Dr. John Travis
Mississippi College
CHAPTER TWO
Defn: A measure space will describe a set X, a s-algebra of subsets A and a measure m, often denoted simply as (X,A,m). Notice, for Lebesgue measure, the space X=Rn. If m(X)<oo , the space is finite and s-finite if m is s-finite.
Note: We will often want to consider the space X=[-oo, oo], called the extended reals. To do so, topologically we declaring the following sets to be open: (a,b), [-oo,a), (a,oo] and any union of segments of this type.
Defn: An extended real-valued function f is measurable if for any open set M in R, f -1(M) = {x | f(x)eM } is a measurable set. This implies that if f is measurable, the inverse image of any open set in the range R is a measurable set in the domain X.
Theorem 2.1.1: Suppose f is an extended real-valued function defined on a measure space X.
Then, f is measurable if and only if f -1{ (-oo,c) }, f -1{-oo} and f -1{oo} are measurable.
Proof (=>): Suppose f is measurable. Then, certainly each of the sets (-oo,c), {y = -oo} and {y = oo} are open sets. So, by the definition of measurable functions, the inverse image of each of these must be a measurable set.
Proof (<=): Suppose the sets f -1{ (-oo,c) }, f -1{-oo} and f -1{oo} are measurable for any real number c. Notice,
f -1{ (-oo,c] } = U f -1{ (-oo,c+1/n) }
which is measurable, being the infinite union of measurable sets. But,
f -1{ (a,b) } = f -1{ (-oo,b] } - f -1{ (-oo,a] },
which is measurable, being the difference of measurable sets. Since each open set M is a countable union of open interavals, then f -1{M} = U f -1{ (a,b) }, which is therefore open.
HOMEWORK: page 30, #4, 5
Defn: A real-valued function f defined on the metric space X is continuous if the inverse image of any open set in R is open in X.
Theorem 2.1.2: If f is continuous, then f is measurable. (Notice, the s-algebra on R is created using the open sets.) Pf: Uses Corollary 1.8.3
Theorem 2.1.3: Let X be a measure space. If f is a measurable function on X, then for any Borel set B of R, f -1{B} is measurable.
Pf: Consider the class D of all sets E on the real line for which f -1(E) is measurable. Then, D is a s-ring and D contains all the open sets. Hence, it also contains all the Borel sets
HOMEWORK: page 31 #6, 8, 9(very important), 10
Defns: In the above Homework, the definitons of the following functions are made:
Lemma 2.2.1: If f and g are measurable functions, then the set E = {x | f(x) < g(x) } is a measurable set.
Pf: The set of all rational numbers is a countable set. Hence, we can write the rationals as an ordered set {rn}. Set
En = {x | f(x) < rn } ^ {x | rn < g(x) } = f -1(-oo,rn) ^ g -1(rn,oo).
By theorem 2.1.1, the first of these two sets is measurable and the complement of the second is by looking at homework problem 2.1.5. Hence, En is itself measurable for every n and so E = U En is measurable.
Notation: By writing f(x) + g(x), f(x) g(x) or f(x)/g(x), we will assume no indeterminant forms arise.
Theorem 2.2.2: The sum, difference and product of measurable functions are measurable functions.
Pf: Suppose f and g are measurable functions. For any constant K, notice K - g(x) is measurable. Indeed, {x | K - g(x) < c } = { x | g(x) > K - c } = g -1(K-c,oo), which is measurable using homework problem 2.1.5.
To show that f(x) + g(x) is measurable, we must show E = {x | f(x) + g(x) < c } is measurable. But,
E = {x | f(x) < c - g(x) }.
Lemma 2.2.1 implies E is measurable. Further, the inverse image involving infinity yields measurable sets. Hence, Theorem 2.1.1 implies f(x)+g(x) is measurable. Similarly for f(x) - g(x).
Notice, from problem 2.1.9, | f - g |2 and | f + g |2 are measurable. But f(x) g(x) = { | f + g |2 - | f - g |2 }/4, which is measurable.
