EXERCISES 2E

9 SupposeF1, . . . ,Fnare disjoint closed subsets ofR. Prove that if g:F1∪ · · · ∪Fn →^R

is a function such thatg|F_k is a continuous function for eachk ∈ {^{1, . . . ,}n}^, thengis a continuous function.

10 SupposeF ⊂ Ris such that every continuous function fromF toRcan be extended to a continuous function fromRtoR. Prove thatFis a closed subset ofR.

11 Prove or give a counterexample: IfF⊂Ris such that every bounded continuous function fromFtoRcan be extended to a continuous function fromRtoR, thenFis a closed subset ofR.

12 Give an example of a Borel measurable functionf fromRtoRsuch that there does not exist a setB ⊂ ^{Rsuch that}|^R\B| = 0and f|B is a continuous function onB.

13 Prove or give a counterexample: Ifft:R→^Ris a Borel measurable function for eacht∈^Rand f:R→(−∞,∞]is defined by

f(_x) =sup{ft(_x):t∈^R}^, thenf is a Borel measurable function.

14 Supposeb1,b2, . . .is a sequence of real numbers. Define f:R→[0,∞]by

f(_x) =







∑

∞ k=1

1

4^k|x−bk| ^ifx∈ {^/ b1,b2, . . .}^,

∞ ifx∈ {b₁,b2, . . .}^. Prove that|{x∈^R: f(_x)<1}|=_∞.

[This exercise is a variation of a problem originally considered by Borel. If b1,b2, . . .contains all the rational numbers, then it is not even obvious that {x∈^R: f(_x)<_∞} ̸=_∅.]

15 SupposeBis a Borel set and f:B → ^Ris a Lebesgue measurable function.

Show that there exists a Borel measurable functiong:B→^{Rsuch that}

|{x∈B:g(x)̸= f(x)}|=0.

Integration

To remedy deficiencies of Riemann integration that were discussed in Section1B, in the last chapter we developed measure theory as an extension of the notion of the length of an interval. Having proved the fundamental results about measures, we are now ready to use measures to develop integration with respect to a measure.

As we will see, this new method of integration fixes many of the problems with Riemann integration. In particular, we will develop good theorems for interchanging limits and integrals.

Statue in Milan of Maria Gaetana Agnesi, who in 1748 published one of the first calculus textbooks.

A translation of her book into English was published in 1801.

In this chapter, we develop a method of integration more powerful than methods contemplated by the pioneers of calculus.

©Giovanni Dall’Orto

73

3A Integration with Respect to a Measure

Integration of Nonnegative Functions

We will first define the integral of a nonnegative function with respect to a measure.

Then by writing a real-valued function as the difference of two nonnegative functions, we will define the integral of a real-valued function with respect to a measure. We begin this process with the following definition.

3.1 Definition S^-partition

SupposeS ^{is a}σ-algebra on a setX. AnS^-partitionofXis a finite collection A1, . . . ,Amof disjoint sets inS ^{such that}A1∪ · · · ∪Am=_X.

We adopt the convention that0·∞ and∞·⁰should both be interpreted to be0.

The next definition should remind you of the definition of the lower Riemann sum (see 1.3). However, now we are working with an arbitrary measure and

thusXneed not be a subset ofR. More importantly, even in the case whenXis a closed interval[_a,b]inRandµis Lebesgue measure on the Borel subsets of[_a,b], the setsA1, . . . ,Amin the definition below do not need to be subintervals of[_a,b]as they do for the lower Riemann sum—they need only be Borel sets.

3.2 Definition lower Lebesgue sum

Suppose(_X,S^,µ) is a measure space, f: X → [0,∞] is an S-measurable function, andPis anS^-partitionA1, . . . ,Am ofX. Thelower Lebesgue sum L(_f,P)is defined by

L(_f,P) =

∑

m j=1

µ(_Aj)inf

Aj f.

Suppose(_X,S^,µ)is a measure space. We will denote the integral of an S^- measurable function f with respect toµ byR

fdµ. Our basic requirements for an integral are that we want R

χ_Edµto equalµ(_E) for allE ∈ S, and we want R(f+g)dµ=R

fdµ+R

gdµ. As we will see, the following definition satisfies both of those requirements (although this is not obvious). Think about why the following definition is reasonable in terms of the integral equaling the area under the graph of the function (in the special case of Lebesgue measure on an interval ofR).

