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Exercise Solutions to Real Analysis
Note: References refer to H. L. Royden, “Real Analysis”
Exersize 1. Given any set A and any > 0, there is an open set O such that
A ⊂ O and m∗ O ≤ m∗ A + .
Solution 1. If m∗ A = ∞, then there is nothing to prove (take O = R). Suppose
now that > 0 is given and m∗ A < ∞. By definition, we have,
)
(
X
∗
l(an , bn ) : A ⊂ ∪n∈N (an , bn )
m A = inf
n∈N
So, by approximation for infima, we know that there exist (an , bn ) ⊂ R such that
A ⊂ ∪n∈N (an , bn ) and
X
l(an , bn ) ≤ m∗ A + n∈N
So take O = ∪n∈N (an , bn ). Then O is open, A ⊂ O and by subadditivity we have:
X
m∗ O ≤
l(an , bn ) ≤ m∗ A + n∈N
Exersize 2. Suppose A, B ⊂ R and m∗ A = 0, then m∗ (A ∪ B) = m∗ B
Solution 2. We have from subadditivity,
m∗ B ≤ m∗ (A ∪ B) ≤ m∗ A + m∗ B = 0 + m∗ B = m∗ B
The result follows.
Exersize 3. If A is measurable then, for every > 0, there is an open set O
containing A such that m(O \ A) < Solution 3. Suppose that > 0 is given. Then we know by problem 1 that there
is an open set O containing A such that m∗ O < m∗ A + . In particular, since
A is measurable and O is open (hence also measurable) we may replace m∗ with
m: mO − mA < . On the other hand, by measurability of A and the identities
O \ A = O ∩ Ac , O ∩ A = A we get:
m(O) = m(A ∩ O) + m(Ac ∩ O) = m(A) + m(O \ A)
So > m(O) − m(A) = m(O \ A)
Exersize 4. If A is measurable then, for every > 0, there is a closed set C
contained in A such that m(A \ C) < 1
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Solution 4. Since A is measurable, so is Ac . So, given > 0, there is O open,
containing Ac , such that m(O \ Ac ) < . But then Oc is closed and is contained in
A. Further, A \ Oc = A ∩ O = O \ Ac . So m(A \ Oc ) = m(O \ Ac ) < .
Exersize 5. Show that if E is any measurable set, then
−ν − E ≤ νE ≤ ν + E
and
|νE| ≤ |ν|(E)
Solution 5. Let A, B be the Hahn decomposition, A positive and B negative.
ν + (E) = ν(A ∩ E) ≥ 0 and ν − (E) = ν(B ∩ E) ≤ 0. Since A and B are disjoint, we
have
ν(E) = ν + (E) − ν − (E)
It follows immediately that
−ν − E ≤ νE ≤ ν + E
Now, |νE| = |ν + E − ν − E| ≤ |ν + E| + |ν − E| = ν + E + ν − E = |ν|(E), since ν + E
and ν − E are positive.
Exersize 6. Show that if ν1 and ν2 are any two finite signed measures, then so is
αν1 + βν2 , where α and β are real numbers. Show that
|αν| = |α||ν|
and
|ν1 + ν2 | ≤ |nu1 | + |ν2 |
Solution 6. Set, for convenience, µ = αν1 +βν2 . It is clear that µ(∅) = 0. Further,
since ν1 and ν2 are finite measures, µ never admits +∞ or −∞. Now, suppose that
{Ai } is a collection of disjoint sets. Then we have:
∞
∞
µ (∪∞
i=1 Ai ) = αν1 (∪i=1 Ai ) + βν2 (∪i=1 Ai )
Now, since ν1 and ν2 are signed measures, either
∞
X
γνj (Ai )
i=1
converges absolutely, or diverges properly (with j = 1, 2, γ = α, β respectively).
Now, sum of absolutely convergent series is again absolutely convergent. Further,
if one is properly divergent, then so is the sum. In both cases, the sum of the series
can be taken term-by-term. Hence we have:
∞
∞
∞
∞
X
X
X
X
∞
αν1 (Ai ) +
βν2 (Ai ) =
(αν1 (Ai ) + βν2 (Ai )) =
µ(Ai )
µ(∪i=1 Ai ) =
i=1
i=1
i=1
i=1
Finitude of µ is immediate by triangle inequality for real numbers (since both ν1
and ν2 are finite).