General Result: If f and g are measurable and real-valued and H is real and continuous on R2, then h(x) = H(f(x),g(x)) is measurable.
Pf: Let Ga = { (u,v) : H(u,v) < a }, which is an open subset of R2. Hence, we can write Ga as a union of intervals In, where In = { (u,v) : an < u < bn, cn < v < dn }.
Notice, since f and g are measurable, so are
Hence, the composite mapping h(x) satisfies
{ x : h(x) < a } = { x : (f(x),g(x)) e Ga } = U { x : (f(x),g(x)) e In } , which is measurable by above.
Corollary: The following are measurable:
Defn : For a given sequence {fn}, define sup fn(x), lim sup fn(x), inf fn(x) and lim inf fn(x) as on page 33. Notice, inf fn(x) = - sup fn(-x); lim sup fn(x) = inf { sup fn(x) } and lim inf fn(x) = sup { inf fn(x) }. So, proving results for sup's often yields results for inf's and together yield results for lim sup's and lim inf's.
Theorem 2.2.3: If {fn} is a sequence of measurable functions, then
are all measurable.
Pf: Notice, {x | sup fn(x) < c } = ^ {x | fn(x) < c } = ^ f -1(-oo,c], which are all measurable by homework problem 2.1.4. Looking at inverse images of infinity yields measurable sets and hence, sup fn(x) is measurable. It follows that inf fn(x) is measurable. By combining these as in the definition above, we get the rest of the functions are measurable.
Defn: A property P is said to be true almost everywhere (a.e.) if the set of points E for which P is not true has measure zero.
Corollary 2.2.4: If the sequence {fn} of measurable functions converges to the function g, then g is measurable. If X is complete, then the convergence need only be almost everywhere.
Pf: Apply the previous theorem noting if lim fn(x) exist, it is equal to lim sup fn(x).
Defn: A function f is a simple function if there exist sets E1, E2, ..., En and real numbers a1, a2, ..., an such that for xeEk, f(x) = ak (for some k) and f(x)=0 otherwise. We write f(x) = S ak cEk(x).
Theorem 2.2.5: Let f be a nonnegative measurable functions. Then, there exists a monotone-increasing sequence { fn } of simple nonnegative functions such that lim fn(x) = f(x) a.e.
Pf: For n=1,2,3,... divide the y-axis up into "didactic" intervals with endpoints , for k=0,1,...,n2n. For any given value of x, define:
If f(x) > n, define fn(x) = n.
If f(x) < n and (k-1)/2n < f(x) < k/2n, for some k, define
fn(x) = (k-1)/2n = greatest lower bound endpoint below the actual value of f(x)
Then, fn(x) is a simple function and fn+1(x) > fn(x).
Case 1: If f(x) < oo for a given x, then 0 < f(x) - fn(x) < 2-n, which approaches zero as n-->oo.
Case 2: If f(x)=oo for a given x, then fn(x)=n.
Hence, fn(x) approaches f(x) as n-->oo.
HOMEWORK: page 35 #3, 6, 7
Remark: The idea of a sequence of functions being "close" to another function can be defined in several ways. Usually, given an x-value, one thinks of close as the sequence of y-values fn(x) getting closer to its limit f(x).
Defn: A measurable function f is said to be a.e. real-valued is the set {x | |f(x)| = oo } has measure zero.
Defn: A sequence {fn} of a.e. real-valued, measurable functions is said to converge almost uniformly to a measurable function f if for any e>0, there exists a measurable set E such that m(E) < e and {fn} converges to f uniformly on X - E.
Theorem 2.3.1: If a sequence {fn} of a.e. real-valued, measurable functions converges almost uniformly to a measurable function f, then {fn} converges to f a.e.
Pf: Since {fn} converges almost uniformly to f, for any integer m>0 there is a set Em such that m(Em) < 1/m and {fn} converges to f uniformly on X - Em. Hence, {fn} converges to f on F = U (X-Em) = X - ^ Em.
But m(X - F) = m( ^ Em) < m(Em) < 1/m for any positive integer m. Thus m(X - F) = 0 and so {fn} converges to f a.e.