3.3 Definition integral of a nonnegative function

Suppose(X,S^,µ)is a measure space and f: X→[0,∞]is anS-measurable function. Theintegralof f with respect toµ, denotedR

fdµ, is defined by Z fdµ=sup{L(f,P):Pis anS-partition ofX}^.

Suppose(_X,S^,µ)is a measure space and f: X → [0,∞]is anS-measurable function. EachS^-partitionA1, . . . ,AmofXleads to an approximation of f from below by theS-measurable simple function∑^m_j=1 inf

Aj f

χ_Aj. This suggests that

∑

m j=1

µ(_Aj)inf

A_j f

should be an approximation from below of our intuitive notion ofR

fdµ. Taking the supremum of these approximations leads to our definition ofR

fdµ.

The following result gives our first example of evaluating an integral.

3.4 integral of a characteristic function

Suppose(_X,S^,µ)is a measure space andE∈ S^{. Then} Z

χ_Edµ=µ(_E).

The symbolR din the expression fdµhas no independent meaning, but it often usefully separatesf from µ. Because thedinR

fdµdoes not represent another object, some mathematicians prefer typesetting an uprightdin this situation, producingR

fdµ. However, the uprightdlooks jarring to some readers who are accustomed to italicized symbols. This book takes the compromise position of using slanteddinstead of math-mode italicizeddin integrals.

Proof IfPis theS-partition ofXcon- sisting of E and its complement X\E, then clearly L(χ_E,P) = µ(E). Thus R χ_Edµ≥^µ(_E).

To prove the inequality in the other direction, suppose P is an S^-partition A1, . . . ,Am of X. Then µ(_Aj)inf

A_j χ_E equalsµ(_Aj) if A_j ⊂ E and equals0 otherwise. Thus

L(χ_E,P) =

∑

{j:A_j⊂E}

µ(_Aj)

=µ _[

{j:Aj⊂E}

Aj

≤µ(E). ThusR

χ_Edµ≤^µ(_E), completing the proof.

3.5 Example integrals of χ_Qandχ_{[0, 1]\Q}

Supposeλis Lebesgue measure onR. As a special case of the result above, we haveR

χ_Qdλ=0(because|^Q|=0). Recall thatχ_Qis not Riemann integrable on [0, 1]. Thus even at this early stage in our development of integration with respect to a measure, we have fixed one of the deficiencies of Riemann integration.

Note also that3.4implies thatR

χ_{[0, 1]\Q}dλ = 1 (because|[0, 1]\^Q| = 1),

which is what we want. In contrast, the lower Riemann integral ofχ_{[0, 1]\Q}on[0, 1] equals0, which is not what we want.

3.6 Example integration with respect to counting measure is summation Supposeµis counting measure onZ⁺andb1,b2, . . .is a sequence of nonnegative numbers. Think ofbas the function fromZ⁺to[0,∞)defined byb(_k) =_bk. Then

Z bdµ=

∑

∞ k=1

bk, as you should verify.

Integration with respect to a measure can be calledLebesgue integration. The next result shows that Lebesgue integration behaves as expected on simple functions represented as linear combinations of characteristic functions of disjoint sets.

3.7 integral of a simple function

Suppose(_X,S^,µ) is a measure space,E1, . . . ,En are disjoint sets in S^{, and} c1, . . . ,cn ∈[0,∞]. Then

Z ⁿ

k=1

∑

c_kχ_E

k

dµ=

∑

n k=1

c_kµ(E_k).

Proof Without loss of generality, we can assume thatE1, . . . ,Enis anS-partition of X[by replacingnbyn+1and settingE_n+1=X\(E₁∪^{. . .}∪En)andc_n+1=0].

IfPis theS-partitionE₁, . . . ,EnofX, thenL ∑ⁿ_k=1c_kχ_Ek,P

=_∑ⁿ_k=1c_kµ(E_k).

Thus _Z

∑

n k=1

ckχ_E

k

dµ≥

∑

n k=1

ckµ(_Ek).