Now, if νE is ∞ or −∞, equality holds for every nonzero α: |ανE| = |α||νE|. On
the other hand, since νE is a finite real number, we clearly have |ανE| = |α||νE|.
Now, assuming that ν1 + ν2 is never of the form ∞ − ∞, if |ν1 + ν2 | = ∞, then
clearly either ν1 = ±∞ or ν2 = ±∞. In either case, equality holds: |ν1 + ν2 | =
|ν1 | + |ν2 |. If, however, |ν1 + ν2 | is finite, then both ν1 and ν2 are finite. So by
triangle inequality for real numbers we get |ν1 + ν2 | ≤ |ν1 | + |ν2 |
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(of course, above when we speak about the numeric value of ν1 and ν2 , we essentially
mean ν1 E, ν2 E for a random measurable set E).
Exersize 7. We define integration with respectt to a signed measure ν by defining
Z
Z
Z
f dν = f dν + − f dν −
If |f | ≤ M ,
Z
f dν ≤ M |ν|(E)
E
Moreover, there is a measurable function f with |f | ≤ 1 such that
Z
f dν = |ν|(E)
E
Solution 7. Note, that since ν is a signed measure, it does not admit both +∞
and −∞. In this case the integral if well-defined, since there doesn’t exist a set E
with ν + E = ∞ and ν − E = ∞.
So, if |f | ≤ M ,
Z
Z
Z
Z
Z
−
f dν + + f dν − f dν = f dν + −
≤
f
dν
E
E
E
E
E
−
+
Now, since ν and ν are both positive measures, the integrals with respect to
those measures are defined in the usual sense. Hence all the standard theorems
apply. Thus:
Z
Z
Z
Z
+
f dν + + f dν − ≤
|f |dν +
|f |dν − ≤ M (ν + E + ν − E) = M |ν|(E)
E
E
E
E
Now, let A and B be the Hahn decomposition. Define f (x) = XA∩E − XB∩E .
Then we have:
Z
Z
Z
Z
Z
+
−
+
f dν =
f dν −
f dν =
XA∩E dν +
XB∩E dν − =
E
E
E
E
E
−
+
ν (A ∩ E) + ν (B ∩ E) = |ν|(E)
Also, it is clear that |f | ≤ 1.
Exersize 8. (a) Show that the Radon-Nikodym Theorem for a finite measure µ
implies the theorem for a σ-finite measure µ.
(b) Show the uniqueness of the function f in the Radon-Nikodym Theorem.
Solution 8. (a) Suppose that the theorem holds for finite measures. Suppose now
that µ is a σ-finite measure on X. Write X = ∪i Xi of µ-finite, disjoint sets Xi .
Then we know that there exist functions fi on Xi such that
Z
νE =
fi dµ
E
P
for every measurable subset E of Xi . Now, define f =
i fi XXi . Clearly f is
nonnegative and measurable. Now, let E be a µ-measurable subset of X. Let Ei
= E ∩ Xi . Ei form a family of disjoint measurable sets and E = ∪i Ei . Further,
X
XZ
νE =
νEi =
fi dµ
i
i
Ei
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Since each fi is nonnegative,
XZ
fi dµ =
Z X
Ei
i
X
Z
f dµ
fi dµ =
X
i
Uniqueness of such f is forced by its existence (see (b) below).
(b) We’ll use the decomposition from (a). Observe that each fi is unique (a.e. [µ])
on each Xi , for suppose that gi is another family of functions with
Z
Z
gi dµ
fi dµ =
Xi
Xi
But then
Z
(fi − gi )dµ = 0
Xi
forcing fi = gi a.e. on Xi . So, by uniqueness of each fi on Xi , we conclude that f
is also unique (a.e.).
Exersize 9. Radon-Nikodym derivatives. Let µ, ν, λ be σ-finite. Show that the
Radon-Nikodym derivative [dν/dµ] has the following properties:
(a) If ν µ and f is a nonnegative measurable function, then
Z
Z dν
f dν = f
dµ.
dµ
(b)
d(ν1 + dν2 )
dν1
dν2
=
+
.
dµ
dµ
dµ
(c) If ν µ λ, then
dν
dν
dµ
=
.
dλ
dµ dλ
(d) If ν µ and µ ν, then
Solution 9. Let g =
h
dν
dµ
i
−1
dµ
dν
=
.
dµ
dν
.
