Recall the following which summarizes Theorems 7.10.1 on Page 505 and 7.8.3 on Page
486.
Theorem 18.6.1Let F be an increasing function on ℝ. Then there is an outermeasure μ and a σ algebra ℱ on which μ is a measure such that ℱ contains the Borelsets. This measure μ satisfies
μ
([a,b])
= F
(b+)
− F
(a− )
,
μ
((a,b))
= F
(b− )
− F
(a+ )
μ
((a,b])
= F
(b+)
− F
(a+)
μ
([a,b))
= F
(b− )
− F
(a− )
.
Furthermore, if E is anyset in ℱ
μ (E ) = sup{μ (K) : K compact, K ⊆ E } (18.17)
(18.17)
μ(E) = inf{μ (V) : V is an open set V ⊇ E} (18.18)
(18.18)
Now from Theorem 18.4.3 it follows that for a.e. x,
lim μ((x−-h,x-+-h))- = lim F (x-+-h)−-F (x-−-h)
h→0+ 2h h→0+ 2h
F ((x+-h)−-)−-F-((x-−-h)+)
= hli→m0+ 2h ≡ Dm μ(x)
and there is a Borel set of m measure zero N such that
μ ⊥(E) = μ(E ∩ N)
In case μ ≪ m it follows from Theorem 18.5.2 that for all E Borel,
∫
μ(E ) = Dm μ (x)dm.
E
This begs the following question. What properties on F are equivalent to this measure μ
being absolutely continuous with respect to m? Here is a definition of what it means for a
function to be absolutely continuous.
Definition 18.6.2Let
[a,b]
be a closed and bounded interval and let F :
[a,b]
→ ℝ. Then F is said to be absolutely continuousif for every ε > 0 there existsδ > 0 such that if∑_{i=1}^{m}
|yi − xi|
< δ where the intervals
(xi,yi)
are non-overlapping,then∑_{i=1}^{m}
|F (yi)− F (xi)|
< ε.
Then the following theorem gives the desired equivalence between properties of the
function F and μ ≪ m.
Theorem 18.6.3Let F be an increasing function on
[a,b]
and let μ be themeasure of Theorem 18.6.1. Then μ ≪ m if and only if F is absolutely continuous.
Proof:First suppose that μ ≪ m. Then by Theorem 18.5.2, for all Borel sets
E,
∫
μ (E ) = D μ(x)dm
E m
In particular, F must be continuous because D_{m}μ is in L_{loc}^{1}. Thus
∫ ∫
F (y− )− F (x+) = Dm μ(x)dm = Dm μ(x)dm = F (y+) − F (x+ )
(x,y) (x,y]
showing that for arbitrary y,F
(y− )
= F
(y+ )
and so the function F is continuous as claimed.
Also
∫
F (b) − F (a) = Dm μ(x)dm
[a,b]
so D_{m}μ is in L^{1}
([a,b],m)
.
If the function is not absolutely continuous, then there exists ε > 0 and Borel sets E_{n}
consisting of unions of finitely many non-overlapping open intervals such that if
E_{n} = ∪_{i=1}^{mn}
(xni,yni )
, then ∑_{i=1}^{m}
|yni − xni|
= m
(En)
< 2^{−n} but
∫ ∑mn n n m∑n n n
XEn (x)Dm μ(x)dm = μ (En) = μ (xi,yi ) = (F (yi )− F (xi)) ≥ ε
[a,b] i=1 i=1
(18.19)
(18.19)
However, X_{En}
(x)
→ 0 a.e. because ∑_{n}m
(En )
< ∞ and so, by the Borel Cantelli lemma,
there is a set of measure zero N such that for x
∕∈
N, x is in only finitely many of the E_{n}. In
particular, X_{En}
(x )
= 0 for all n large enough if x
∕∈
N. Then by the dominated convergence
theorem, the inequality 18.19 cannot be valid for all n because the limit of the integral on the
left equals 0. This is a contradiction. Hence F must be absolutely continuous after
all.