To prove the inequality in the other direction, suppose thatPis anS-partition A₁, . . . ,AmofX. Then

L

∑

ⁿ

k=1

c_kχ_Ek,P

=

∑

m j=1

µ(A_j) min

{i:A_j∩E_i̸=∅}c_i

=

∑

m j=1

∑

n k=1

µ(_Aj∩Ek) min

{i:Aj∩Ei̸=∅}ci

≤

∑

^m

j=1

∑

n k=1

µ(A_j∩E_k)c_k

=

∑

n k=1

ck

∑

m j=1

µ(_Aj∩Ek)

=

∑

n k=1

c_kµ(E_k). The inequality above implies thatR

∑ⁿ_k=1ckχ_E

k

dµ≤∑ⁿ_k=1ckµ(_Ek), completing the proof.

The next easy result gives an unsurprising property of integrals.

3.8 integration is order preserving

Suppose(_X,S^,µ)is a measure space and f,g:X→[0,∞]areS-measurable functions such that f(_x)≤g(_x)for allx∈X. ThenR

fdµ≤R gdµ.

Proof SupposePis anS-partitionA₁, . . . ,AmofX. Then infA_j f ≤^inf

A_j g

for eachj=1, . . . ,m. ThusL(_f,P)≤ L(_g,P). HenceR

fdµ≤R gdµ.

Monotone Convergence Theorem

For the proof of the Monotone Convergence Theorem (and several other results), we will need to use the following mild restatement of the definition of the integral of a nonnegative function.

3.9 integrals via finite simple functions

Suppose(_X,S^,µ)is a measure space andf:X→[0,∞]isS-measurable. Then Z fdµ=supn^m

j=1

∑

cjµ(_Aj):A1, . . . ,Amare disjoint sets inS^, 3.10

c1, . . . ,cm∈[0,∞), and f(_x)≥

∑

m j=1

cjχ_A

j(_x)for everyx ∈Xo . Proof First note that the left side of3.10is bigger than or equal to the right side by 3.7and3.8.

To prove that the right side of3.10is bigger than or equal to the left side, first assume thatinf

A f <∞for everyA∈ S ^withµ(_A)>0. Then forPanS^-partition A1, . . . ,Amof nonempty subsets ofX, takecj =inf

A_j f, which shows thatL(_f,P)is in the set on the right side of3.10. Thus the definition ofR

fdµshows that the right side of3.10is bigger than or equal to the left side.

The only remaining case to consider is when there exists a setA∈ S^{such that} µ(_A)>0andinf

A f =∞[which implies that f(_x) =∞for allx∈A]. In this case, for arbitraryt ∈(0,∞)we can takem=1,A1 = _{A, andc1}= t. These choices show that the right side of3.10is at leasttµ(_A). Becausetis an arbitrary positive number, this shows that the right side of3.10equals∞, which of course is greater than or equal to the left side, completing the proof.

The next result allows us to interchange limits and integrals in certain circum- stances. We will see more theorems of this nature in the next section.

3.11 Monotone Convergence Theorem

Suppose(X,S^,µ)is a measure space and0≤ f₁≤ f2≤ · · · is an increasing sequence ofS-measurable functions. Define f:X→[0,∞]by

f(x) = lim

k→∞f_k(x). Then

k→∞lim

Z fkdµ=

Z fdµ.

Proof The functionf isS-measurable by2.53.

Because f_k(_x) ≤ f(_x)for everyx ∈ X, we haveR

f_kdµ ≤ R

fdµfor each k∈^{Z+(by3.8). Thuslim}k→∞R

fkdµ≤R fdµ.

To prove the inequality in the other direction, supposeA1, . . . ,Amare disjoint sets inS ^andc1, . . . ,cm∈[0,∞)are such that

3.12 f(_x)≥

∑

m j=1

cjχ_A

j(_x) for everyx∈X.

Lett∈(0, 1). Fork∈^{Z+, let}

Ek=ⁿ_x∈X: fk(_x)≥t

∑

^m

j=1cjχ_A

j(_x)^o.

ThenE1⊂E2⊂ · · · is an increasing sequence of sets inS whose union equalsX.

Thuslim_k→∞µ(A_j∩E_k) =µ(A_j)for eachj∈ {^{1, . . . ,}m}(by2.59).

Ifk∈Z⁺, then

f_k(_x)≥

∑

m j=1

tcjχ_A

j∩Ek(_x) for everyx ∈X. Thus (by3.9)

Z fkdµ≥t

∑

^m

j=1cjµ(_Aj∩Ek).

