a h, continuous and of bounded support
such that R |f − h| < /2M . It follows that h uniformly continuous and bounded by
some constant N . Let {tn } be any sequence tn → 0 and define hn (x) = |h(x)−h(x+tn )|.
Then hn → 0 pointwise and 0 ≤ hn (x) ≤ 2N. It follows by the bounded convergence
theorem that
Z
lim
|f (x) − f (x + tn )| = 0.
n→∞
R
3
Therefore
Z
Z
g(x) · [f (x) − f (x + tn )] dx ≤ M
|f (x) − h(x)| dx
R
R
Z
Z
+M
|f (x + tn ) − h(x + tn )| dx + M
|f (x) − f (x + tn )| dx
R
R
By part i) and the way that h was chosen, the sum of the first two terms on the
righthand side above are bounded by . Taking the lim supn of both sides, we obtain
Z
Z
lim sup g(x) · [f (x) − f (x + t)] dx < + M lim
|f (x) − f (x + tn )| dx = .
n
n→∞
R
R
Since this is true for every > 0 and every {tn }, tn → 0, we conclude
Z
lim g(x) · [f (x) − f (x + t)] dx.
t→0
R
Problem 50: Let F be a family of functions, each which is integrable over E if and only if
for each > 0, there is a δ > 0 such that for each f ∈ F,
Z if A ⊆ E is measurable and m(A) < δ, then f < .
A
R R
Solution: Clearly by the triangle inequality if A |f | < , then A f and so uniform
integrability
R easily
implies
this condition. However, for any A measurable we may write
R
R
+ f + − f where A+ = {x ∈ A : f ≥ 0} and A− = {x ∈ A : f < 0} are
|f
|
=
A
A
A
disjoint. It follows that there exists δ + and δ − such that if m(A+ ) < δ + and m(A− ) < δ − ,
then
Z
Z
f < /2 and f < /2.
A−
A+
Choosing δ = min{δ + , δ − }, we conclude that if m(A) < δ, then
Z
|f | < .
A
Therefore this condition implies uniform integrability.
Problem 51: Let F be a family of functions, each of which is integrable over E. Show that
F is uniformly integrable over E if and only if for each > 0, there is a δ > 0 such that for
each f ∈ F, if U is open and m(E ∩ U) < δ,
Z
|f | < .
E∩U
4
Proof: Clearly if F is uniformly integrable, it follows trivially tha there exists a δ > 0 such
that if U is open and m(E ∩ U) < δ, then
Z
|f | < .
E∩U
Now assume
the other property, and let δ > 0 be such that if U is open and m(E ∩ U) < δ,
R
then E∩U |f | < . If A ⊆ E is measurable such that m(A) < δ, then we know that there is
an open set O containing A such that m(O ∼ A) < δ − m(A). Therefore
m(E ∩ O) ≤ m(O) = m(A) + m(O ∼ A) < δ,
and since A ⊆ E ∩ O, it follows
Z
Z
|f | ≤
A
|f | < .
E∩O
5