Theorem 2.3.2: (Egoroff's Theorem) Let X be a finite measure space. If a sequence {fn} of a.e. real-valued, measurable functions converges a.e. to f, then {fn} converges to f almost uniformly.
Pf: Since f and {fn} are real-valued a.e., then it is sufficient to assume that all functions are real-valued everywhere. For positive integers k and n, define
En,k = ^ m=n...{x | |fm(x) - f(x) | < 1/k }.
Notice, as n increases, the sets En,k form a monotone increasing sequence of sets. Since {fn} converges a.e. to f, then lim En,k = E, where E is a set such that X-E has measure zero. By Theorem 1.2.1,
lim m(x- En,k) = m(X-E) = 0.
Hence, for any e>0 there is an integer nk such that m(X - En,k) < e/2k, if n > nk.
If F = ^ En,k, then F is measurable and
m(X - F) = m(X - ^ En,k) = m( U (X-En,k) ) < S m( X-En,kk ) < e.
Hence, on the set F, the sequence {fn} is uniformly convergent to f.
Defn: A sequence {fn} is convergent in measure if there is a measurable function f such that for any e>0,
lim m[ { x : |fn(x) -f(x)| > e } ] = 0.
Theorem: If {fn} converges in measure to both f and g, then f=g a.e. and both are real-valued a.e.
Theorem 2.4.1: If a sequence {fn} of a.e. real-valued measurable functions converges almost uniformly to a measurable function f, then {fn} converges in measure to f.
Pf: For and e>0 and d>0, there is a set E with m(E)<d such that |fn(x) - f(x)| < e, for all x e X-E and n sufficiently large. This implies convergence in measure.
Corollary 2.4.2: If m(X) < oo, then any sequence {fn} of a.e. real-valued, measurable functions that converges a.e. to an a.e. real-valued, measurable function f is also convergent to f in measure.
Pf: Apply Theorem 2.4.1 and Egoroff's Theorem.
HOMEWORK: page 39 #1, 2
Defn: A given simple function f(x) = S ak cEk(x) is said to be integrable if m(Ek) < oo, for all k such that ak is nonzero. The integral over X is given by ,/` f(x) dm = S ak m(Ek), where we use the convention that (0)(oo) = 0. The integral is independent of the (several equivalent) representations of f(x). (See text, page 40.)
If E is any measurable set, then the Integral of f over E is given by ,/`E f(x) dm = S ak m(E ^ Ek).
Theorem 2.5.1: Let f and g be integrable simple functions and a and b be real numbers. Then,
Defn: A sequence of measurable functions fn is said to be a Cauchy sequence in the mean if
,/` | fn - fm| dm-->0 as n,m-->oo .
Lemma 2.5.2: If fn is a sequence of integrable simple functions that is Cauchy in the mean, then there is an a.e. real-valued, measurable function f such that fn converges in measure to f.
Pf: Let e>0. Choose En,m= { x | |fn(x) - fm(x)| > e |.
Since fn(x) - fm(x) is integrable, En,m has finite measure.
Theorem 2.5.1 implies ,/` | fn - fm| dm > e m(En,m) > 0
Since the sequence is Cauchy, the integral approaches zero and hence m(En,m)-->0 as n,m-->oo .
Hence fn is Cauchy in measure and so by Corollary 2.4.4, fn is convergent in measure.
HOMEWORK: page 42, #1, 2, 3
Consider the following conditions:
Defn 2.6.1: f is said to be integrable if there exists a sequence {fn} of integrable simple functions such that C1 and C2 hold.
Theorem 2.6.1: f is integrable if and only if C1 and C3 hold.
Pf: Suppose C1 and C2 hold. Lemma 2.5.2 implies {fn} converges in measure to an a.e. real-valued, measurable function g.
Theorem 2.4.3 and 2.3.1 imply that there is a subsequence {fn,k} of {fn} that converges a.e. to g. C2 implies g=f, a.e. Hence, {fn} converges in measure to f.
