FUNDAMNTALS OF RAL ANALYSIS by Doğan Çömez IV. DIFFRNTIATION AND SIGND MASURS IV.1. Differentiation of Monotonic Functions Question: Can we calculate f easily? More eplicitly, can we hope for a result like the Fundamental Theorem of Calculus for Riemann Integral? Answer: Yes, but with lots of work and patience. The following fact is well-known: Proposition. Let f : [a, b] R be a monotonic function. Then the set of discontinuities of f is at most countable (hence has measure 0. ercise. Prove that if be a countable subset of [a, b], then there eists f : (a, b R which is continuous only on (a, b \. Convention: Throughout this chapter, [a, b] will be an interval with < a < b <. Definition. A collection F of closed and bounded intervals is called a Vitali cover for a set R if for all, for all ɛ > 0 there eists I F such that I and l(i < ɛ. ample. Let R be a bounded set (say [c, d]. Then the collection F = [α, β] : c α < β d, α, β Q} is a Vitali Cover for. Net theorem is an important one! Theorem.1 (Vitali Covering Lemma Let R with m ( < and let F be a Vitali cover for. Then for all ɛ > 0 there eists a finite disjoint sub-collection I k } n F such that n m ( \ I k < ɛ. Comments: (a One can always find an open set O such that O and m(o <. (b Without loss of generality, we can assume that ( I F I O m Ik = m(i k m(o <. }} =l(i k Hence l(i k 0 if the collection F is infinite. (c For any interval I, Ĩ will be the closed interval having the same midpoint as I and has length 5 times the length of I. Proof (of Theorem.1 We will consider two cases. Case 1: There eists a disjoint sub-collection I k } n ( n I k. 1 F such that for all I F, I
2 In this case we claim that n I k (and hence the theorem is proven. If the claim is false, then there eists a point \ n I k. Then I k for all 1 k n. Since F is a Vitali cover, there eists I F such that I. By assumption I I k0 for some 1 k 0 k. Without loss of generality we can assume that I I k = for k k 0. First of all I I k0, hence l(i \ I k0 > 0. Let ɛ < 1l(I \ I 2 k, then there eists J F such that J and l(j < ɛ. Thus ( n J I k =. This contradicts our assumption, proving the claim. Case 2: No such finite family eists. That is, given any finite disjoint collection I k } n, there eists I F such that I ( n I k =. [In this case we will construct (inductively a countably disjoint sub-collection I k } F such that \ k I k is as small as we wish.] Let I 1 be arbitrary. By assumption there eists I (there may be others such that I I 1 =. Pick I 1 F such that I 2 I 1 = and l(i 2 > 1S 2 1 where S 1 = l(i. Now assume we have a disjoint collection I k } n F constructed in this manner. Then let S n = supl(i : I F, I I k =, k = 1,... n}. Pick I n+1 F such that I n+1 I k = for all k = 1,... n and l(i n+1 > 1S 2 n. So I k } n+1 is a disjoint family. Hence, by induction we have a countably disjoint family I k } F such that each I k O for all k, where O is an open set such that O and m(o <. Then l(i k 0 must hold. Claim: This collection has the property that for all n 1 \ n I k k=n+1 To prove this claim let n 1 and let \ n I k. Then there eists I F such that I and ( n I I k =. Then I I m for some m n + 1 must hold (otherwise, if I I m = for all m, then l(i m > 1 2 l(i for all m, which implies that l( m 0. So now we have l(i m 1 2 S m. Let c m be the center of I m. So Ĩ k. c m < 1 2 l(i m + l(i < 1 2 l(i n + 2l(I m. Hence must belong to Ĩm. Therefore k=n+1 Ĩk. This implies This now proves the above claim. \ n I k k=n+1 Ĩ k.