Since
Z
Z
+
f dν −
f dν =
Z
f dν −
by definition, it will suffice to prove the assertion for
Z
f dν +
and
Z
f dν −
By symmetry since ν + ⊥ ν − and in fact both are positive measures, it will suffice
to prove the assertion only for, say, ν + . The other case will follow similarly. Set
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η = ν + . Suppose that φ is a nonnegative simple function φ(x) =
that φ ≤ f . We have:
Z
Z
X Z
X
gdµ = φgdµ
ai
ai ηEi =
φdη =
And by definition we have
Z
Z
φgdµ =
i
ai XEi such
Ei
i
i
P
+
φgdµ −
Z
φgdµ−
Taking the supremum over all such φ and observing that µ+ ⊥ µ− , both are finite,
and φg is measurable, nonnegative and φg ≤ f , we obtain:
Z
Z
Z
φgdµ+ − φgdµ−
sup φgdµ = sup
Z
= sup
φgdµ+ − sup
Z
φgdµ− =
Z
f gdµ− −
Z
f gdµ+ =
Z
f gdµ
(b) This is straightforward. We have, for every measurable E,
Z Z Z dν1
dν2
dν1
dν2
(ν1 + ν2 )E =
dµ +
dµ =
+
dµ
dµ
dµ
dµ
dµ
E
E
E
So by uniqueness,
dν1
dν2
d(ν1 + dν2 )
=
+
.
dµ
dµ
dµ
(c) This one is also straightforward: Observe that for each measurable E,
Z dν
dµ
νE =
E dµ
But by part (a) we have:
Z E
Z dν
dν
dµ
dµ =
dλ
dµ
dµ
dλ
E
So by uniqueness we obtain:
dν
dµ
dν
=
.
dλ
dµ dλ
(d) Well, since µ ν, for each measurable E we have
Z dµ
µE =
dν
E dν
By part (a) we get
Z Z dµ
dµ dν
dν =
dµ
dµ
E dν
E dν
Since µ µ and from definition of the integral,
Z
µE =
1dµ
E
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for each measurable E, by the uniqueness of the Rodon-Nykodim derivative we
obtain
dµ dν
1=
dν
dµ
a.e. Hence
−1
dν
dµ
=
dν
dµ
a.e.
Exersize 10. (a) Show that if ν is a signed measure such that ν ⊥ µ and ν µ,
then ν = 0.
(b) Show that if ν1 and ν2 are sigular with respect to µ, then so is αν1 + βν2 .
(c) Show that if ν1 and ν2 are absolutely continuous with respect to µ, so is αν1 +
βν2 .
(d) Prove the unigueness assertion in the Lebesgue decomposition.
Solution 10. (a) Let X = A ∪ B such that νA = µB = 0. Then clearly ν is
zero for every subset of A, since ν ⊥ µ is taken to mean |ν| ⊥ |µ|. That is to say,
|ν|A = |µ|B = 0. Further, since ν µ and for each subset of C of B, µC = 0,
we observe that νC = 0. Hence, if E is any measurable subset of X, writing
E = (A ∩ E) ∪ (B ∩ B), we conclude that nuE = 0.
(b) There exist decompositions A1 , B1 and A2 , B2 , Ai , Bi disjoint, such that
X = Ai ∪ Bi and νi Ai = µBi = 0. Let E = A1 ∩ A2 . Then E c = Ac1 ∪ Ac2 , and
E, E c for a decomposition of X with ν1 E = ν2 E = 0 and µE c = 0. Hence,
(c1 ν1 + c2 ν2 )E = µE c = 0
(c) Let E be a µ-measurable subset of X such that µE = 0. Then by hypothesis
ν1 E = ν2 E = 0. Hence (αν1 + βν2 )E = αν1 E + βν2 E = 0.
(d) Suppose, by the Lebesgue decomposition, ν = ν0 + ν1 and ν = ν˜0 + ν˜1 . Then
clearly 0 = (ν0 − ν˜0 ) + (ν1 − ν˜1 ). Observe that by parts (b) and (c), (ν0 − ν˜0 ) µ
and (ν1 − ν˜1 ) ⊥ µ. Hence there exists a decomposition A, B such that (ν1 − ν˜1 )A =
µB = 0. So for each subset E of B, by absolute continuity, (ν0 − ν˜0 )E = 0. Hence
(ν1 − ν˜1 )E = 0, so that ν1 E = ν˜1 E. On the other hand, for each subset E of A,
(ν1 − ν˜1 )E = 0, so that (ν0 − ν˜0 )E = 0, hence ν0 E = ν˜0 E. This shows uniqueness
of ν0 and ν1 .
