Real Analysis HW 10 Solutions

Problem 47: Show that a function f is absolutely continuous on [a, b] if and only if foreach ε > 0, there is a δ > 0 such that for every finite disjoint collection {(ak, bk)}nk=1 of openintervals in (a, b), ∣∣∣∣∣

n∑k=1

[f(bk)− f(ak)]

∣∣∣∣∣ < ε, ifn∑k=1

[bk − ak] < δ.

Solution: Clearly, by the triangle inequality the usual definition of absolute continuityimplies this one. It remains to show the converse. To see this, let {(ak, bk)}nk=1 be a collectionof disjoint open intervals in [a, b], we denote by the invervals {(a+k , b

+k )}n+

k=1 the intervals suchthat f(b+k )− f(a+k ) ≥ 0 and {(a−k , b

−k )}n−

k=1 the intervals such that f(b−k )− f(a−k ) < 0, and letδ+ and δ− be such that∣∣∣∣∣

n+∑k=1

[f(b+k )− f(a+k )]

∣∣∣∣∣ =n+∑k=1

|f(b+k )− f(a+k )| < ε/2, ifn+∑k=1

[b+k − a+k ] < δ+

and ∣∣∣∣∣n−∑k=1

[f(b−k )− f(a−k )]

∣∣∣∣∣ =n−∑k=1

|f(b−k )− f(a−k )| < ε/2, ifn−∑k=1

[b−k − a−k ] < δ−.

Summing the two inequalities above it follows that

n∑k=1

|f(bk)− f(ak)| < ε ifn∑k=1

[bk − ak] < min{δ+, δ−}.

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Problem 48: The Cantor Lebesgue function ϕ is continuous and increasing on [0, 1]. Con-clude from Theorem 10 that ϕ is not absolutely continuous on [0, 1]. Compare this reasoningwith that proposed in Problem 40.

Solution: The Cantor Lebesgue function has the property that ϕ′(x) = 0 a.e. and ϕ(1) −ϕ(0) = 1. If it were continuous, this would imply that

0 =

∫[0,1]

ϕ′ = ϕ(1)− ϕ(0) = 1,

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which is clearly a contradiction. This differs from the reasoning proposed in problem 40since we are using here a characterization of absolute continuity in terms of the Fundamentaltheorem of calculus instead of the definition of absolute continuity. �

Problem 49: Let f be continuous on [a, b] and differentiable almost everywhere on (a, b).Show that ∫ b

a

f ′ = f(b)− f(a)

if and only if∫ b

a

[limn→∞

Diff1/nf]

= limn→∞

[∫ b

a

Diff1/nf

].

Solution: Since f is continuous and differentiable a.e. on (a, b), we have that

limn→∞

[∫ b

a

Diff1/nf

]= lim

n→∞

(Av1/nf(b)−Av1/nf(a)

)= f(b)− f(a),

and ∫ b

a

[limn→∞

Diff1/nf]

=

∫ b

a

f ′

for almost all x. The result then follows immediately. �

Problem 52: Let f and g be absolutely continuous on [a, b]. Show that∫ b

a

f · g′ = f(b)g(b)− f(a)g(a)−∫ b

a

f ′ · g.

Solution: Since g is absolutely continuous {Diff1/ng} is uniformly integrable and therefore{(Diff1/ng) · f} is uniformly integrable. Thus∫ b

a

f · g′ = limn→∞

∫ b

a

f ·Diff1/ng.

however by direct computation we see that∫ b

a

f ·Diff1/ng = n

[∫ b

a

f(x)g(x+ 1/n)dx−∫ b−1/n

a−1/nf(x+ 1/n)g(x+ 1/n)dx

]

= n

∫ b

a

(f(x)− f(x+ 1/n))g(x+ 1/n)dx+ n

∫ b

b−1/nf(x+ 1/n)g(x+ 1/n)dx

− n∫ a

a−1/nf(x+ 1/n)g(x+ 1/n)dx

= Av1/n(f · g)(b)−Av1/n(f · g)(a)−∫ b

a

Diff1/nf · g.

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Since f and g are continuous,

Av1/n(f · g)(b)→ f(b)g(b), Av(f · g)(a)→ f(a)g(a),

and f is absolutly continuous so {Diff1/nf · g} is uniformly integrable, therefore∫ b

a

Diff1/nf · g →∫ b

a

f ′ · g.