So given ɛ > 0, since I k } are disjoint, m( I k = l(i k < m(o. Hence, we can pick n such that k=n+1 l(i k < ɛ. Then 5 ( n m ( \ I k m Ĩ k l(ĩk < ɛ. k=n+1 k=n+1 Observe: if f : [a, b] R is continuous and differentiable on (a, b, by the Mean Value Theorem, if f ( > α > 0 on (c, d (a, b, then f(d f(c α(d c. ercise.1 Let f : [a, b] R be a continuous function which is differentiable on (a, b. Let = (a, b : f ( > α} and let Then the collection F is a Vitali Cover for. ( F = [c, d] (a, b : (* holds}. Definition. Let f : [a, b] R. The upper derivative of f at c (a, b is the value (possibly + defined by [ ] f( + t f( Df(c = lim. h 0 t sup 0< t h Similarly, the lower derivative of f at c (a, b is [ Df(c = lim h 0 inf 0< t h f( + t f( t If Df(c = Df(c and is <, then we say that f is differentiable at c, and call this common value as the derivative of f at c, which is denoted by f (c. ercise.2 (i Let f : [ 1, 1] R be given as f( = if 0 Q 0 if 0 R \ Q if < 0. 2 Find Df(0 and Df(0. (ii Construct f : [a, b] R continuous such that Df(c Df(c, for some c (a, b. (iii Construct f : [a, b] R such that Df(c = whereas Df(c = 0, for some c (a, b. Fact.1 Let f : [a, b] R be increasing. Then for all α > 0 (a m ( (a, b : Df( α} 1 [f(b f(a] and α (b m ( (a, b : Df( = + } = 0. Proof. (a Let α, ɛ > 0 be arbitrary. Let α = (a, b : Df( α}. Choose 0 < β < α, and let F = [c, d] (a, b : f(d f(c β(d c}. Then F is a Vitali Cover for α. Then by the Vitali Covering Lemma, ] [ n n ] ( n m ( α = m ([ α \ [c k, d k ] [c k, d k ] < ɛ + m [c k, d k ] ]. 3
4 for some disjoint sub-collection [c k, d k ]} n F. Hence ( n ɛ + m [c k, d k ] ɛ + (d k c k < ɛ + 1 β [f(d k f(c k ] ɛ + 1 β [f(b f(a] m ( α < 1 [f(b f(a]. α For = (a, b : Df( = + }, N for any N 1. Therefore, m ( m ( N 1 [f(d f(c] 0 as N, implying (b. N Corollary. If f is monotonic and differentiable, then f < almost everywhere. Question: Which functions are differentiable? (In particular, are monotonic functions differentiable? Theorem.2 (Lebesgue Differentiation Theorem Let f : (a, b R be monotonic. Then f is differentiable almost everywhere on (a, b. Proof. We will assume that f is increasing on (a, b. Let α,β = (a, b : Df( > β > α > Df(}. Note that f fails to be differentiable on points at which Df( < Df(. Then the family α,β } α,β Q is a countable collection of sets on which f is not differentiable. Hence, if = α,β, α,β Q α<β we need to show that m ( = 0, which will be the case if we show that m ( α,β = 0 for each α and β. So, fi α < β and let F = α,β. Given ɛ > 0, pick O open such that F O (a, b and m(o < m (F + ɛ. Let F = [c, d] : f(d f(c < α(d c}. Then F is a Vitali Cover for F by ercise.1. Hence by the Vitali Covering Lemma there eists a finite collection [c k, d k ]} n F such that n m (F \ [c k, d k ] < ɛ. Then [f(d k f(c k ] < α ( n (d k c k = αm [c k, d k ] αm(o < α(m (F + ɛ. Now focus on the interval (c k, d k. f (ck,d k is increasing. By Fact.1, m (F (c k, d k < 1 β [f(d k f(c k ]. Thus β m (F (c k, d k < α(m (F + ɛ.
5 Now we have Hence where c = ( n ( m (F m (F [c k, d k ] + m F \ n [c k, d k ] m (F (c k, d k + ɛ α β (m (F + ɛ. m (F α β m (F + α β ɛ m (F < α. Since ɛ is arbitrary, it follows that β α m (F = 0. ( ( α β ɛ 1 1 α m (F < cɛ β Corollary. Let f : [a, b] R be increasing. Then f is integrable over [a.b] with f f(b f(a. Remark. The inequality above can be strict. ample. Let f : [0, 1] R be the Cantor-Lebesgue function. Then f is differentiable almost everywhere on [0, 1] with f = 0 almost everywhere. However, f(1 f(0 = 1 and f = 0 f(1 f(0. Proof (of Corollary. Define, for n 1, a sequence of functions g n : [a, b] R by g n ( = f( + 1 f( n 1. n Then each g n is a measurable function. Furthermore, since f is differentiable almost everywhere, g n f a.e. on [a, b]. Hence f is measurable (we will assume that f( = f(b for b. Then by Fatou s Lemma f lim inf g n. Now [ ] [ ] g n = n f( + 1/n f( = n f( f( [a+ 1 n,b+ 1 n ] [ ] = n f( f( = n [f(b 1n ] f( [b,b+ 1 n ] [a,a+ 1 n ] [a,a+ 1 n [ ] n f(b 1 n f(a 1 ] = f(b f(a, since f(a f(, a. n Remark. If f : (a, b R is increasing, then, following the same line of proof as above, we have f sup f( inf f(. (a,b (a,b (a,b ercises
