SOLUTIONS OF HW6
MINGFENG ZHAO
March 31, 2011
1. (Problem 32, in Page 63) Suppose µ(X) < ∞. If f and g are complex-valued measurable functions
on X, define
Z
ρ(f, g) =
X
|f − g|
dµ.
1 + |f − g|
Then ρ is a metric on the space of measurable functions if we identify functions that are equal a.e,
and fn → f with respect to this measure iff fn → f in measure.
Proof. Let Fµ = {all complex-valued µ-measurable functions on X}.
(1). (a). Well-defined: For any f, g ∈ Fµ , we know that
0≤
|f (x) − g(x)|
≤1
1 + |f (x) − g(x)|
for all x ∈ X.
So we have
Z
X
|f − g|
dµ ≤
1 + |f − g|
Z
1dµ
=
µ(X)
<
∞.
This means, 0 ≤ ρ(f, g) ≤ µ(X) < ∞.
(b) Positive-definite: If ρ(f, g) = 0, that is,
Z
X
Since 0 ≤
|f (x)−g(x)|
1+|f (x)−g(x)|
|f − g|
dµ = 0.
1 + |f − g|
for all x ∈ X, then we have |f (x) − g(x)| = 0 a.e x ∈ X, which implies that
f (x) = g(x) for all x ∈ X. In our sense, we have f = g.
1
2
MINGFENG ZHAO
(c) Symmetry: For all x ∈ X, we have
|f (x) − g(x)|
|g(x) − f (x)|
=
.
1 + |f (x) − g(x)|
1 + |g(x) − f (x)|
So we have
Z
X
|f (x) − g(x)|
dµ =
1 + |f (x) − g(x)|
Z
X
|g(x) − f (x)|
dµ.
1 + |g(x) − f (x)|
That is, ρ(f, g) = ρ(g, f ).
(d) Triangle inequality: For any f, g, h ∈ Fµ . So for all x ∈ X, we have
|f (x) − g(x)| ≤ |f (x) − h(x)| + |h(x) − g(x)|.
Since F (t) =
t
1+t
=1−
|f (x) − g(x)|
1 + |f (x) − g(x)|
1
1+t
is increasing on [0, ∞), then we can get for all x ∈ X, we have
≤
|f (x) − h(x)| + |h(x) − g(x)|
1 + |f (x) − h(x)| + |h(x) − g(x)|
=
|f (x) − h(x)|
|f (x) − h(x)|
+
1 + |f (x) − h(x)| + |h(x) − g(x)| 1 + |f (x) − h(x)| + |h(x) − g(x)|
≤
|f (x) − h(x)|
|f (x) − h(x)|
+
1 + |f (x) − h(x)| 1 + |h(x) − g(x)|
So we have
Z
X
Z |f (x) − g(x)|
dµ ≤
1 + |f (x) − g(x)|
X
Z
=
X
|f (x) − h(x)|
|f (x) − h(x)|
+
dµ
1 + |f (x) − h(x)| 1 + |h(x) − g(x)|
Z
|f (x) − h(x)|
|f (x) − h(x)|
dµ +
dµ
1 + |f (x) − h(x)|
X 1 + |h(x) − g(x)|
That is, ρ(f, g) ≤ ρ(f, h) + ρ(h, g).
Therefore, by (a),(b),(c) and (d), we know ρ is a metric on Fµ .
(2). (=⇒) If fn −→ f with respect to ρ, as n → ∞. That is,
Z
lim
n→∞
X
|fn (x) − f (x)|
dµ = 0.
1 + |fn (x) − f (x)|
For any > 0 and fix , we have
µ({x ∈ X : |fn (x) − f (x)| > }) ·
1+
SOLUTIONS OF HW6
Z
=
{x∈X: |fn (x)−f (x)|>}
Z
≤
{x∈X: |fn (x)−f (x)|>}
Z
≤
X
3
dµ
1+
|fn (x) − f (x)|
dµ Since |fn (x) − f (x)| > 1 + |fn (x) − f (x)|
|fn (x) − f (x)|
dµ.
1 + |fn (x) − f (x)|
So we have
1+
µ({x ∈ X : |fn (x) − f (x)| > }) ≤
Z
X
|fn (x) − f (x)|
dµ.
1 + |fn (x) − f (x)|
Take the supremum, as n → ∞, we get
lim sup µ({x ∈ X : |fn (x) − f (x)| > }) ≤
n→∞
1+
lim
n→∞
Z
X
|fn (x) − f (x)|
dµ = 0.
1 + |fn (x) − f (x)|
Therefore, by the arbitrary of > 0, we have fn → f in measure, as n → ∞.
(⇐=) Assume fn −→ f in measure, as n → ∞, that is, for any > 0, we have
lim µ({x ∈ X : |fn (x) − f (x)| > }) = 0.
n→∞
Then we have
Z
ρ(fn , f )
=
X
|fn (x) − f (x)|
dµ
1 + |fn (x) − f (x)|
Z
=
{x∈X:
Z
≤
|fn (x) − f (x)|
dµ +
|fn (x)−f (x)|>} 1 + |fn (x) − f (x)|
Z
1dµ +
{x∈X: |fn (x)−f (x)|>}
Z
{x∈X: |fn (x)−f (x)|≤}
{x∈X: |fn (x)−f (x)|≤}
dµ
1+
Z
= µ({x ∈ X : |fn (x) − f (x)| > }) +
{x∈X: |fn (x)−f (x)|≤}
Z
≤ µ({x ∈ X : |fn (x) − f (x)| > }) +
dµ
X
= µ({x ∈ X : |fn (x) − f (x)| > }) + µ(X).
dµ
1+
|fn (x) − f (x)|
dµ
1 + |fn (x) − f (x)|
4
MINGFENG ZHAO
For any δ > 0, we take 0 < <
δ
2µ(X) ,
then there exists N ∈ N such that whenever n ≥ N , we
have
µ({x ∈ X : |fn (x) − f (x)| > }) <
δ
.
