7.9 Measures From Outer Measures
Earlier in Theorem 7.8.3 an outer measure on P
was constructed. This can be used to
obtain a measure defined on
. However, the procedure for doing so is a special case of a
general approach due to Caratheodory in about 1918.
Definition 7.9.1 Let Ω be a nonempty set and let μ : P(Ω) → [0,∞] be an outer
measure. For E ⊆ Ω, E is μ measurable if for all S ⊆ Ω,
To help in remembering 7.10, think of a measurable set E, as a process which divides a given
set into two pieces, the part in E and the part not in E as in 7.10. In the Bible, there are several
incidents recorded in which a process of division resulted in more stuff than was originally
Measurable sets are exactly those which are incapable of such a miracle. You might think of
the measurable sets as the non-miraculous sets. The idea is to show that they form a σ
algebra on which the outer measure μ is a measure.
First here is a definition and a lemma.
Definition 7.9.2 (μ⌊S)(A) ≡ μ(S ∩ A) for all A ⊆ Ω. Thus μ⌊S is the name
of a new outer measure, called μ restricted to S.
The next lemma indicates that the property of measurability is not lost by considering this
Lemma 7.9.3 If A is μ measurable, then A is μ⌊S measurable.
Proof: Suppose A is μ measurable. It is desired to to show that for all T ⊆ Ω,
Thus it is desired to show
But 7.11 holds because A is μ measurable. Apply Definition 7.9.1 to S ∩ T instead of S.
If A is μ⌊S measurable, it does not follow that A is μ measurable. Indeed, if you believe in
the existence of non measurable sets, you could let A = S for such a μ non measurable set and
verify that S is μ⌊S measurable.
The next theorem is the main result on outer measures which shows that starting with an
outer measure you can obtain a measure.
Theorem 7.9.4 Let Ω be a set and let μ be an outer measure on P
collection of μ measurable sets S, forms a σ algebra and
Fn ⊆ Fn+1 ⊆
, then if F
= ∪n=1∞Fn and Fn ∈S, it follows that
Fn ⊇ Fn+1 ⊇
, and if F
= ∩n=1∞Fn for Fn ∈S then if μ
) < ∞,
This measure space is also complete which means that if μ
for some F ∈S then if
G ⊆ F, it follows G ∈S also.
Proof: First note that ∅ and Ω are obviously in S. Now suppose A,B ∈S. I will show
A ∖ B ≡ A ∩ BC is in S. To do so, consider the following picture.
It is required to show that
First consider S ∖
From the picture, it equals
and so this shows that A ∖ B ∈S
whenever A,B ∈S
Since Ω ∈S, this shows that A ∈S if and only if AC ∈S. Now if A,B ∈S,
A∪B = (AC ∩BC)C = (AC ∖B)C ∈S. By induction, if A1,
, then so is ∪i=1nAi
If A,B ∈S
, with A ∩ B
By induction, if Ai ∩ Aj = ∅ and Ai ∈S,
Now let A = ∪i=1∞Ai where Ai ∩ Aj = ∅ for i≠j.
Since this holds for all n, you can take the limit as n →∞ and conclude,
which establishes 7.12.
Consider part 7.13. Without loss of generality μ
for all k
since otherwise there is
nothing to show. Suppose
is an increasing sequence of sets of
. Then letting F0 ≡∅,
is a sequence of disjoint sets of S
since it was shown above that the
difference of two sets of S
is in S
. Also note that from 7.15
and so if μ
which also equals
it follows from part 7.12 just shown that
In order to establish 7.14, let the Fn be as given there. Then, since
The problem is, I don’t know F ∈S and so it is not clear that μ
But since F ⊆ Fn,
and this establishes 7.14. Note that it was assumed μ
from both sides.
It remains to show S is closed under countable unions. Recall that if A ∈S, then AC ∈S
and S is closed under finite unions. Let Ai ∈S, A = ∪i=1∞Ai, Bn = ∪i=1nAi. Then
By Lemma 7.9.3 Bn
) measurable and so is BnC
. I want to show
) ≥ μ
(S ∖ A
) + μ
(S ∩ A
). If μ
) = ∞
, there is nothing to prove. Assume μ
) < ∞
Then apply Parts 7.14
to the outer measure μ⌊S
and let n →∞
and this yields μ(S) = (μ⌊S)(A) + (μ⌊S)(AC) = μ(S ∩ A) + μ(S ∖ A).
Therefore A ∈S and this proves Parts 7.12, 7.13, and 7.14.
It only remains to verify the assertion about completeness. Letting G and F be as
described above, let S ⊆ Ω. I need to verify
because by assumption, μ
Corollary 7.9.5 Completeness is the same as saying that if
N ∈ℱ and μ
, then if E ∈ℱ, it follows that E′∈ℱ also.
Proof: If the new condition holds, then suppose G ⊆ F where μ
is given to equal 0. Therefore,
Now suppose the earlier version of completeness and let
= 0 and
. Then we know
and all have measure zero. It follows E ∖
E ∩ E′∈ℱ
The measure μ which results from the outer measure of Theorem 7.8.3 is called the
Lebesgue Stieltjes measure associated with the integrator function F. Its properties will be
discussed more in the next section.
class=”left” align=”middle”(ℝ)7.10. ONE DIMENSIONAL LEBESGUE STIELTJES