Taking the limit ask→^∞of both sides of the inequality above gives

klim→∞

Z f_kdµ≥t

∑

^m

j=1

cjµ(_Aj). Now taking the limit astincreases to1shows that

k→∞lim

Z f_kdµ≥

∑

^m

j=1

c_jµ(A_j).

Taking the supremum of the inequality above over all S-partitions A1, . . . ,Am

ofXand allc1, . . . ,cm ∈ [0,∞)satisfying 3.12shows (using3.9) that we have lim_k→∞R

fkdµ≥R

fdµ, completing the proof.

The proof that the integral is additive will use the Monotone Convergence Theorem and our next result. The representation of a simple functionh: X→[0,∞]in the form∑ⁿ_k=1ckχ_E

kis not unique. Requiring the numbersc1, . . . ,cnto be distinct and E₁, . . . ,En to be nonempty and disjoint withE₁∪ · · · ∪En = Xproduces what is called thestandard representationof a simple function [takeE_k = h⁻¹({c_k}), wherec1, . . . ,cnare the distinct values ofh]. The following lemma shows that all representations (including representations with sets that are not disjoint) of a simple measurable function give the same sum that we expect from integration.

3.13 integral-type sums for simple functions

Suppose(_X,S^,µ)is a measure space. Supposea1, . . . ,am,b1, . . . ,bn ∈[0,∞] andA1, . . . ,Am,B1, . . . ,Bn ∈ Sare such that∑^m_j=1ajχ_A

j=_∑ⁿ_k=1bkχ_B

k. Then

∑

m j=1

a_jµ(A_j) =

∑

n k=1

b_kµ(B_k).

Proof We assumeA1∪ · · · ∪Am=_X(otherwise add the term0χ_X\(A

1∪ · · · ∪Am)).

SupposeA1andA2are not disjoint. Then we can write 3.14 a1χ_A

1+_a2χ_A

2=_a1χ_A

1\A2+_a2χ_A

2\A1+ (_a1+_a2)χ_A

1∩A2, where the three sets appearing on the right side of the equation above are disjoint.

NowA1 = (A1\A2)∪(A1∩A2)andA2 = (A2\A1)∪(A1∩A2); each of these unions is a disjoint union. Thusµ(A1) =µ(A1\A2) +µ(A1∩A2)and µ(_A2) =µ(_A2\A1) +µ(_A1∩A2). Hence

a1µ(A1) +a2µ(A2) =a1µ(A1\A2) +a2µ(A2\A1) + (a1+a2)µ(A1∩A2). The equation above, in conjunction with3.14, shows that if we replace the two sets A1,A2 by the three disjoint sets A1\A2,A2\A1,A1∩A2 and make the appropriate adjustments to the coefficientsa1, . . . ,am, then the value of the sum

∑^m_j=1ajµ(_Aj)is unchanged (althoughmhas increased by1).

Repeating this process with all pairs of subsets among A1, . . . ,Am that are not disjoint after each step, in a finite number of steps we can convert the ini- tial listA1, . . . ,Am into a disjoint list of subsets without changing the value of

∑^mj=1ajµ(_Aj).

The next step is to make the numbersa₁, . . . ,amdistinct. This is done by replacing the sets corresponding to eacha_jby the union of those sets, and using finite additivity of the measureµto show that the value of the sum∑^m_j=1a_jµ(A_j)does not change.

Finally, drop any terms for whichAj = ∅, getting the standard representation for a simple function. We have now shown that the original value of∑^m_j=1ajµ(_Aj) is equal to the value if we use the standard representation of the simple function

∑^m_j=1ajχ_A

j. The same procedure can be used with the representation∑ⁿ_k=1bkχ_B

kto show that∑ⁿ_k=1bkµ(_Bk)equals what we would get with the standard representation.

Thus the equality of the functions∑^m_j=1ajχ_A

jand∑ⁿ_k=1bkχ_B

kimplies the equality

∑^m_j=1ajµ(_Aj) =_∑ⁿ_k=1bkµ(_Bk).

If we had already proved that integration is linear, then we could quickly get the conclusion of the previous result by integrating both sides of the equation

∑^m_j=1ajχ_A

j=_∑ⁿ_k=1bkχ_B

kwith

respect toµ. However, we need the previous result to prove the next result, which is used in our proof that integration is linear.