6 1. Let R be a bounded set (say [a, b]. If V = [α, β] : a α < β b, α, β Q}, show that V is a Vitali cover for. 2. Let f : [a, b] R be a continuous function which is differentiable on (a, b, and let = (a, b : f ( > α}. If V = [c, d] (a, b : f(d f(c α(d c}, show that V is a Vitali cover for. 3. a Find a function f : [a, b] R such that Df(c =, and Df(c = 0 for some c (a, b. b Determine Df(0 and Df(0 for sin( 1 if 0, f( = 0 if = 0. c Determine Df( and Df( for the function f : [0, 1] R given by f( = χ Q (. 4. Consider functions f, g : [ 1, 1] R given by f( = 2 cos( 1 if 0, 2 0 if = 0, g( = Find Df(0, Df(0, Dg(0 and Dg(0. 2 cos( 1 if 0, 0 if = 0. IV.2. Functions of Bounded Variation and Absolute Continuity In this section and the net, we will improve the assertion of the Corollary above. Since any monotone function is measurable, our aim is to generalize Lebesgue Differentiation Theorem (and hence the Corollary above to largest possible class of measurable functions. Definition. Let f : [a, b] R be a function. Let P = [ i, i+1 ]} n i=1 be a partition of [a, b]. Then the quantity Va b (f, P := f( i f( i+1 is called as the variation of f over [a, b] with respect to P. The quantity i=1 T b a(f := supv b a (f, P : P is a partition of [a, b]} is called the total variation of f. If T b a(f <, then f is called a function of bounded variation. ample. (1 Any increasing (decreasing function is of bounded variation. (2 Lipschitz continuous functions ( c > 0, such that f( f(y c y,, [a, b] are of bounded variation. (3 Let f be defined as sin( 1 if (0, 1] f( = 0 if = 0. Then f is not of bounded variation. Fact.2 Let f : [a, b] R. Then
7 (a V b a (f, P V b a (f, P where P is a refinement of P. (b For all a c b, T b a(f = T c a(f + T b c (f. (c The function T a (f is a non-decreasing function. (d The function f( + T a (f is a non-decreasing function. Proof. We will leave (a and (b as an eercise. (c Assume < y. Then we have T y a (f = T a (f + T y c (f by (b. Thus, by the definition, T y (f 0. Hence T y a (f T a (f. (d Assume that < y. Then [f(y + T y a (f] [f( + T a (f] = [f(y f(] + [T y a (f T a (f] = [f(y f(] + T y (f. We know this is nonnegative since T y (f f(y f( = V b a (f, P where P = [, y]}. Theorem.3 (Jordan Decomposition of a function A function f : [a, b] R is of bounded variation if and only if f = g h, where g and h are increasing on [a, b]. Proof. ( Write f( = f(+t a (f T a (f. Then, letting g = f(+t a (f and h = T a (f, the assertion follows from Fact.2(c and (d. ( Let P = a = 0 < 1 < < n 1 < n = b} be a partition of [a, b]. Then Va b (f, P = f( k f( k 1 = [g( k g( k 1 ] [h( k h( k 1 ] g( k g( k 1 + h( k h( k 1 = [g(b g(a] + [h(b h(a] <. Corollary. Any function of bounded variation is differentiable almost everywhere (and f is integrable. Proof. By Theorem.3, write f = g h, where g, h : [a, b] R are monotonic. Apply Lebesgue Differentiation Theorem. Question: Is it true that f = f(b f(a? Note that the Cantor-Lebesgue function is of bounded variation by the previous theorem. Also ϕ = 0 almost everywhere. Hence, the Cantor-Lebesgue function, although is of bounded variation, fails to satisfy the Fundamental Theorem of Calculus since ϕ(1 ϕ(0 = 1 0 = f. Definition. A function f : [a, b] : R is called absolutely continuous on [a, b] if for all ɛ > 0 there eists δ > 0 such that for any finite disjoint collection [a k, b k ]} n of subintervals of [a, b] b k a k < δ f(b k f(a k < ɛ. Remark. The Cantor-Lebesgue function is not absolutely continuous. [0,1]
8 amples. 1. Lipschitz continuous functions are absolutely continuous on any compact interval. 2. is absolutely continuous on [0, 1]. Question: Are absolutely continuous functions differentiable? Answer YS!! Fact.3 if f : [a, b] R is absolutely continuous, then its of bounded variation (hence differentiable almost everywhere. Proof. Let f : [a, b] R be absolutely continuous, let ɛ = 1 and δ > 0 be the appropriate real number satisfying the definition. Pick N Z + such that b a < δ. Now let N a = 0 < 1 < < n 1 < n = b be such that k k 1 < δ, 1 k n. Then take T k+1 k (f = sup V k+1 k (f, P. Now, for any partition P of [ k, k+1 ], say P = k = t 0 < t 1 < < t m 1 < t m = k+1 }, V k+1 k (f, P = m f(t i f(t i 1. Since t i t i 1 k k 1 < δ and since f is absolutely continuous,, f(t i f(t i 1 < 1 = ɛ. n 1 So T k+1 k (f 1. Therefore, Ta(f b = T k+1 k (f = n <. i=1 k=0 i=1 Theorem.4 Let f : [a, b] R be absolutely continuous. If f ( = 0 almost everywhere on [a, b], then f is constant. Proof. It is enough to prove that f(t = f(a for all t [a, b]. So given ɛ > 0, let δ > 0 be chosen to satisfy the absolute continuity of f. Let = (a, t : f ( = 0}. Then F and m( = t a. Given any, we can find h > 0 such that f( + h f( < ɛh t a (since f ( = 0. Then the collection of intervals of the form [, + h ] forms a Vitali Cover. So, by Vitali Covering Lemma, there eists a finite disjoint sub-collection, say [ i, i+1 ]} n i=1 such that ( n m \ [ i, i + h i ] < δ. i=1 Hence, letting F = \ n i=1 [ i, i + h i ] and let a = 0, t = n+1, and h 0 = 0, h i = h i, we have F = [a, 1 ] [ 1 + h 1, 2 ] [ 2 + h 2, 3 ] [ n 1 + h n 1, n ] [ n + h n, t].