Exersize 11. Use the following example to show that the hypothesis in the RadonNikodym Theorem that µ is σ-finite cannot be omitted. Let X = [0, 1], B the class
of Lebesgue measurable subsets of X, and take ν to be Lebesgue measure and µ
to be the counting measure on B. Then ν is finite and absolutely
continuous with
R
respect to µ, but there is no function f such that νE = E f dµ for all E ∈ B. At
what point does the proof of Theorem 23 break down for this example?
Solution 11. Suppose that f is any nonnegative µ-measurable functions on [0, 1].
Now, given E ⊂ [0, 1], we see immediately that
Z
f dµ = α ∈ R∗+
E
(the extended positive real numbers) where α is zero if f is zero (but this is a
pathological case which we exclude immediately); otherwise:
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(i) α is finite provided that E is a finite set and f is bounded. But if E is a
finite set, νE = 0, forcing f to be identically zero on every finite subset of [0, 1],
hence on [0, 1]. We already excluded this case.
(ii) α is infinite, then E is infinite or E is finite but f is not bounded on it. In the
former case, νE is still finite, in the latter case, νE is zero.
We see that Radon-Nykodim fails in this case.
Well, observe that the theorem is first proved for µ finite, and then extended to
the σ-finite case. However, the extension is valid only when X can be decomposed
into a countable collection of µ-finite sets (see problem 4). However, [0, 1] cannot
be decomposed into a countable collection of µ-finite sets. In a sense, by solution
to problem 4, the extension will break down since an uncountable sum is always
properly divergent (not to mention that we don’t know how to take an integral of
an uncountable sum, or to interchange the integral and the limit in this case). Exersize 12. (a) Let L1 (X, m) be the metric space with distance
Z
d(f, g) =
tan−1 |f − g|dm
X
Define convergence in measure as before. Prove that convergence in measure coincides with convergence in this metric space.
(b) do the same for
Z
|f − g|
dm
d(f, g) =
X 1 + |f − g|
(and prove that it is a metric, too).
Extra: Show that either one of these metric spaces is complete.
Solution 12. (a) In general, convergence in measure only implies convergence of a
subsequence a.e. But in this case, we show that if kfn − f k → 0, then d(fn , f ) → 0.
1
0
Observe first that if φ(x) = x−tan−1 (x), then φ0 (x) = 1− 1+x
2 . Hence φ (x) > 0
for all x. Thus φ(x) is strictly increasing. Further, φ(0) = 0. This shows that
φ(x) ≥ 0 for all x, hence x ≥ tan−1 (x) for all x. Using this, we have:
Z
Z
−1
d(fn , f ) =
tan |fn − f |dm ≤
|fn − f |dm = kfn − f k
X
X
Thus, since kfn − f k → 0, we must have d(fn , f ) → 0.
R
|f −g|
(b) If we set d(f, g) = X 1+|f
−g| dm, only the triangle inequality requires attenx
tion (the other two properties of a metric are obvious). Set φ(x) = 1+x
. Then
1
φ0 (x) = (1+x)2 which is clearly everywhere (on the nonnegative axis) positive.
Hence φ is an increasing function. So, φ(|a + b|) ≤ φ(|a| + |b|) =
|a|
1+|a|
|b|
1+|b|
|a|+|b|
1+|a|+|b|
≤
+
= φ(|a|) + φ(|b|). Substituting |f − g| for |a| and |g − h| for |b|, we get
φ(|f − h|) ≤ φ(|f − g|) + φ(|g − h|). Therefore,
Z
Z
Z
Z
φ(|f −h|)dm ≤
φ(|f −g|)+φ(|g −h|)dm =
φ(|f −g|)dm+
φ(|g −h|)dm
X
X
X
X
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So d(f, h) ≤ d(f, g) + d(g, h).
Now, observe again that |fn − f | ≥ d(fn , f ). This is obvious. Now using the
argument from part (a), we see that convergence in measure implies convergence
in the metric d.
|f −g|
We show now completeness with respect to the measure d(f, g) = 1+|f
.