Now we can show that our definition of integration does the right thing with simple measurable functions that might not be expressed in the standard represen- tation. The result below differs from3.7 mainly because the setsE1, . . . ,Enin the result below are not required to be dis- joint. Like the previous result, the next result would follow immediately from the linearity of integration if that property had already been proved.

3.15 integral of a linear combination of characteristic functions

Suppose(_X,S^,µ)is a measure space,E1, . . . ,En ∈ S^{, and}c1, . . . ,cn ∈[0,∞].

Then _Z

ⁿ

k=1

∑

ckχ_E

k

dµ=

∑

n k=1

ckµ(_Ek).

Proof The desired result follows from writing the simple function∑ⁿ_k=1c_kχ_E

kin the standard representation for a simple function and then using3.7and3.13.

Now we can prove that integration is additive on nonnegative functions.

3.16 additivity of integration

Suppose(_X,S^,µ)is a measure space and f,g:X→[0,∞]areS-measurable functions. Then _Z

(f +g)dµ=

Z fdµ+

Z gdµ.

Proof The desired result holds for simple nonnegativeS-measurable functions (by 3.15). Thus we approximate by such functions.

Specifically, let f1,f2, . . . andg1,g2, . . . be increasing sequences of simple non- negativeS-measurable functions such that

klim→∞f_k(_x) = _f(_x) and lim

k→∞g_k(_x) =_g(_x)

for allx∈X(see2.89for the existence of such increasing sequences). Then Z

(_f+_g)dµ= lim

k→∞

Z

(_fk+_gk)dµ

= lim

k→∞

Z fkdµ+ lim

k→∞

Z gkdµ

=

Z fdµ+

Z gdµ,

where the first and third equalities follow from the Monotone Convergence Theorem and the second equality holds by3.15.

The lower Riemann integral is not additive, even for bounded nonnegative measur- able functions. For example, iff =χ_{Q∩[0, 1]}andg=χ_{[0, 1]\Q}, then

L(_f,[0, 1]) =0 and L(_g,[0, 1]) =0 but L(_f +_g,[0, 1]) =1.

In contrast, ifλis Lebesgue measure on the Borel subsets of[0, 1], then Z fdλ=0 and ^Z gdλ=1 and ^Z (_f +_g)dλ=1.

More generally, we have just proved that R

(_f +_g)dµ = R

fdµ+R

gdµfor every measureµand for all nonnegative measurable functions f andg. Recall that integration with respect to a measure is defined via lower Lebesgue sums in a similar fashion to the definition of the lower Riemann integral via lower Riemann sums (with the big exception of allowing measurable sets instead of just intervals in the partitions). However, we have just seen that the integral with respect to a measure (which could have been called the lower Lebesgue integral) has considerably nicer behavior (additivity!) than the lower Riemann integral.

Integration of Real-Valued Functions

The following definition gives us a standard way to write an arbitrary real-valued function as the difference of two nonnegative functions.

3.17 Definition f⁺; f⁻

Supposef: X→[−∞,∞]is a function. Define functions f⁺and f⁻fromXto [0,∞]by

f⁺(_x) =

(f(_x) if f(_x)≥^0,

0 if f(x)<0 and f⁻(_x) =

(0 if f(_x)≥^0,

−f(x) if f(x)<0.

Note that iff:X→[−∞,∞]is a function, then

f = _f⁺−f⁻ and |f|= _f⁺+_f⁻.

The decomposition above allows us to extend our definition of integration to functions that take on negative as well as positive values.

3.18 Definition integral of a real-valued function;R fdµ

Suppose(X,S^,µ)is a measure space andf:X→[−∞,∞]is anS-measurable function such that at least one ofR

f⁺dµandR

f⁻dµis finite. Theintegralof f with respect toµ, denotedR

fdµ, is defined by Z fdµ=

Z f⁺dµ−

Z f⁻dµ.

If f ≥ ^{0, then} f⁺ = _f and f⁻ =0; thus this definition is consistent with the previous definition of the integral of a nonnegative function.

The conditionR

|f|^dµ < _∞is equivalent to the conditionR

f⁺dµ < _∞and R f⁻dµ<_∞(because|f|= _f⁺+_f⁻).

3.19 Example a function whose integral is not defined

Supposeλis Lebesgue measure onRandf: R→^Ris the function defined by f(_x) =

(1 ifx≥^0,

−^{1 if}x<0.

