Một số vấn đề về không gian độ đo Đề tài Nghiên cứu khoa học

A σ P A
A
∈ A A µ A
A → R
+
∪ {+∞}
µ(∅) = 0
µ(

B
n
) =

µB
n
,
{B
n
: n = 1, } ⊆ A A µ
A µ
σ A ⊆ P(X)
∀B ∈ A, δ
x
B =



0, x /∈ B
1, x ∈ B.
A δ
x

δ
x
A µ → R A µ
∀α ∈ R, {x : f(x) > α} ∈ A,
∀α ∈ R, {x : f(x) ≥ α} ∈ A,
∀α ∈ R, {x : f(x) < α} ∈ A,
∀α ∈ R, {x : f(x) ≤ α} ∈ A,
∀U ⊆ R, f
−1
(U) ∈ A f
−1
(±∞) ∈ A.
A µ µ
µ ∞ µ σ {B
n
: n = 1, } ⊆ A
µB
n
< ∞

B
n
{B
n
: n = 1, }
A µ
→ R
n

1

a
j
χ
B
j
B
j
∈ A, a
j
∈ R, χ
B
j
B
j

fdµ =

a
j
µB
j
B
j
= {x : f(x) = a
j
} µB
j
< ∞

B

fdµ =

χ
B
fdµ
∈ A
f : X → R, µX < ∞ µ µ

fdµ = inf{

hdµ : f ≤ h }.
µ
L
1
µ
(X)
A
µ : A → R
∗
C)
µ(∅) = 0
µ(
∞

1
B
n
) =
∞


1
µB
n
{B
n
: n = 1, } µ
µ
A
A µ µ
+∞ −∞
B
1
, B
2
∈ A µ(B
1
) = +∞, µ(B
2
) = −∞
µ(B
1
∪ B
2
) = µ(B
1
) + µ(B
2
) = +∞ − (+∞)
A µ ∈ A
∀E ⊆ B ∈ A , µE ≥ 0 µE ≤ 0 .

∈ A
µ
+
B = sup{µE : E ⊆ B, E ∈ A}
µ
−
B = sup{−µE : E ⊆ B, E ∈ A}.
| µ |= µ
+
B + µ
−
B.
µ ν µ ν µ  ν
∀B ∈ A, | ν | (B) = 0 ⇒ µB = 0.
A µ
A | µ |
∈ A µ µ
+
B µB = −µ
−
B
µ µ
+
B < ∞ −µB = µ
−
B < ∞
A µ
+
A µ
−

µ
±
(∅) = 0 µ(∅) = 0
∈ A ∅ ⊆ A µ
±
B ≥ 0
{B
n
: n = 1, } ⊆ A µ
+
(

B
n
) =

µ
+
B
n
µ
−
⊆ ∈ A
µ
+
B ≤ µ
+
D.
⊆ ∈ A µE ≤ µ
+

D µ
+
D ⊆
µ
+
B ≤ µ
+
D
µ
+
B
n
= ∞ µ
+
B
n
≤ µ
+
(

B
j
) µ
+
(

B
j
) = ∞
µ

+
≥ 0 µ
+
B
n
= ∞ µ
+
µ
+
B
+
< ∞
ε > 0 E
n
⊆ B
n
E
n
∈ A µE
n
≥ µ
+
B
n
− (
ε
2
n
) {E
n

: n = 1, }
{B
n
: n = 1, } µ
+
µ(

E
n
) =

µE
n
≥

µ
+
B
n
− ε;
µ
+
µ
+
(

B
n
) ≥


µ
+
B
n
.
λ < µ
+
(

B
n
) λ <

µ
+
B
n
⊆

B
n
µE > λ µ
µE =

µ(E ∩ B
n
).

µ(E ∩ B
n

) ≤ µ
+
B
n
λ < µE ≤

µ
+
B
n
.
µ
+
(

B
n
) =

µ
+
B
n
µD ≥ 0 ∈ A ⊆ B\ ⊆
µD ≤ µB µ
+
, µ
+
B ≤ µB
µ

+
B = µB.
⊆ ∈ A µD < 0
µ \ ∪ µB = µD + µ(B\D) µB < ∞
µB < µ(B\D) µD < 0
B\D ⊆ B µ
+
B ≥ µ(B\D) µ
+
B > µB
A µ
∈ A \
µ
+
X < ∞ λ {µ
+
B : B ∈ A} µ
−
B = 0
µ
+
µ
+
X < ∞ 0 ≤ λ < ∞
{B
n
: n = 1, } ⊆ A µ
−
B
n

= 0 µ
+
B
n
→ λ.

