证明不存在仅含可数成员的无限σ-代数

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16 ABSTRACT INTEGRATION

13. Show that proposition 1.24(c) is also true when c = +∞.

Proof: We want to prove that if f ≥ 0 and c = +∞ (f is measurable), then

cfdµ = c

fdµ

In fact, since f is nonnegative measurable, so 0 ≤

fdµ always exists. If

fdµ = 0, then f ∈ L

(µ). According to Theorem 1.39 and ∀ E ∈ M, we

have

0 ≤

fdµ ≤

fdµ = 0

it follows that f = 0 a.e on X, then ∞f(x) = 0 a.e on X. Thanks to

Theorem 1.24(d), ∃ A ∈ M s.t. µ(A) = 0, ∞f(x) = 0 on X − A, and

X−A

∞f(x)dµ = 0.

Note that by Theorem 1.24(e), one has

∞·f(x)dµ =

f(x)∞dµ = 0.

Then,

∞·f(x)dµ =

X−A

∞·f(x)dµ +

∞·f(x)dµ = 0 + 0 = 0 = ∞

f(x)dµ

If 0 <

fdµ, using the facts {x : f(x) > 0} =

+∞

n=1

{x : f(x) ≥

} and

µ{x : f(x) > 0} > 0, there ∃ n s.t. µ{x : f(x) ≥

} > 0, so

∞ · f(x)dµ ≥

{x:f(x)≥

}

∞ · f(x)dµ = ∞µ{x : f(x) ≥

} = ∞.

Moreover, ∞ ·

fdµ = ∞. Hence we get

∞ ·

fdµ =

∞ · fdµ.

We obtain the desired conclusion. 2

18 POSITIVE BOREL MEASURES

is open, we get f is lower semicontinuous on X.

As for the the second ⇔, the ⇒ is obvious and we only need to prove

the part ⇐. If y

∈ {x : f(x) > f(x

) − ε}, using the condition, ∃ open

set B(y

) ⊂ {x : f(x) > f(y

) − [f(y

) − (f(x

) − ε)]}, i.e. ∃ open set

B(y

) ⊂ {x : f(x) > f(x

) − ε}, so {x : f(x) > f(x

) − ε} is open set,

according to the above result we get f is lower semicontinuous on X.

(b) is true. In fact, if f

and f

are l.s.c, then ∀ x

∈ X, ∀ ε > 0,

∃ open sets V and W , x

∈ V ∩ W , s.t. V ⊂ {x : f

(x) > f

) −

W ⊂ {x : f

(x) > f

) −

}. Note V ∩ W is also open, x

∈ V ∩ W , then

V ∩ W ⊂ {x : f

(x) > f

) −

} ∩ {x : f

(x) > f

) −

}

⊂ {x : f

(x) + f

(x) > f

) + f

) − ε}

According to the above proposition, we conclude that f

+ f

is l.s.c. In an

analogous way, we can prove that (a) is true. Moreover we must notice that

(a) and (b) hold without the hypothesis that f

and f

are nonnegative.

(d) is true. Set S

(x) =

i=1

, since f

is l.s.c, according to (b), S

l.s.c on X, so ∀ a ∈ R

, {x : S

(x) ≤ a} = X − {x : S

(x) > a} is closed set,

then g(x) =

∞

i=1

(x) = lim

n→∞

(x) always exists. The following is well

known: g(x) : X → [0, ∞], g(x) l.s.c on X ⇔ ∀ a ∈ R

, {x : g(x) ≤ a} is a

closed set. We will show that

{x : g(x) ≤ a} =

+∞

n=1

{x : S

(x) ≤ a}

On one hand, ∀ x

∈ {x : g(x) ≤ a}, then a ≥ g(x

) = lim

n→∞

) ≥

), ∀ m ∈ N, since f

≥ 0. So ∀ m ∈ N, one has x

∈ {x : S

(x) ≤ a},

then x

∈

+∞

n=1

{x : S

(x) ≤ a}.

On the other hand, if x

∈

+∞

n=1

{x : S

(x) ≤ a}, we get ∀ n, S

) ≤ a,

hence g(x

) = lim

n→∞

) ≤ a, x

∈ {x : g(x) ≤ a}. Thus {x : g(x) ≤

a} =

+∞

n=1

{x : S

(x) ≤ a}. Note that {x : S

(x) ≤ a} is closed, then

{x : g(x) ≤ a} is closed, which entails

∞

is lower semicontinuous.

If f

≤ 0, u.s.c on X, then −f

≥ 0 l.s.c on X. hence

+∞

n=1

(−f

) =

−

+∞

n=1

l.s.c, then

+∞

n=1

u.s.c.

≥ 0 u.s.c, g(x) =

+∞

n=1

(x). Note that g(x) u.s.c ⇔ ∀

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