2
Hence, for all n ≥ N , we have
ρ(fn , f ) ≤ µ({x ∈ X : |fn (x) − f (x)| > }) + µ(X)
≤
δ
δ
+
· µ(X)
2 2µ(X)
= δ
By the arbitrary of δ > 0, we know ρ(fn , f ) → ∞, as n → ∞, that is, fn −→ f with respect to ρ,
as n → ∞.
2. (Problem 44, in Page 64) (Lusin’s theorem). If f : [a, b] → C is Lebesgue measurable and > 0,
there is a compact set E ⊂ [a, b] such that µ(E c ) < and f |E is continuous.
Proof. (1). If there exists M > 0 such that |f (x)| ≤ M a.e x ∈ [a, b], then we have f ∈ L1 ([a, b]). We
extend f trivially on R by:
F (x) =




 f (x), if x ∈ [a, b]



 0,
if x 6∈ [a, b].
Then we know F ∈ L1 (R), F (x) = f (x) for all x ∈ [a, b], and
R
R
F dµ =
proof of Theorem 2.26, in Page 55, there exists Gn ∈ C(R) such that
Z
|F (x) − Gn (x)|dµ = 0.
lim
n→∞
R
Since
Z
Z
|F (x) − Gn (x)|dµ ≤
[a,b]
|F (x) − Gn (x)|dµ.
R
R
[a,b]
f dµ. Then by the
SOLUTIONS OF HW6
5
So we have
Z
|F (x) − Gn (x)|dµ ≤ lim
lim
n→∞
Z
n→∞
[a,b]
|F (x) − Gn (x)|dµ = 0.
R
So
Z
|F (x) − Gn (x)|dµ = 0.
lim
n→∞
[a,b]
Let gn (x) = Gn (x) for all x ∈ [a, b]. Since Gn is continuous on R, then gn is continuous on [a, b].
Since F (x) = f (x) for all x ∈ [a, b], so we have
Z
|f (x) − gn (x)|dµ = 0.
lim
n→∞
[a,b]
By Corollary 2.32, in Page 62, we know there exists a subsequence gnk (x) = fk (x) such that
fk (x) → f (x) a.e x ∈ [a, b], as k → ∞.
On the other hand, since µ([a, b]) = b − a < ∞, by Egoroff’s theorem, Theorem 2.33 in Page 62, for
any > 0, there exists a µ-measurable subset F ⊂ [a, b] such that µ(F ) < 2 , and fk → f uniformly
on F c , as k → ∞.
By the exterior regularity theorem of Lebesgue measure, Theorem 1.18, in Page 36, there exists an
open subset U ⊂ [0, 1] such that F ⊂ U , and
µ(U \F ) <
.
2
So we have µ(U ) ≤ µ(F ) + µ(U \F ) < , and U c ⊂ F c . So fk → f uniformly on U c , as k → ∞. Let
E = U c , then we have fk → f uniformly on E, as k → ∞, and µ(E c ) = µ(U ) < . So f is continuous
on E.
(2). If f is not bounded on [a, b]. For any n ≥ 1, we let fn (x) = f (x) · 1{x∈[a,b]:
all n ≥ 1, we have |fn (x)| ≤ n for all x ∈ [a, b], and for all x ∈ [a, b], we have
lim fn (x) = f (x)
n→∞
|f (x)|≤n} (x).
Then for
6
MINGFENG ZHAO
By the result of (1), for any > 0, there exists a compact set En ⊂ [a, b] such that µ(Enc ) <
2n+1 ,
and fn is continuous on En . Let E0 = ∩∞
n=1 En , then we can get
µ(E0c )
c
µ(∪∞
n=1 En )
=
∞
X
≤
µ(Enc )
n=1
∞
X
≤
n+1
2
n=1
=
.
2
And fn is continuous on E0 for all n ≥ 1. Since µ(E0 ) ≤ µ([a, b]) < ∞, and fn (x) → f (x), as
n → ∞, for all x ∈ E0 . By Egoroff’s theorem, Theorem 2.33 in Page 62, for the > 0, there exists a
µ-measurable subset F ⊂ E0 such that µ(F ) < 8 , and fn → f uniformly on E0 \F , as n → ∞. Since
fn is continuous on E0 \F for all n ≥ 1, by the uniformly convergence, we have f is continuous on
E0 \F .
By the exterior regularity theorem of Lebesgue measure, Theorem 1.18, in Page 36, there exists an
open subset U ⊂ [0, 1] such that F ⊂ U , and
µ(U \F ) <
So we have µ(U ) ≤ µ(F ) + µ(U \F ) <
8
+
4
=
3
8 ,
.
4
and U c ⊂ F c . Let E = U c ∩ (E0 \F ), so E is
compact, and f is continuous on E. And
µ(E c )
= µ(U ∪ (E0 \F )c )
= µ(U ∪ E0c ∪ F )
≤ µ(U ) + µ(E0c ) + µ(F )
<
3
+ +
8
2 8
= .
SOLUTIONS OF HW6
7
In a summary, if f : [a, b] → C is Lebesgue measurable and > 0, there is a compact set E ⊂ [a, b]
such that µ(E c ) < and f |E is continuous.
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009
E-mail address: mingfeng.zhao@uconn.edu