B
n
µ
−
P = 0 µ
−
P ≤

µ
−
B
n
B
n
⊆ P µ
+
P ≥ µ
+
B
n
µ
+
≥ λ
∈ A µ

−
P = 0 λ ≥ µ
+
P λ = µ
+
P.
\
⊆ µD > 0 ⊆ µ
+
E = µE ≥ µD > 0
µ
+
X < ∞
∩P = ∅
µ
+
(P ∪ E) = µ
+
P + µ
+
E = λ + µ
+
E > λ.
µ
−
P = 0
µ
−
(P ∪ E) = µ
−

P + µ
−
E = µ
−
E = 0
⊆ µF ≥ 0
0 ≤ µ
−
E = sup(−µF : F ⊆ E) ≤ 0 ⇒ µ
−
E = 0.
λ ∪
A µ
µ ν
A µ  ν f ∈ L
1
µ
(X)
∀B ∈ A, µB =

B
fdν.
∈ µ − rν P
r
N
r
µ − rν
µB ≤ rνB ≤ sνB B ⊆ N
r
sνB ≤ µB B ⊆ P

s
νB = µB = 0 B ⊆ N
r
\N
s
N
r
\N
s
⊆ X\N
s
= P
s
µ  ν
N
r
(µ − rν)B ≤ 0 µB ≤ rνB ν
rνB ≤ sνB
P
s
(µ − sν)B ≥ 0 sνB ≤ µB

s∈Q

r<s,r∈Q
(N
r
\N
s
),

R
r
= N
r
\E,
G = X\(∪R
r
\ ∩ R
s
).
νE = 0
⇒ R
r
⊆ R
s
,
ν(

r∈Q
R
r
) = 0,
ν(X\

r∈Q
R
r
) = ν(

R

c
r
) = 0 R
c
r
R
r
νG = 0
νE = 0
R
r
∀s ∈ Q,

r∈Q
R
r
⊆

N
r
⊆ N
s
,
(µ − sν)(

r∈Q
R
r
) ≤ 0
ν(


r∈Q
R
r
) > 0, s
n
→ ∞ µ(

R
r
) = −∞
µ
X\ ∪ R
r
=

(X\(N
r
\E)) =

((X\N
r
)

E) = E

(

N
c

r
).
P
r
⊆ P
s
ν(

P
r
) > 0 ν(

P
r
) = 0
G = (

R
r
\

R
s
)
c
= [(

R
r
) ∩ ((


R
s
)
c
)]
c
= [(

R
r
) ∩ (

R
c
s
)]
c
= (

R
c
r

(

R
s
).
νG = 0

f =



0, x ∈ G
sup{s ∈ Q : x /∈ R
s
}, x /∈ G.
ν α ≤ 0
{x : f(x) < α} =

r<α,r∈Q
R
r
,
α > 0
{x : f(x) < α} = G

(

r<α,r∈Q
R
r
).
⊆ R
s
\R
r
|


B
fdν − µB |≤ (s − r)νB. (1)
R
r
\R
s
⊆ ((∪R
t
)\

u
R
u
) B ∩ G = ∅
∈ R
t
f(x) ≥ t
r ≤ f(B) ≤ s,
rνB ≤

B
f(x)dν(x) ≤ sνB
R
r
\R
s
= (N
s
\E)\(N
r

\E) µB ≤ sνB B ⊆ N
s
B ⊆ P
r
R
r
\R
s
⊆ (N
s
\E)\N
r
⊆ X\N
r
µB ≥ sνB
−sνB ≤ −µB ≤ −rνB
B ⊆ R
n+1
\R
n
µB =

B
f(x)dν(x). (3)
B
j
= B ∩ (R
n+(j\p)
\R
n+[(j−1)\p]

),
{B
j
: j = 1, } B =

B
j
|µB −

B
fdν| ≤
p

j=1
|µB
j
−

B
j
fdν| ≤
p

j=1
νB
j
=
1
p
νB,

∈ A
B = (B ∩ G)

(

j
B
j
),
{B
j
: j = 1, }
B ⊆ R
n+1
\R
n
µB = µ(B ∪ G) + µ(

j
B
j
),
µB = µ(B ∪ G) +

B
fdν
µ
νG = 0 µ  ν
µB =


B
fdν.
ν

B
fdν ∈ A f ∈ L
1
µ
(X)
A µ ν
µ ∈ M(X)
ν σ
µ  ν f ∈ L
1
µ
(X)
∀B ∈ A, µB =

B
fdν