Suppose
−g|
that fn is a Cauchy sequence. Set f = limfn . Now, given > 0, we know that
there exists N such that for all n, m ≥ N , we have d(fn , fm ) < /2. So,
Z
|fn − fm |
dm < /2
X 1 + |fn − fm |
But then, by Fatou’s lemma, we get:
Z
Z
Z
|fn − fm |
|fn − fm |
|fn − f |
dm =
limm
dm ≤ limm
dm ≤ /2 < 1 + |fn − fm |
X
X 1 + |fn − fm |
X 1 + |fn − f |
for all n ≥ N . This shows that d(fn , f ) < for all n ≥ N . Hence fn converges to
f . Further, since f = limfn , and each fn is in L1 (X, m), we know that f is also in
L1 (X, m). This establishes the result.
Exersize 13. Let ν : B → R (no infinities allowed!) be a signed measure. Prove
that supA∈B ν(A) < ∞, inf A∈B ν(A) > −∞.
Solution 13. Suppose that for each natural m > 0 there exists a measurable set
Am with ν(Am ) > m. Now, set B1 = A1 and Bn>1 = An \ (A1 ∪ · · · ∪ An−1 ).
Clearly Bi form a collection of disjoint sets and, in fact, An = ∪nk=1 Bk . This shows
that,
m < ν(Am ) = ν(∪m
k=1 Bk )
Therefore,
∞
X
∞
m<
ν(Bk ) = ν(∪∞
k=1 Bk ) = ∪k=1 Ak
k=1
for all natural m > 0. Hence ν(∪∞
k=1 Ak ) = ∞. But this cannot be by hypothesis!
Now, given that ν is a measure, by problem 2, we see that −ν is also a measure.
Further, since ν does not admit ±∞, neither does −ν. Now apply the above result
to −ν. We see that supA∈B −ν(A) < ∞. Hence inf A∈B ν(A) > −∞.
Exersize 14. Prove that there exist A1 and A2 so that ν(A1 ) = supA∈B ν(A) and
ν(A2 ) = inf A∈B ν(A).
Solution 14. First of all, observe that since ν(∅) = 0, the infimum is at most 0
and the supremum is at least 0.
Well, set S = supA∈B ν(A) and I = inf A∈B ν(A). If S = ∞, then for each natural
m > 0 there is a measurable Am with ν(Am ) > m. Now apply the construction in
the first part of problem 9 above to obtain a set A with ν(A) = ∞.
Similarly, if I = −∞, then consider −ν, whose supremum is ∞. Apply the argument above to get a set A such that −ν(A) = ∞. Then ν(A) = −∞.
Suppose now that S, I are finite real numbers. Then by approximation for the
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suprema, for each natural n > 0, there exists An such that S − 1/n ≤ ν(An ) ≤ S.
Consider Bi as in the first part of problem 9. Set A = ∪∞
k=1 Bk . Then clearly (see
first part of problem 9)
ν(A) ≥ S − 1/n
for all n. But then ν(A) ≥ S. So ν(A) = S.
Now consider −ν and apply the construction above to get a set A such that
−ν(A) = −I (−I is the supremum in this case). But then ν(A) = I.
Exersize 15. Extend the Radon-Nikodym Theorem to the case of signed measures.
Solution 15.
Exersize 16. Complex Measures.
(a) Show that eacy complex measure ν may be expressed as ν = µ1 − µ2 + iµ3 + iµ4 ,
where µi are finite measures.
(b) Show that for each complex measure ν there is a measure µ and a complexvalued measurable function φ with |φ| = 1 such that for each set E in B,
Z
νE =
φdµ
E
Solution 16.
Lemma 0.1. Let (X, B, µ) be a finite measure space and g an integrable function
such that for some constant M ,
Z
gφdµ ≤ M kφkp
for all simple functions φ. Then g ∈ Lq .
Exersize 17. Prove Lemma 0.1.
Solution 17.
Exersize 18. Show that Lemma 0.1 remains true if we only assume µ to be σ-finite.
Solution 18.
Lemma 0.2. Let hEn i be a sequence of disjoint measurable sets, and for
each n
P∞
let fn be a function in Lp (1 P
≤ p < ∞) that vanishes outside P
En . Set f = n=1 fn .
kfn kp < ∞. In this case f =
fn in Lp ; that is,
Then f ∈ Lp if and only if
∞
X
kf −
fi k p → 0
i=1
and
kf kp =
∞
X
kfn kp
n=1
Exersize 19. Prove Lemma 0.2
Solution 19.