Next suppose the function F is absolutely continuous. Suppose m
(E)
= 0. Does it follow
that μ
(E)
= 0? Let ε > 0 be given. Let δ correspond to ε∕2 in the definition of absolute
continuity. Let E ⊆ V where V is an open set such that m
which converges to 0 as h → 0 since x is a Lebesgue point. Similarly, at each Lebesgue
point,
F (x)− F (x− h)
lhim→0--------------- = Dm μ(x)
h
Thus F is differentiable at each Lebesgue point and the derivative equals D_{m}μ at these
points. Now 18.20 yields the desired result that the function can be recovered from integrating
its derivative.
Next suppose F
(y)
−F
(x)
= ∫_{x}^{y}F^{′}
(t)
dm where F^{′}
(t)
exists a.e. and F^{′} is in L^{1}. Then
showing that F is absolutely continuous is just a repeat of the first part of Theorem 18.6.3.
■
Definition 18.6.5A finite subset, P of
[a,b]
is called a partition of
[x,y]
⊆
[a,b]
ifP =
{x0,x1,⋅⋅⋅,xn}
where
x = x0 < x1 < ⋅⋅⋅,< xn = y.
For f :
[a,b]
→ ℝ and P =
{x0,x1,⋅⋅⋅,xn}
define
∑n
VP [x,y] ≡ |f (xi)− f (xi−1)|.
i=1
Denoting by P
[x,y]
the set of all partitions of
[x,y]
define
V [x,y] ≡ sup VP [x,y].
P∈P[x,y]
For simplicity, V
[a,x]
will be denoted by V
(x)
. It is called thetotal variation of the function,f.
There are some simple facts about the total variation of an absolutely continuous function
f which are contained in the next lemma.
Lemma 18.6.6Let f be an absolutely continuous function defined on
[a,b]
and let V beits total variation function as described above. Then V is an increasing bounded function. Alsoif P and Q are two partitions of
[x,y]
with P ⊆ Q, then V_{P}
[x,y]
≤ V_{Q}
[x,y]
and if
[x,y]
⊆
[z,w]
,
V [x,y] ≤ V [z,w ] (18.21)
(18.21)
If P =
{x0,x1,⋅⋅⋅,xn}
is a partition of
[x,y]
, then
n
V [x,y] = ∑ V [x ,x ]. (18.22)
i=1 i i−1
(18.22)
Also if y > x,
V (y)− V (x) ≥ |f (y)− f (x)| (18.23)
(18.23)
and the function, x → V
(x)
−f
(x)
is increasing. The total variation function V is absolutelycontinuous.
Proof:The claim that V is increasing is obvious as is the next claim about P ⊆ Q
leading to V_{P}
[x,y]
≤ V_{Q}
[x,y]
. To verify this, simply add in one point at a time and verify
that from the triangle inequality, the sum involved gets no smaller. The claim that V is
increasing consistent with set inclusion of intervals is also clearly true and follows directly
from the definition.
Now let t < V
[x,y]
where P_{0} =
{x ,x ,⋅⋅⋅,x }
0 1 n
is a partition of
[x,y]
. There exists a
partition, P of
[x,y]
such that t < V_{P}
[x,y]
. Without loss of generality it can be assumed
that
{x ,x ,⋅⋅⋅,x }
0 1 n
⊆ P since if not, you can simply add in the points of P_{0} and the
resulting sum for the total variation will get no smaller. Let P_{i} be those points of P which are
contained in
[x ,x ]
i−1 i
. Then
∑n ∑n
t < Vp[x,y] = VPi [xi−1,xi] ≤ V [xi− 1,xi].
i=1 i=1
Since t < V
[x,y]
is arbitrary,
n
V [x,y] ≤ ∑ V [xi,xi− 1] (18.24)
i=1
(18.24)
Note that 18.24 does not depend on f being absolutely continuous.
Suppose now that f is absolutely continuous. Let δ correspond to ε = 1. Then if
[x,y]
is an interval of length no larger than δ, the definition of absolute continuity
implies