9 Since m(f < δ, by absolute continuity of f we have f( k+1 f( k + h k < ɛ Then k=0 f( 1 f(a + f( k f( k 1 + f(t f( n + h n. f(t f(a = f(a + f( 1 f( 1 + f( 1 + h 1 f( 1 + h 1 f( 2 + f( 2 +... f( n + f( n f( n + h n + f( n + h n f(t f( 1 f(a + f( k f( k 1 + h k 1 + f(t f( n + h n + f( k + h k f( k < ɛ + Since ɛ > 0 is arbitrary, we have f(t = f(a for all t [a, b]. ɛh k t a < 2ɛ. ercises 1. Consider function sin( 1 if 0, f( = 0 if = 0. a Determine if f a function of bounded variation on [0, 1]. b Show that f uniformly continuous on on [0, 1]. 2. Consider functions f, g : [ 1, 1] R given by f( = 2 cos( 1 if 0, 2 0 if = 0, g( = 2 cos( 1 if 0, 0 if = 0. Determine if f and g are of bounded variation on [ 1, 1]. 3. Let f, g : [0, 1] R be given by f( = 2 sin( 1 if 0, 0 if = 0, g( = Determine if f and g are absolutely continuous. sin( 1 2 if 0, 0 if = 0. 4. For any R, let BV ( denote the collection of all R-valued functions of bounded variation on. Show that if f, g BV ( and α R, then f +g, αf BV (. Is fg BV (? (Prove or provide a countereample.
10 5. Let f BV ([a, b]. Show that there eists a countable family P n } n 1 of partitions of [a, b] such that lim Va b (f, P n = T b n a(f. 6. Show that for any Lebesgue integrable function f : [a, b] R, the function g( = f(ydy is absolutely continuous. [a, 7. Let f, g : [0, 1] R be functions be given by f( = 2 sin( 1 if > 0 0 if = 0, Show that g( =. a f and g are absolutely continuous on [0, 1]. b f g is absolutely continuous on [0, 1]. c g f is not absolutely continuous on [0, 1]. IV.3. Fundamental Theorem for Lebesgue Integration Definition. Let f : [a, b] R be integrable. The indefinite integral of f is defined as F ( = f + c, [a, b]. [a,] Recall: If f is Riemann integrable with F ( = f(, then a f d = [F ( F (a]. Question: How nice is F ( if f is Lebesgue integrable? Fact.4 Let f : [a, b] R be integrable. Then the indefinite integral F of f is uniformly continuous on [a, b] and is of bounded variation (hence differentiable almost everywhere. Proof. Observe if y >, then F (y F ( f. Since f is integrable, given ɛ > 0, [,y] there eists δ > 0 such that whenever A F with m(a < δ, then f < ɛ. A So pick and y such that y < δ. Then m([, y] < δ. So f < ɛ and F is uniformly [,y] continuous. Let a = 0 < 1 < < n 1 < n = b be a partition of [a, b], call it P. Then F ( i F ( i 1 f = f <. i=1 i=1 [ i 1, i ] So V b a (f, P <. Therefore T b a(f < ; and hence, F is of bounded variation. Fact.5 Let f : [a, b] R be integrable. If [a,] f = 0 for all [a, b], then f = 0 almost everywhere.