Conversely, suppose C1 and C3 hold. Then, there is a sequence {gn} of integrable simple functions such that {gn} is Cauchy in the mean and converges in measure to g.
Theorems 2.4.3 and 2.3.1 imply there is a subsequence {gn,k} of {gn} that is convergent to f a.e. Denote this subsequence by {fk}. Then, {fk} satisfies C1 and C2.
Result: If f is integrable, then f is a.e. real-valued. (See result preceding Theorem 2.4.1.)
Result: Let {fn} be a sequence of simple functions convergent to f. Then, lim ,/` fn dm exists.
Pf: Consider | ,/` fn dm - ,/` fm dm | < ,/` |fn - fm| dm which approaches zero since convergent implies Cauchy.
Defn 2.6.2: Let f be an integrable function and let C1 and C2 hold. The integral of f is defined to be the number lim ,/` fn dm and is denoted by ,/` f dm . Hence, ,/` f dm = lim ,/` fn dm .
Theorem 2.6.2: The definition of ,/` f dm is independent of the sequence {fn} chosen.
Proof: See lemmas in text, pages 43-46.
Defns (2.6.2, 2.6.5, 2.6.6): Suppose E is a measurable set and f an integrable function. Then, the integral of f over E is defined by
,/`E f dm = lim ,/` cE fn dm
If we are using a Lebesque measure space, we often denote the Lebesque integral of f as
,/`E f(x) dx
If f is a non-negative measurable function and not integrable on the set E, then we say that
,/`E f dm = oo.
HOMEWORK: page 47 #1, 2, 3, 7
Theorem 2.7.1: Let f and g be integrable functions and a and b be real numbers. Then,
(i) ,/` (a f + b g) dm = a,/` f dm + b,/` g dm
Pf: Take limits in 2.5.1.
(ii) If f>0 a.e., then ,/` f dm > 0.
(iii) If f>g a.e., then ,/` f dm > ,/` g dm .
(iv) | f | is integrable and | ,/` f dm | < ,/` | f | dm
Pf: Note f < | f | and -f < | f |. Apply (iii).
(v) ,/` | f + g | dm < ,/` | f | dm + ,/` | g | dm
(vi) m < f < M a.e. on a measurable set E with m(E) < oo yields m m(E) < ,/`E f dm < M m(E).
(vii) If f > 0 a.e. and E and F are measurable sets such that ECF, then ,/`E f dm < ,/`F f dm .
(viii) If f > m > 0 on a measurable set E, then m(E)<oo .
Pf: Assume m(E) is infinite. By problem 2.6.2, E is s-finite. Hence, there exist a monotone-increasing sequence of sets Ek with m(Ek)<oo and lim Ek = E.. By (vii), ,/`E f dm > ,/`Ek f dm > m m(Ek) which approaches oo. Contradiction.
Defn 2.7.1: A sequence {fn} of integrable functions is said to be a Cauchy sequence in the mean if ,/` | fn - fm| dm approaches zero as n and m get large. If there is an integrable function f such that ,/` | fn - f| dm approaches zero as n gets large, then we say that {fn} converges in the mean to f.
Result: If {fn} is convergent in the mean to f, then it is also Cauchy in the mean.
Theorem 2.7.2: If {fn} is a sequence of integrable functions that converges in the mean to an integrable function f, then {fn} converges in measure to f.
Pf: Let e>0 be given and define En = { x | |fn(x) - f(x) | > e }. By Theorem 2.7.1, m(En)<oo and
,/`E | fn - f | dm > ,/`En | fn - f | dm > m m(En).
Hence, m(En) approaches zero as n gets large.
Theorem 2.7.3: If f is an a.e. nonnegative, integrable function, then ,/`E f dm = 0 if and only if f=0 a.e. on E.
Pf: If f=0 a.e., then HW 2.6.1 with g=0 everywhere yields ,/`E f dm = 0.
On the other hand, if ,/`E f dm = 0, then there exists a sequence {fn} that is Cauchy in the mean and this is convergent in measure to f. Since f > 0, then the same is true for |fn|. So, lim ,/`E fn dm = ,/`E f dm = 0. Hence, {fn} converges in the mean to zero. Theorem 2.7.2 implies that {fn} converges in measure to zero and thus f=0 a.e.