Exersize 20. For g ∈ Lq , let F be the linear functional on Lp defined by
Z
F (f ) = f gdµ
Show that kF k = kgkq .
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Solution 20.
Exersize 21. Let µ be the counting measure on a countable set X. Show that
Lp (µ) = lp .
Solution 21.
Exersize 22. Let µ1 , µP
of Lebesgue measurable
2 be defined on the σ-algebra
P
subsets of R by µ1 (A) = n∈A 21n and µ2 (A) = n∈A 1/n. Here n is taken to run
through the natural numbers.
Exhibit a Hahn decomposition for 3µ1 − µ2 .
Solution 22. We show first that µ1 and µ2 are measures, µ1 finite and µ2 σ-finite,
which will establish that 3µ1 − µ2 is a signed, σ-finite measure.
Well, notice that since n runs over the set of natural numbers, we may restrict
our attention only to subsets A of (0, ∞). Clearly both are zero on the empty set.
σ-additivity follows from definition of infinite series and the fact that all terms are
positive. Further, since µ1 (0, ∞) is simply a geometric series, it is finite. Setting
(0, ∞) = ∪n [n − 1, n], we see that µ2 is σ-finite.
Now, observe that for all n ≥ 4, 3/2n < 1/n. Conversely, for all n < 4, 3/2n > 1/n.
Hence we see that for every subset of E of (0, 4), 3µ1 (E) − µ2 (E) ≥ 0. On the
other hand, for every subset E of [4, ∞), 3µ1 (E) − µ2 (E) ≤ 0. Therefore we have
the following Hahn decomposition: A = (0, 4) and B = [4, ∞).
Exersize 23. Find the Lebesgue decomposition for µ1 w.r.t. µ2 and for µ2 w.r.t.
µ1 . Find corresponding Radon-Nikodym derivatives for the a.c. parts.
Solution 23. Set ν0 = µ1 and ν1 = 0. Let A = N and B = (0, ∞) \ N. Then
clearly ν1 ⊥ µ2 , since ν1 (A) = µ2 (B) = 0. Now, if µ2 (E) = 0, then E ∩ N = ∅. But
in this case µ1 (E) = 0. Hence we see that ν0 = µ1 µ2 . Hence hour Lebesgue
decomposition of µ1 w.r.t. µ2 is:
µ1 = ν0 + ν1
Repeat the process as above: set ν0 = µ2 and ν1 = 0. Then the Lebesgue decomposition of µ2 w.r.t. µ1 is
µ2 = ν0 + ν1
Since µ1 is the absolutely continuous part of the Lebesgue decomposition w.r.t. µ2 ,
and vice versa, we must find the Rodon-Nikodym derivatives f, g, such that:
Z
µ1 (E) =
f dµ2
E
and
Z
µ2 (E) =
gdµ1
E
Well, define
∞
X
k
X(k−1,k]
f (x) =
2k
k=1
Then we have:
Z
f dµ2 =
E
∞ Z
X
k=1
E∩(k−1,k]
f dµ2 =
∞
X
k
µ2 (E ∩ (k − 1, k])
2k
k=1
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Observe that µ2 (E ∩ (k − 1, k]) = 0 if E ∩ (k − 1, k] = ∅ and 1/k otherwise. In which
case, from above we obtain:
k
0 if E ∩ (k − 1, k] = ∅
µ2 (E ∩ (k − 1, k]) =
1/2k otherwise
2k
Hence,
Z
f dµ2
µ2 (E) =
E
Similarly, setting
g(x) =
∞
X
2k
k
k=1
one obtains
X(k−1,k]
Z
µ1 (E) =
gdµ1
E
Exersize 24. Let
Z
µ1 (E) =
ZE
1
dx + 5δ1 + 2δπ
1 + x2
µ2 (E) =
dx + πδ1 + 2δe
E∩[0,1]
Find the Lebesgue decomposition for µ1 w.r.t. µ2 and for µ2 w.r.t. µ1 . Find
corresponding Radon-Nikodym derivatives for the a.c. parts.