Proof. Observe that the hypothesis states that f = 0 for any open or closed [a, b]. Assume that assertion is false. That is there eists F F, F [a, b] such that m(f > 0 and f > 0 (or f < 0 on F. Then there eists an open set O [a, b] \ F with m(o > b a. Then [a, b] \ O is closed and [a, b] \ O F with m([a, b] \ O > 0. Further f > 0 on [a, b] \ O. Hence f > 0. This is a contradiction. \O Theorem.5 Let f : [a, b] R be integrable and F ( = F (a + [a,] f. Then F = f almost everywhere on [a, b]. Proof. Assume that f is bounded, say f M. F is of bounded variation (Fact.4; and hence, differentiable a.e., implying that F ( eists. Define g n : [a, b] R by Then g n ( = F ( + 1 F ( n 1. n g n ( = n F ( + 1/n F ( = n [,+ 1 n ] f n f M. [,+ 1 n ] Also g n ( F ( almost everywhere. Hence the Bounded Convergence Theorem now implies g n = F = F (b F (a = f, lim n where the the last equality follows from the definition of F and the second equality follows from the fact that F is Riemann-integrable (since it is continuous by Fact.4. Therefore, (F f = 0; and hence, by Fact.5, F = f almost everywhere. When f is an arbitrary integrable function, do the same for f m = f M, for any m Z +. Theorem.6 (F.T.C. for Lebesgue Integral Let f : [a, b] R be integrable. The function f is absolutely continuous on [a, b] if and only if f is integrable where f( = f(a + f [a, b]. [a,] Proof. ( Since f is integrable, by Corollary.2 of DCT, given ɛ > 0, there eists δ > 0 such that whenever A F with m(a < δ, we have f dm < ɛ. Let [ A i, i+1 ]} n i=0 be a collection of sub-intervals of [a, b]. Then, from the assumption we have f(y f( = f for any [,y], y [a, b]; and hence, f( i+1 f( i = f f f, i=0 i=0 [ i, i+1 ] i=0 [ i, i+1 ] where = n i=0 [ i, i+1 ]. Therefore, for any collection [ i, i+1 ]} n i=0 such that i+1 i < δ, we have i=0 f( i+1 f( i i=0 Hence, f must be absolutely continuous. f < ɛ. 11
12 ( Assume that f is absolutely continuous. differentiable and f is integrable. So let g( = [a,] Then, by the Corollary of Theorem.3, f is Then f = g almost everywhere by Theorem.5. Now look at the function h := f g. Note that h is absolutely continuous and h = 0 almost everywhere. Thus h must be constant by Theorem.4. So f( = h( + g( = h(a + g( = h(a + f = f(a + f. f. [a,] [a,] Corollary.1 (Lebesgue Decomposition of a function Let f : [a, b] R be of bounded variation. Then there eists functions of bounded variation g, h : [a, b] R such that g is absolutely continuous on [a, b], h = 0 almost everywhere on [a, b] and f = g h almost everywhere. Proof. Set g( = f and h = f g. [a,] Corollary.2 very function of bounded variation is the indefinite integral of its derivative. Proof. ercise. Question. Can we claim that FTC holds in the setting of Corollary.1 and Corollary.2? ercise. Show that f : [a, b] R is absolutely continuous if and only if f = T b a(f. IV.4 Signed Measures and Radon-Nikodým Theorem We will let (X, A be a measurable space and (X, A, µ be a measure space. Recall that if f : X R + is (µ measurable and f 0, then the set function ν = f dµ, A, defines a measure on A. Question: What if non-negativity of f is dropped? Then f = f + f where f + and f are non-negative. So ν( = f dµ = f + dµ f dµ. Note that now ν is not a measure anymore; however, it is still a nice set function since: ν( = 0 ν( is well defined (as an etended real number. If n } n A is a disjoint collection, then ( ν n = f = f + n=1 n n n n = f + f = n n n=1 n=1 n n f n=1 n f = ν( n. Definition. Let (X, A be a measurable space. A set function ν : A R # such that n=1
13 ν( = 0 ν takes at most one of either + or as a value. If n } A is a disjoint collection of measurable sets, then ( ν n = ν( n n=1 n=1 and the series is absolutely convergent if ν( n n <, is called a signed measure (on A or X. Question: Can we characterize signed measures in terms of simple (ordinary measures? Definition. Let (X, A be a measurable space and ν a signed measure on A. A set A A is called ν-positive (ν-negative, ν-null if for every measurable A, ν( 0 (ν( 0, ν( = 0, resp.. Remark. (1 If A B and ν(b <, then ν(a <. (2 very ν-null set has ν-measure zero; however a set with ν-measure zero may not be ν-null. (ercise: Find an eample of such a set. Fact.6 Let ν be a signed measure on (X, A. Then (a Any measurable subset of a ν-positive set is ν-positive. (b Any countable union of ν-positive sets is a ν-positive set. Proof. ercise. Fact.7 (Continuity of signed measures Let ν be a signed measure on (X, A and A k } k A be a collection of subsets of X. (a If A k A k+1 for all k 1, then ν( 1 A k = lim k ν(a k. (b If A k A k+1 for all k 1, then ν( 1 A k = lim k ν(a k. Proof. ercise. [Hint: Mimic the proof of Theorem.3 in Chapter II. Net, we will show that given a measurable space (X, A and a signed measure on it, one can decompose X into a disjoint union of a positive and a negative set w.r.t. the signed measure. For, we need the following statement. Lemma. If ν is a signed measure on (X, A and A such that 0 < ν( <, then there eists a measurable set A which is positive with ν(a > 0. Proof. See p: 344 of Royden. Theorem.7 (Hahn Decomposition If ν is a signed measure on (X, A, then there eist a positive set P and a negative set N such that P N = X and P N =. Proof. By the definition of signed measure we can assume that ν does not assume + (for otherwise, consider ν. Define λ = supν( : is a positive set}.