Theorem 2.7.4: Let f be measurable and E a set of measure zero. Then, f is integrable on E and ,/`E f dm = 0.
Pf: Notice that cE f = 0 a.e. By Problem 2.6.1, cE f is integrable and ,/` cE f dm = 0.
Theorem 2.7.5: Let f be an integrable function that is positive everywhere on a measurable set E.
If ,/`E f dm = 0, then m(E)=0.
Pf: Let En = {xeE | f(x) > 1/n }. Then, {En} is a monotone increasing sequence of sets and E - U En has measure zero. Hence, m(E) = lim m(En).
Since m(En) is finite, 0 = ,/`E f dm > ,/`En f dm > m(En)/n > 0. Thus, m(En)=0 for all n and so m(E)=0.
Theorem 2.7.6: Let f be an integrable function. If ,/`E f dm = 0 for every measurable set E, then f=0 a.e.
Pf. By Theorem 2.7.5, the set where f(x)>0 has measure zero. Similarly, the set where f(x)<0 has measure zero. Hence, f=0 a.e.
HOMEWORK: page 50 #2, 3, 4, 6
Lemma 2.8.1: If {fn} is a sequence of integrable simple functions yielding the integrable function f, then limn-->oo ,/` | fn - f | dm = 0
Pf: Consider the sequence {gk} = { | fn - fk | }, for any positive integer n. Notice that | |a| - |b| | < | a - b |.
Hence, ,/` | gm - gk | dm = ,/` | | fn - fm | - | fn - fk | | dm < ,/` | fn - fk | dm -->0 as m and k get large, since {fn} is Cauchy in the mean. Thus, {gk} is also Cauchy in the mean. Further, gk converges to | fn - f | a.e.
Applying the definition of integral yields the result.
Theorem 2.8.2: If {fn} is a sequence of integrable functions which are Cauchy in the mean such that lim fn = f, an integrable function, then limn-->oo ,/` fn dm = ,/` f dm.
Pf: By Lemma 2.8.1, for each n, there is a sequence of integrable simple functions {fn,k} such that limk-->oo ,/` | fn - fn,k | dm = 0.
Hence, for each n, there is a term`fn of the sequence fn,k such that ,/` | fn - `fn | dm < 1/n2.
The proof of Theorem 2.7.2 with e=1/n yields m{ x | | fn(x) -`fn(x) | > 1/n } < 1/n. Hence, {`fn } is Cauchy in the mean and converges in measure to f. Therefore, f is integrable and
,/` f dm = limn-->oo ,/` `fn dm = limn-->oo .,/` fn dm
Theorem 2.8.3: If {fn} is a sequence of integrable functions that is Cauchy in the mean, then there is an integrable function f such that {fn} converges in the mean to f.
Pf: Extending the proof of Lemma 2.5.2 applied to any integrable function, we conclude that {fn} is convergent in measure to a measurable function f. Theorem 2.8.2 implies that f is integrable. Hence, the sequence {| f - fn |} is a sequence of integrable functions that is Cauchy in the mean and that converges in measure to 0. Theorem 2.8.1 implies limn-->oo ,/` | f - fn | dm = 0
Defn 2.8.1: A real-valued set function l is said to be absolutely continuous is for any e>0, there exists a number d>0 such that for any measurable set E with m(E)<d, |l(E)|<e.
Theorem 2.8.4: Let f be an integrable function and let l be the set function defined by l(E) = for all the measurable sets E. Then, l is completely additive and absolutely continuous and is called the indefinite integral of f.
Pf: see text.
Theorem 2.9.1: (Lebesque's Bounded Convergence Theorem - LBCT) Let {fn} be a sequence of integrable functions that converges either in measure or a.e. to a measurable function f. Suppose there exists an integrable function g such that |fn(x)| < g(x) a.e. for all n. Then, f is integrable and limn-->oo ,/` | f - fn | dm = 0.