Solution 24. Set
Z
ν0 (E) =
dx + πδ1
E∩[0,1]
and
ν1 (E) = 2δe
Set also A = R\{e} and B = {e}. Observe that A∪B = R and ν1 (A) = µ1 (B) = 0.
So ν1 ⊥ µ1 . Further, suppose that µ1 (E) = 0. Then we see that
Z
1
XE dx = 0
1 + x2
and 1, π ∈
/ E. But this means that m(E) = 0 (where m is the Lebesgue measure) and πδ1 (E) = 0. Hence ν0 (E) = 0, showing that ν1 µ1 . It is clear that
µ2 = ν0 + ν1 .
For the other case, take
Z
E∩[0,1]
1
dx + 5δ1 (E)
1 + x2
E∩[0,1]c
1
dx + 2δπ (E)
1 + x2
ν̃0 (E) =
and
Z
ν̃1 (E) =
It’s easy to see that µ1 = ν̃0 + ν̃1 and that ν̃0 µ2 . Now, take A = [0, 1] ∪ {e} and
B = Ac . Then we observe that ν̃1 (A) = µ2 (B) = 0, so that ν̃1 ⊥ µ2 . This gives the
Lebesgue decomposition.
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Now we find the Rodon-Nikodym derivatives in each case. So we must find measurable f, g such that:
Z
f dµ1
ν0 (E) =
E
and
Z
ν̃0 (E) =
gdµ2
E
To this end the following result will be useful.
Proposition 0.3. Suppose that µ1 , µ2 , . . . , µn are positive measures on a measure
space (B, X) and a1 , a2 , . . . , an are nonnegative constants. Suppose further that f
is a µ1 -measurable, nonnegative function. Then
(i) If λ = a1 µ1 + a2 µ2 + · · · an µn , and g is a λ-measurable functions, then
Z
Z
Z
Z
gdλ = a1 gdµ1 + a2 gdµ2 + · · · + an gdµn
(ii) If ν(E) =
R
E
f dµ1 and g is a ν-measurable function, then
Z
Z
gdν = gf dµ
Proof. The second assertion (ii) was proved as part of a previous homework (p.
268 #22(b)). The first assertion (i) follows easily by first considering g to be a
simple functions, in which case the result holds. Then extending the result by the
Monotone Convergence Theorem, by considering a sequence φn of simple functions
that increase to g.
Now take
f = (1 + x2 )X[0,1) +
π
X{1}
5
and
1
5
X[0,1) + X{1}
1 + x2
π
Then by Proposition 0.3, it follows that
Z
ν0 (E) =
gdµ1
g=
E
and
Z
ν̃0 (E) =
f dµ2
E
Exersize 25. Show that the product of two semi-rings is a semi-ring, but the
product of two rings is not necessarily a ring.
Solution 25.
Exersize 26. Assume that hEi i is a sequence of disjoint measurable sets and
E = ∪Ei . Then for any set A we have
X
µ∗ (A ∩ E) =
µ∗ (A ∩ Ei )
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Solution 26. This is true when n = 1. Suppose the result holds for some n.
Consider n + 1 measurable, disjoint sets E1 , . . . En , En+1 . Now we have, by measurability of En+1 and the fact that all are disjoint:
˜ )
µ∗ (A ∩ ∪n+1
Ei ) = µ∗ (A ∩ ∪n+1
Ei ∩ En+1 ) + µ∗ (A ∩ ∪n+1
Ei ∩ En+1
i
i
i
= µ∗ (A ∩ [∪ni Ei ]) + µ∗ (A ∩ En+1 )
And by induction hypothesis from above we obtain:
n
n+1
X
X
∗
∗
)
=
µ
(A
∩
E
)
+
µ
(A
∩
E
)
=
µ∗ (A ∩ Ei )
µ∗ (A ∩ ∪n+1
E
i
n+1
i
i
i
i
Now,
µ∗ (A ∩ E) ≥ µ∗ (A ∩ [∪ni Ei ]) =
n
X
µ∗ (A ∩ Ei )
i
After taking the limit we obtain
µ∗ (A ∩ E) ≥
∞
X
µ∗ (A ∩ Ei )
i
Hence
µ∗ (A ∩ E) =
∞
X
µ∗ (A ∩ Ei )
i
Exersize 27. Let A be a collection of sets which is closed under finite unions and
finite intersections; an algebra of sets, for example.
(a) Show that Aσ is closed under countable unions and finite intersections.