Hence we have our result. �

Problem 55: Let f be of bounded variation on [a, b] and define v(x) = TV (f[a,x]) for allx ∈ [a, b].

(i) Show that |f ′| ≤ v′ a.e. on [a, b], and infer from this that∫ b

a

|f ′| ≤ TV (f).

(ii) Show that the above is an equlity if and only if f is absolutely continuous on [a, b].

(iii) Compare parts (i) and (ii) with Corollaries 4 and 12, repectively.

Solution:

(i) Suppose we take a partition P = {u, v}, with u, v ∈ [a, b]

|f(v)− f(u)| = V (f, P ) ≤ TV (f[u,v]) = T (f[a,v])− T (f[a,u])

therefore|f(v)− f(u)|

v − u≤T (f[a,v])− T (f[a,u])

v − u,

since f is of bounded variation, and TV (f[a,x]) is increasing, the limits as u→ v existspointwise a.e., so |f ′| ≤ v′ a.e. However,∫ b

a

|f ′| ≤∫ b

a

v′ ≤ v(b)− v(a) = TV (f).

(ii) If f is absolutely continuous, then we know that for any (u, v) ⊆ [a, b]∫ v

u

f ′ = f(v)− f(u).

Also f is of bounded variation, therefore for any partition P of [a, b],

V (f, P ) =n∑k=1

|f(bk)− f(ak)| =n∑k=1

∣∣∣∣∫ bk

ak

f ′∣∣∣∣ ≤ n∑

k=1

∫ bk

ak

|f ′| =∫ b

a

|f ′|.

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Therefore

TV (f) ≤∫ b

a

|f ′|

and so by the previous inequality,

TV (f) =

∫ b

a

|f ′|.

To show the coverse, suppose that we have equality. We can see from the work in part(i) that this must imply that ∫ b

a

v′ = v(b)− v(a).

Since v is strictly increasing, this means that v is absolutely continuous on [a, b]. Letδ > 0 be such that if {(ak, bk)} is a collection of disjoint open intervals then

n∑k=1

|TV (f[a,bk])− TV (f[a,ak])| =n∑k=1

TV (f[ak,bk]) < ε

whenever∑n

k=1[bk − ak] < δ. By the definition of TV (f[ak,bk]) have

|f(bk)− f(ak)| ≤ TV (f[ak,bk]).

Thereforen∑k=1

|f(bk)− f(ak)| ≤n∑k=1

TV (f[ak,bk]) < ε

whenever∑n

k=1[bk − ak] < δ.

(iii) If f is increasing as in Corollaries 4 and 12, then we know that total variation isTV (f) = f(b)− f(a), and f ′ is positive, Part (i) says that∫ b

a

f ′ ≤ f(b)− f(a)

which is equivalent to Corollary 4, and part (ii) says that f is absolutely continuous ifand only if ∫ b

a

f ′ = f(b)− f(a)

which is equivalent to Corollary 12.

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Problem 56: Let g be strictly increasing and absolutely continuous on [a, b].

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(i) Show that for any open subset O of (a, b),

m(g(O)) =

∫Og′(x)dx.

(ii) Show that for any Gδ subset E of (a, b),

m(g(E)) =

∫E

g′(x)dx.

(iii) Show that for any subset E of [a, b] that has measure 0, its image g(E) also has measure0, so that

m(g(E)) = 0 =

∫E

g′(x)dx.

(iv) Show that for any measurable subset A of [a, b],

m(g(A)) =

∫A

g′(x)dx.

(v) Let c = g(a), and d = g(b). Show that for any simple function φ on [c, d],∫ d

c

φ(y)dy =

∫ b

a

φ(g(x))g′(x)dx.

(vi) Show that for any nonnegative integrable function f over [c, d].∫ d

c

f(y)dy =

∫ b

a

f(g(x))g′(x)dx.

(vii) Show that part (i) follows from (vi) in the case that f is the characteristic function ofO and the composition is defined.

Solution:

(i) Write O countable collection of disjoint open intervals, O =⋃∞k=1(ak, bk), since g

is increasing and continuous, we can also write g(O) =⋃∞k=1(g(ak), g(bk)), where

{(g(ak), g(bk))} is also a countable collection of disjoint open intervals. Thus we see

m(g(O)) =∞∑k=1

[g(bk)− g(ak)]

which by the absolute continuity of g gives,

m(g(O)) =∞∑k=1

[∫ bk

ak

g′]

=

∫Og′.