Pf:
Case I: Suppose {fn} converges to f in measure. Show {fn} is a Cauchy sequence in the mean. If so, then by Theorem 2.8.3 an integrable function h exists such that limn-->oo ,/` | h - fn | dm = 0. Theorem 2.7.2 implies then that {fn} converges in measure to h. Since f is also the limit in measure of {fn}, we have f=h a.e. and so f is integrable. Replacing h with f gives the result.
To show {fn} is a Cauchy sequence in the mean, see text, page 55.
Case II: Suppose {fn} converges a.e. to f. Show {fn} also converges in measure to f and apply Case I.
Set N = {x | |f(x)|>g(x) or |fn(x)|>g(x) } and for any e>0, set En = U k=n.. { x | |fk(x) - f(x) | > e}.
Notice, { x | |fn(x) - f(x) | > e } C En.
Then, En C {x | g(x) > e/2 } U N.
Since g is integrable, HW 2.7.2 implies that m(En) is finite. Since fn converges to f a.e., m(En)=0.
Therefore, by Theorem 1.2.1, lim m(En)=0 and so {fn} converges to f in measure
HOMEWORK: page 56 #1, 4
Theorem 2.10.1: Let f and g be measurable. If |f| < g a.e. and g is integrable, then f is integrable.
Pf: HW problem 2.6.3 implies that f is integrable if and only if |f| is integrable. Indeed,
=> Assume |f| is integrable. Then, | ,/` f dm | < ,/` |f| dm, which is finite by assumption.
<= Assume f is integrable. Then, ,/` |f| dm = ,/`A |f| dm + ,/`B |f| dm < /`A f+ dm + ,/`B f - dm
both of which are finite, where A={x: f(x)>0} and B={x:f(x)<0}.
So, if we can show |f| is integrable, then so is f.
However, by T2.2.5, we can approximate any measurable function f with an increasing sequence {hn} of simple functions. So, hn<|f|<g implies {hn} < g. Since g is integrable, using HW 2.7.2 implies the {hn} are also integrable. Apply the LBCT to complete.
Theorem 2.10.4: (Lebesque Monotone Convergence Theorem = LMCT) Let {fn} be a monotone increasing sequence of non-negative integrable functions and let f(x) = lim fn(x). Then, limn-->oo ,/` fn dm = ,/` f dm .
Pf: If f is integrable, then fn < f easily implies ,/` fn dm < ,/` f dm, for all n and so limn-->oo ,/` fn dm < ,/` f dm.
If f is not integrable, then is infinite and so the inequality still holds. It remains to show that equality holds.
If limn-->oo ,/` fn dm is infinite, then equality will hold. So consider the case when this limit is finite. Show C1 and C2 hold. By hypothesis C2 holds. Hence, we must show the the sequence fn is Cauchy in the mean. Consequently, consider fn and fm, where we'll assume that m>n.
By monotonicity, fm > fn, and so fm - fn > 0.
Hence, ,/` | fm - fn | dm = ,/` ( fm - fn ) dm = ,/` fm dm - ,/` fn dm which approaches zero as m and n get large.
Thus, C1 and C2 hold. By Theorem 2.8.2, f is integrable and the result holds.
Theorem 2.10.5: (Fatou's Lemma) Let {fn} be a sequence of nonnegative integrable functions and let f(x) = lim inf fn(x). Then lim infn-->oo ,/` fn dm > ,/` f dm. Thus, if lim infn-->oo ,/` fn dm is finite, f is integrable.
Pf: If lim infn-->oo ,/` fn dm is infinite, the inequality is obviously true. So, suppose it is finite.
Set gn = infj>n fj(x). Then, {gn} is a monotone-increasing sequence of non-negative integrable functions and gn < fn. Hence, lim ,/` gn dm < lim inf ,/` fn dm which is finite.
Since lim gn = lim fn = f, apply the LMCT to see that f is integrable and limn-->oo ,/` gn dm = ,/` f dm. Combine with the above inequality.
HOMEWORK: page 59 #3, 12