14 Since is a positive set, λ 0. Let A k } be a countable collection of positive sets such that λ = lim k ν(a k. If P = k A k, then P is a positive set with ν(p = λ. Since ν does not assume +, it d follows that λ <. Net, let N = X \ P ; we will prove that N is a negative set. For, assume that N is not a negative set. Hence, B N with positive measure. Then, by Lemma above, there is a positive set A B with positive measure. It follows that P A is a positive set and ν(p A = ν(p + ν(a > λ, which is a contradiction. Remark. Hahn decomposition is not unique; however, it is unique up to a set of measure zero. Observe that, given a measure space (X, A, µ, one can find functions f, g : X R with A = supp(f, B = supp(g, where A B =, and A B = X. Hence, if ν 1 (. = fdµ and ν 2 (. = gdµ, then A is ν 2 -null and B is ν 1 -null; i.e., ν 1 and ν 2 live on disjoint sets. More generally, Definition. Let (X, A be a measurable space. The signed measures ν 1 and ν 2 on A are called singular (or ν 1 (ν 2 is singular with respect to ν 2 (ν 1 if there eists 1, 2 A such that 1 2 =, 1 2 = X and i is ν i -null, i = 1, 2. Notation. ν 1 ν 2. Theorem.8 (Jordan Decomposition of a measure If ν is a signed measure on a measurable space (X, A, then there eist unique positive measures ν + and ν such that ν = ν + ν and ν + ν. Proof. The eistence follows from Hahn decomposition by letting ν + ( = ν( P and ν ( = ν( N for any A. Then ν = ν + ν and ν + ν. For uniqueness, if ν = µ + µ and µ + µ is another decomposition, then pick A, B A with A B =, A B = X and µ + (B = 0, µ (A = 0. Then this gives another Hahn decomposition for ν, and consequently, P A will be ν-null. Hence, for any A, i.e., µ + = ν +. Similarly, µ = ν. µ + ( = µ + ( A = ν( A = ν( P = ν + (, Observe the similarity between the decomposition in Theorem.8 and the Jordan decomposition of a function of bounded variation (into difference of two monotonic functions. Hence, the following definition is appropriate. Definition. The measures ν + and ν are called the positive and negative variations of ν; ν = ν + ν is called the Jordan decomposition of ν, and for the measure ν defined by ν = ν + + ν, ν := ν (X is called the total variation of ν. ample. Let f : R R be Lebesgue integrable function, and let ν( = fdm for any F. Then ν is a signed measure on (R, F. Let P = R : f( 0}, N = R : f( < 0} and define, for F, ν + ( = fdm and ν ( = fdm. P N Then P, N} is a Hahn decomposition of R w.r.t. ν, and ν = ν + ν is Jordan decomposition of ν. ercises. 1 ν (X = sup n ν( k : k } n 1 A disjoint collection}.