(b) Show that each set Aσδ is the intersection of a decreasing sequence of sets in
Aσ .
Solution 27. (a) Suppose that hAi i is a countable union of sets in Aσ . Then by
definition each Ai is in turn a countable union of sets in A, say Ai = ∪j Bji . But
then
∪i Ai = ∪i ∪j Bji
is a countable union of sets Bji i,j∈N , hence is in Aσ .
Suppose that A1 , A2 , . . . , An is a finite sequence of sets
in Aσ . Then again, each is
in turn a countable union of sets from A, say Ai = Bji . Now we have:
n i
∩i Ai = ∩i ∪j Bji = ∪∞
j=1 ∩i Bj
Now, since A is by hypothesis closed under finite intersections, ∩ni Bji is in A.
Hence above we end up with a countable union of sets from A, which is in Aσ .
This completes the proof.
(b) Suppose that B ∈ Aσδ . But then B is a countable intersection of sets hAi i
in Aσ . Let B1 = A1 and for n ≥ 2, Bn = ∩ni Ai . Then by part (a), Bi ∈ Aσ for
each i and
B = ∩j Bj
with Bn+1 ⊂ Bn . This completes the proof.
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Exersize 28. Let µ be a finite measure on an algebra A, and µ∗ the induced outer
measure. Show that a set E is measurable if and only if for each > 0 there is a
set A ∈ Aσ , A ⊂ E, such that µ∗ (E \ A) < .
Solution 28. By Aδ we understand the collection of all countable intersections of
sets from A.
Proposition 0.4. If µ is a finite measure and A, B are µ-measurable sets with B ⊂ A,
then µ∗ (A \ B) = µ∗ (A) − µ∗ (B).
Proof. Given that A, B are measurable, µ∗ (A) = µ∗ (A ∩ B) + µ∗ (A ∩ B c ). Since
B ⊂ A, we have µ∗ (A) = µ∗ (B) + µ∗ (A \ B) =⇒ µ∗ (A \ B) = µ∗ (A) − µ∗ (B). ( =⇒ ) Since E is measurable, so is E c . By approximation for the infimum, there
exists a cover of E c of sets hAi i in A such that (set A = ∪n An )
X
µ(An ) ≥ µ∗ (A)
(∗)
µ∗ (E c ) + >
n
Since E c ⊂ A, we know that Ac ⊂ E. Further,
c
Ac = [∪n An ] = ∩n Ac ⊂ Aδ
Now, by Proposition 0.4, µ∗ (E) = µ∗ (X) − µ∗ (E c ) and µ∗ (Ac ) = µ∗ (X) − µ∗ (A),
where X is our total space. In addition from (∗) above we get:
µ∗ (A) − µ∗ (E c ) < hence
µ∗ (X) − µ∗ (X) + µ∗ (A) − µ∗ (E c ) < Hence
µ∗ (E) − µ∗ (Ac ) < And again from Proposition 1, since certainly Ac is measurable, we get
µ∗ (E \ Ac ) = µ∗ (E) − µ∗ (Ac ) < This completes the proof in one direction.
(⇐=) Suppose that given E, for all > 0 there exists A ∈ Aδ such that
µ∗ (E \ A) < For convenience let us set C = E \ A. Now, for any set B we have:
µ∗ (E ∩ E) + µ∗ (B ∩ E c ) ≤ µ∗ (B ∩ E) + µ∗ (B ∩ Ac )
since A ⊂ E implies E c ⊂ Ac . Further,
µ∗ (B ∩ E) + µ∗ (B ∩ Ac ) = µ∗ ([B ∩ E] ∪ [B ∩ C]) + µ∗ (B ∩ Ac )
≤ µ∗ (B ∩ E) + µ∗ (B ∩ C) + µ∗ (B ∩ Ac ) ≤ µ∗ ([B ∩ E]) + + µ∗ (B ∩ Ac )
Since A is measurable, we have
µ∗ (B) = µ∗ (B ∩ A) + µ∗ (B ∩ Ac )
so we obtain from above
µ∗ (E ∩ E) + µ∗ (B ∩ E c ) ≤ µ∗ (B) + Neither the left nor the right hand side depends on . Hence
µ∗ (E ∩ E) + µ∗ (B ∩ E c ) ≤ µ∗ (B)
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From which we deduce that E is measurable. This completes the proof in the other
direction.