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(ii) Write aGδ sets E as the intersection of a countable collection of open sets E =⋂∞k=1Ok.

Since for any n, En =⋂nk=1Ok is open and descending, we have

m(g(En)) =

∫En

g′.

Since g is strictly increasing and continuous, we know g(En) is also open and descend-ing, therefore, by continuity of measure and continuity of integration

m(g(E)) =

∫E

g′.

(iii) Problem 40.

(iv) Since A is measurable, we may write it as A = E ∪ F , a union of a Gδ set E and F aset of measure 0. Since g(E ∪ F ) = g(E) ∪ g(F ), and g(F ) is measure 0 and g(E) isGδ,

m(g(A)) = m(g(E)) =

∫E

g′ =

∫A

g′.

(v)* We first show this for an indicator function χE of a measurable set E ⊂ [c, d]. That is,we would like to show that

m(E) =

∫ b

a

(χE ◦ g) · g′.

Note, the difficulty here is that we do not know the measurability of χE ◦ g ≡ χg−1(E),since the inverse of a strictly increasing absolutely continuous function may not mapmeasurable sets to measurable sets. This can be seen in a counter-example by SilviaSpataru where an absolutely continuous strictly increasing fucntion f is constructedso that f ′ = 0 a.e. on a ‘fat’ cantor set B, and f maps B to a set of measure 0. Henceg−1 behaves like the Cantor-Lebesgue function, mapping a set of measure 0 to a set ofpositive measure. In particular this means that there f−1 maps a measure zero set toa non-measurable set.

However, it turns out that while χE◦g may not be measurable, (χE◦g)·g′ will always bemeasurable. We will show this by approximation with measurable functions. Denoteh := (χE ◦ g) · g′ and note that since g is continuous, then for any open set O andcompact K, (χO ◦ g) and (χK ◦ g) are both measurable functions. In particular thisimplies by (iv) that

m(O) =

∫ b

a

(χO ◦ g) · g′, and m(K) =

∫ b

a

(χK ◦ g) · g′.

We now choose {On} a decreasing sequence of open sets containing E, and {Kn} anincreasing sequence of compact sets contained in E so that,

m(On)→ m(E) and m(Kn)→ m(E).

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Define φn := (χKn ◦ g) · g′ and ψn := (χOn ◦ g) · g′. We see that for every n,

φ1 ≤ . . . ≤ φn ≤ h ≤ ψn ≤ . . . ≤ ψ1.

Since g′ ≥ 0 a.e., {φn} and {ψn} are monotone sequences of functions bounded by g′.Thus they have pointwise limits a.e. In particular since g′ is integrable, dominatedconvergence gives ∫ b

a

limn→∞

(ψn − φn) = m(E)−m(E) = 0.

Since ψn − φn ≥ 0 we conclude

limn→∞

(ψn − φn) = 0

pointwise. Thereforelimn→∞

|h− φn| ≤ limn→∞

(ψn − φn) = 0,

pointwise. So h is measurable, being the pointwise limit of measurable functions φn.We conclude by monotone (or dominated) convergence that

m(E) = limn→∞

∫ b

a

(χKn ◦ g) · g′ =∫ b

a

(χE ◦ g) · g′.

Now suppose that ϕ is a simple function over (c, d) given in canonical form by

ϕ =n∑k=1

skχEk.

where {Ek} is a finite disjoint collection of measurable sets such that⋃nk=1Ek = (c, d).

We find ∫ g(b)

g(a)

ϕ =n∑k=1

sk ·m(Ek) =n∑k=1

sk

∫ b

a

(χEk◦ g) · g′ =

∫ b

a

(ϕ ◦ g) · g′.

(vi) If f is a non-negative integrable function over [c, d] there is a monotone sequence ofsimple functions ϕn that converge to f p.w. Therefore by the previous problem∫ d

c

ϕn =

∫ b

a

(ϕn ◦ g) · g′

so by the monotone convergence theorem,∫ d

c

f =

∫ b

a

(f ◦ g) · g′.

(vii) Choose f to be the characteristic function of g(O), where O ⊆ (a, b). Then

m(g(O)) =

∫ b

a

(χg(O) ◦ g)g′ =

∫ b

a

χOg′ =

∫Og′

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