15 2 A is ν-null if and only if ν ( = 0. 3 ν µ if and only if ν µ if and only if ν + µ and ν µ. 4 If ν omits the value, then ν + (X = ν(p ; hence ν + is a finite measure and ν is bounded above by ν + (X. The same is also valid if ν omits the value (with appropriate changes. Let f = χ P χ N, where X = P N is the Hahn decomposition for ν, and let µ = ν, then ν( = fdµ for any A. Generalizing this idea, one defines integration w.r.t. a signed measure ν by fdν = fdν + fdν for any ν + and ν -integrable function f. In regards to differentiation w.r.t. a signed measure, we will follow similar steps as in the differentiation w.r.t. a measure we did above. Definition. Let ν be a signed measure and µ be a measure on A. ν is called absolutely continuous w.r.t. µ, denoted by ν << µ, if ν( = 0 for all A with µ( = 0. ample. Let µ be a measure on a measurable space (X, A and f be an integrable function. Then the signed measure ν defined by ν( = fdµ, for all A, is absolutely continuous w.r.t. µ. ercises. 1. ν << µ if and only if ν << µ if and only if ν + << µ and ν << µ. 2. If λ 1 µ and λ 2 µ, then λ 1 + λ 2 µ. 3. If ρ 1 << µ and ρ 2 << µ, then ρ 1 + ρ 2 << µ. Theorem.9 Let ν be a finite signed measure and µ be a measure on (X, A. Then ν << µ if and only if for every ɛ > 0 there eists δ > 0 such that ν( < ɛ whenever µ( < δ. Proof. By the eercise above, it is enough to prove the statement for ν = ν is positive. The direction ( is obvious from the definitions. So, we ll prove the converse by contrapositive. For, if ɛ δ condition is not satisfied, then there eists an ɛ > 0 such that for all n N n A with µ( n < 1 and ν( 2 n n ɛ. Letting F k = n=k and F =, we have µ(f k < n k 2 n = 2 1 k ; hence, µ(f = 0. On the other hand, ν(f k ɛ for all k 1; hence, ν(f = lim k ν(f k ɛ. But this contradicts with the fact that ν << µ. Remarks. 1. Compare this theorem with Theorem.12 in Section III.3. 2. Finiteness of ν is essential. For otherwise, if X = (0, 1, A = F, µ = m, and ν is defined by ν(a = 1dm, then the assertion of Theorem.9 does not hold. Another countereample A is given by X = N, A = P(N, µ is defined by µ( = 1 n, N, and ν =counting 2 n measure. Notation. If ν( = fdµ, then this relationship can also be epressed by dν = fdµ. Lemma. Let ν and µ be finite measures on (X, A. Then, either ν µ, or there eists ɛ > 0 and A such that µ( > 0 and is a positive set for the measure ν ɛµ. Proof. ercise. [Hint: Use Hahn decomposition for ν n 1 µ.]
16 Theorem.10 (Radon-Nikodým Let (X, A, µ be a finite measure space and ν : A R be a finite measure with ν << µ. Then there eists a non-negative A-measurable function, unique µ-a.e., such that ν( = fdµ, A. Proof. (Uniqueness If there eists non-negative A-measurable functions f 1 and f 2 such that ν( = f idµ, A, i = 1, 2, then 0 = (f 1 f 2 dµ A. Hence, f 1 = f 2 µ-a.e. on X. (istence If ν( = 0 A, the assertion holds with f 0. Hence, assume that ν does not vanish on all of A. Define G = g : X [0, ] : g ν(, A}. Since 0 G, we see that G. Also, if h, g G, then by letting A = : h( > g(}, for any A, it follows that (h gdµ = hdµ + gdµ ν( A + ν( \ A = ν( A \A implying that h g G. Now, define α = sup g G gdµ. Since α ν(x <, we can pick g n } G such that g n α. Let f n = mag 1,..., g n }, then f n G for all n and f n f pointwise, for some non-negative A-measurable function f. Since f n dµ g n dµ, it follows that f n α; and hence, by MCT, f G with fdµ = α (in particular, f < µ-a.e.. Net, define a new measure η by η( = ν( fdµ for all A. The, by the Lemma above, we have two cases: (i η µ, or (ii ɛ > 0 & A A such that (η ɛµ(a 0 & µ(a > 0. In the first case, since η << µ by construction and ν << µ, it follows that η 0. Hence, ν( = fdµ for all A, proving the assertion. In the second case, we have ν(a fdµ ɛµ(a 0. Thus, ν(a fdµ + ɛµ(a = A A (f + ˆfdµ, where ˆf = ɛχ A A. Since µ(a > 0, we have ˆf 0 and ˆfdµ > 0. Thus, (f + A X ˆfdµ > fdµ = α, contradiction. Hence (ii is not possible, proving the theorem. X Corollary. a Theorem.10 is valid if both µ and ν are σ-finite. b Theorem.10 is valid if and ν is a σ-finite signed measure and µ is a σ-finite measure. Proof. a Let X = i X i and define µ i ( = µ(x i, A, and ν i ( = ν(x i, A. Apply Radon-Nikodým Theorem to µ i, ν i } to obtain f i. Let