is a measure space as above and suppose μ is a measure definedon ℱ. Denote by BV
(Ω;μ)
those finitely additive measures of BV
(Ω)
ν such that ν ≪ μ in theusual 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 14.4.3For s a simple function s
(ω)
= ∑_{k=1}^{n}c_{k}X_{Ek}
(ω)
and ν ∈ 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 5.1.9, to the positive andnegative parts of real and imaginary parts of f, there exists a sequence of simple functions
{sn}
whichconverges uniformly to f off a set of μ measure zero. Then
∫ ∫
fdν ≡ lim sndν
n→∞
Lemma 14.4.4The above definition of the integral with respect to a finitely additive measure inBV
(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
∑ ∑ ∑
= aj ν (Ei) = ckν (Ek)
j i:ci=aj k
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,
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 n ∑p 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
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)
.
∫
Tν (f) ≡ fdν
Lemma 14.4.5For T_{ν}just defined,
|T νf| ≤ ||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)14.5. REPRESENTATIONS FOR POSITIVE LINEAR
FUNCTIONALS ON C_{C}CLASS = ”LEFT”ALIGN = ”MIDDLE”+1