ThenR

fdλis not defined becauseR

f⁺dλ=_∞andR

f⁻dλ=_∞.

The next result says that the integral of a number times a function is exactly what we expect.

3.20 integration is homogeneous

Suppose(X,S^,µ)is a measure space andf: X→[−∞,∞]is a function such thatR

fdµis defined. Ifc∈^{R, then} Z c fdµ=_c

Z fdµ.

Proof First consider the case where f is a nonnegative function andc≥^{0. If}Pis anS-partition ofX, then clearlyL(_{c f,P}) =_cL(_f,P). ThusR

c fdµ=_cR fdµ.

Now consider the general case wheref takes values in[−^∞,∞]. Supposec≥^0.

Then Z

c fdµ= Z

(_{c f})⁺dµ− Z

(_{c f})⁻dµ

=

Z c f⁺dµ−

Z c f⁻dµ

=_c

Z f⁺dµ−

Z f⁻dµ

=c^Z fdµ,

where the third line follows from the first paragraph of this proof.

Finally, now supposec<0(still assuming thatf takes values in[−∞,∞]). Then

−c>0and

Z c fdµ= Z

(_{c f})⁺dµ− Z

(_{c f})⁻dµ

= Z

(−c)f⁻dµ− Z

(−c)f⁺dµ

= (−c)

Z f⁻dµ−

Z f⁺dµ

=_c

Z fdµ, completing the proof.

Now we prove that integration with respect to a measure has the additive property required for a good theory of integration.

3.21 additivity of integration

Suppose(_X,S^,µ) is a measure space and f,g: X → R are S-measurable functions such thatR

|f|^dµ<_∞andR

|g|^dµ<_{∞. Then} Z

(_f +_g)dµ=

Z fdµ+

Z gdµ.

Proof Clearly

(_f+_g)⁺−(_f +_g)⁻ = _f +_g

= f⁺− f⁻+g⁺−g⁻. Thus

(_f+_g)⁺+ _f⁻+_g⁻= (_f +_g)⁻+_f⁺+_g⁺.

Both sides of the equation above are sums of nonnegative functions. Thus integrating both sides with respect toµand using3.16gives

Z

(_f +_g)⁺dµ+

Z f⁻dµ+

Z g⁻dµ= Z

(_f +_g)⁻dµ+

Z f⁺dµ+

Z g⁺dµ.

Rearranging the equation above gives Z

(_f +_g)⁺dµ− Z

(_f +_g)⁻dµ=

Z f⁺dµ−

Z f⁻dµ+

Z g⁺dµ−

Z g⁻dµ, where the left side is not of the form∞−∞because(_f +_g)⁺ ≤ f⁺+_g⁺ and (_f +_g)⁻≤ f⁻+_g⁻. The equation above can be rewritten as

Gottfried Leibniz(1646–1716) invented the symbolR

to denote integration in 1675.

Z

(_f+_g)dµ=

Z fdµ+

Z gdµ,

completing the proof.

The next result resembles3.8, but now the functions are allowed to be real valued.

3.22 integration is order preserving

Suppose(X,S^,µ) is a measure space and f,g: X → R are S-measurable functions such thatR

fdµandR

gdµare defined. Suppose also thatf(x)≤g(x) for allx∈X. ThenR

fdµ≤R gdµ.

Proof The cases whereR

fdµ=±∞orR

gdµ=±∞are left to the reader. Thus we assume thatR

|f|^dµ<_∞andR

|g|^dµ<_∞.

The additivity (3.21) and homogeneity (3.20withc=−1) of integration imply

that _Z

gdµ−

Z fdµ= Z

(g−f)dµ.

The last integral is nonnegative becauseg(x)−f(x)≥⁰for allx∈X.

The inequality in the next result receives frequent use.

3.23 absolute value of integral≤integral of absolute value

Suppose(_X,S^,µ)is a measure space andf: X→[−^∞,∞]is a function such thatR

fdµis defined. Then

Z fdµ≤ Z

|f|^dµ.

Proof BecauseR

fdµis defined,f is anS-measurable function and at least one of R f⁺dµandR

f⁻dµis finite. Thus

Z fdµ=

Z f⁺dµ−

Z f⁻dµ

≤

Z f⁺dµ+

Z f⁻dµ

= Z

(_f⁺+ _f⁻)dµ

= Z

|f|^dµ, as desired.