f = i f i. b Let ν = ν + ν and apply (a to µ, ν + } and µ, ν } to obtain f + and f, respectively. Let f = f + f. Notation. The function f in Radon-Nikodým Theorem and its Corollary is denoted by dν dµ and is called as the Radon-Nikodým derivative of ν with respect to µ. Theorem.11 (Lebesgue Decomposition of a measure Let ν be a σ-finite signed measure and µ be a σ-finite measure on a measurable space (X, A. Then there eists µ-a.e. unique σ-finite signed measures ν 0 and ν 1 on A such that ν 0 µ, ν 1 << µ and ν = ν 0 + ν 1. Proof. WLOG we can assume that both ν and µ are finite. First, let ν be a measure, and let λ = µ + ν, then µ << λ. Hence, by Radon-Nikodým Theorem, there eists a λ-a.e. unique
A-measurable function f such that µ( = fdλ for all A. Observe that f is also µ-a.e. and ν-a.e. A-measurable function. Thus, µ( = fdµ + fdν, A. ( 17 Let X 1 = : f( > 0}, X 0 = : f( = 0}, and define ν 0 = ν X0 have: and ν 1 = ν X1. Then we (i ν 0 and ν 1 are measures on A. (ii ν = ν 0 + ν 1. Also, by construction, X = X 0 X 1, X 0 X 1 =, and ν 0 (X 1 = ν 1 (X 0 = 0. Since by (* µ(x 0 = X 0 fdµ + X 0 fdν, it follows that ν 0 µ. If A A such that µ(a = 0, then fdµ = 0. So, by (*, we have A 0 = fdµ = fdµ + fdν = fdν = fdν + fdν. A A A A A X 0 A X 1 Since f = 0 on A X 0 and f > 0 on A X 1, we have A X 1 fdν = 0, so ν(a X 1 = 0. Thus ν 1 (A = 0, implying that ν 1 << µ. If ν is a signed measure, then ν = ν + ν where ν + and ν are measures. Then apply the above to ν + and ν, respectively, to obtain the decomposition ν ± = ν ± 0 + ν ± 1 with ν ± 0 µ, ν ± 1 << µ. Then, ν = ν 0 + ν 1, where ν 0 = ν + 0 + ν 0 and ν 1 = ν + 1 + ν 1. Remark. It is essential in Theorem.11 that ν and µ are σ-finite. [ercise: Provide countereamples when σ-finiteness in Theorem.11 is dropped.] ercises 1. Let µ be a probability measure on (R, B. Define a measure ν : B R by ν( = 1 if 0 and ν( = 0 otherwise (ν is called the point mass at 0. Find the Jordan decomposition of the signed measure λ = µ ν. 2. Let (X, A, µ be a measure space and ν 1 and ν 1 be signed measures on A. Prove the following. a If ν 1 µ and ν 2 µ, then ν 1 + ν 2 µ. b If ν 1 << µ and ν 2 << µ, then ν 1 + ν 2 << µ. c If ν 1 << µ, then ν 1 << µ; and conversely. d If ν 1 << µ and ν 2 µ, then ν 1 ν 2. e If ν 1 << µ and ν 1 µ, then ν 1 0. 3. Let (X, A, µ be a measure space, k } n A, and c k} n be a collection of real numbers. For A, define ν( = n c k.µ( k. Show that ν << µ and find dν. dµ 4. Let µ n } be a sequence of measures on a measurable space (X, A such that there is a constant C with µ n (X C for all n 1. Define ν : A R by µ n ( ν( =, A. 2 n Show that ν is a measure on A and that µ n << ν for each n 1. n=1
18 5. Consider the measure space ([0, 1], F [0,1], m and let ν be the counting measure on F [0,1]. Show that a m << ν, and b there is no function f : [0, 1] R for which m( = fdν for all F [0,1]. 6. Let a n } n be a fied sequence of real numbers and p n } n be a sequence of positive real numbers. Define a set function ν : F R by ν( = p n, F. a n a Show that ν is a σ-finite measure on F. b Find the Lebesgue decomposition of ν with respect to the Lebesgue measure on F. 7. Let f : R R be given by f( = 2 6 + 5, and define a set function ν : B R by ν( = f(dm, B. a Show that ν is a σ-finite measure on B. b Find the Hahn decomposition of R with respect to ν. c Find the Jordan decomposition of ν. d Find the Lebesgue decomposition of ν with respect to the Lebesgue measure on F. 8. Give an eample in which the assertion of the Radon-Nikodỳm theorem fails (that is, λ << µ but there is no measurable function f such that λ(a = fdµ for all measurable set A. A 9. Let (X, A, µ be a measure space and g be a non-negative measurable function on X. Let a measure λ : A R be defined as λ(a = gdµ, for all A A. Prove that if f : X R is a measurable function then fdλ = fgdµ, X A in the sense that if one integral eists, so does the other, and the two integrals are equal. 10. Consider the measure space ([0, 1], F [0,1], m [0,1]. Define a set function λ on all subintervals [a, b] [0, 1] by λ([a, b] = φ(b φ(a, where φ is the Cantor-Lebesgue function on [0, 1]. If µ is the measure obtained by restricting the outer measure λ to λ -measurable subsets of [0, 1], prove that µ m [0,1]. X