is a measure space as above and suppose μ is ameasure defined on ℱ. Denote by BV
(Ω;μ)
those finitely additive measures of BV
(Ω )
ν such that ν ≪ μ in the usual sense that if μ
(E )
= 0, then ν
(E )
= 0. Then BV
(Ω; μ)
is a closed subspace of BV
(Ω)
.
Proof: It is clear that it is a subspace. Is it closed? Suppose ν_{n}→ ν and each ν_{n} is in
BV
(Ω; μ)
. Then if μ
(E)
= 0, it follows that ν_{n}
(E )
= 0 and so ν
(E )
= 0 also, being the limit
of 0. ■
Definition 25.4.3For s a simple function s
(ω )
= ∑_{k=1}^{n}c_{k}X_{Ek}
(ω)
andν ∈ BV
(Ω)
, define an “integral” with respect to ν as follows.
∫ ∑n
sdν ≡ ckν (Ek ).
k=1
For f function which is in L^{∞}
(Ω;μ)
, define∫fdν as follows. Applying Theorem 7.1.6, to thepositive and negative parts of real and imaginary parts of f, there exists a sequence ofsimple functions
{sn}
which converges uniformly to f off a set of μ measure zero.Then
∫ ∫
fdν ≡ lim s dν
n→∞ n
Lemma 25.4.4The above definition of the integral with respect to a finitely additivemeasure in BV
(Ω;μ)
is well defined.
Proof:First consider the claim about the integral being well defined on the simple
functions. This is clearly true if it is required that the c_{k} are disjoint and the E_{k} also disjoint
having union equal to Ω. Thus define the integral of a simple function in this manner. First
write the simple function as
∑n
ckXEk
k=1
where the c_{k} are the values of the simple function. Then use the above formula to define the
integral. Next suppose the E_{k} are disjoint but the c_{k} are not necessarily distinct. Let the
distinct values of the c_{k} be a_{1},
⋅⋅⋅
,a_{m}
( )
∑ ∑ ∑ ∑ ( )
ckXEk = aj( XEi) = ajν ∪i:ci=ajEi
k j i:ci=aj j
= ∑ a ∑ ν(E ) = ∑ cν (E )
j ji:c=a i k k k
i j
and so the same formula for the integral of a simple function is obtained in this case also.
Now consider two simple functions
∑n ∑m
s = akXEk , t = bjXFj
k=1 j=1
where the a_{k} and b_{j} are the distinct values of the simple functions. Then from what was just
shown,
n m m n
= ∑ ∑ αa ν(E ∩ F )+ ∑ ∑ βb ν(E ∩F )
k=1j=1 k k j j=1k=1 j k j
n m
= ∑ αakν(Ek) + ∑ βbjν (Fj)
k=1 j=1
∫ ∫
= α sdν + β tdν
Thus the integral is linear on simple functions so, in particular, the formula given in the above
definition is well defined regardless.
So what about the definition for f ∈ L^{∞}
(Ω;μ)
? Since f ∈ L^{∞}, there is a set of μ measure
zero N such that on N^{C} there exists a sequence of simple functions which converges
uniformly to f on N^{C}. Consider s_{n} and s_{m}. As in the above, they can be written
as
p p
∑ n ∑ m
ckXEk, ck XEk
k=1 k=1
respectively, where the E_{k} are disjoint having union equal to Ω. Then by uniform
convergence, if m,n are sufficiently large,
n m
|ck − ck |
< ε or else the corresponding E_{k} is
contained in N^{C} a set of ν measure 0 thanks to ν ≪ μ. Hence
| |
||∫ ∫ || ||∑p n m ||
|| sndν − smd ν|| = || (ck − ck )ν(Ek)||
k=p1
∑ n m
≤ |ck − ck ||ν (Ek )| ≤ ε ||ν||
k=1
and so the integrals of these simple functions converge. Similar reasoning shows that the
definition is not dependent on the choice of approximating sequence. ■
Note also that for s simple,
||∫ ||
|| sdν|| ≤ ||s||L∞ |ν|(Ω) = ||s||L∞ ||ν ||
Next the dual space of L^{∞}
(Ω; μ)
will be identified with BV
(Ω;μ)
. First here is a simple
observation. Let ν ∈ BV
(Ω;μ)
. Then define the following for f ∈ L^{∞}
(Ω;μ)
.
∫
T ν (f) ≡ fdν
Lemma 25.4.5For T_{ν}just defined,
|Tνf| ≤ ||f||L∞ ||ν||
Proof:As noted above, the conclusion true if f is simple. Now if f is in L^{∞}, then it is
the uniform limit of simple functions off a set of μ measure zero. Therefore, by the definition
of the T_{ν},
, there exists a sequence of simple functions converging to f
uniformly off a set of μ measure zero and so passing to a limit in the above with s replaced
with s_{n} it follows that
∫
Λ(f) = fdν
and so θ is onto. ■
class=”left” align=”middle”(Ω)25.5. NON σ FINITE CASE