Chapter 5 The Space Meas(X; A) In this chapter, we will study the properties of the space of finite signed measures on a measurable space (X; A). In particular we will show that with respect to a very natural norm that this space is in fact a Banach space. We will then investigate the nature of this space when X = R and A = B(R). 5.1 The Space Meas(X; A) Definition 5.1.1. Let (X; A) be a measurable space. We let Meas(X; A) = fµ j µ is a finite signed measure on (X; A)g It is clear that Meas(X; A) is a vector space over R. For each µ 2 Meas(X; A) define kµkmeas = jµj(X): Observe that if µ, ν 2 Meas(X; A) and if µ = µ+ − µ− and ν = ν+ − ν− are the respective Jordan decomposiitions, then kµ + νkmeas = jµ + νj(X) = (µ + ν)+(X) + (µ + ν)−(X) ≤ (µ+(X) + ν+(X)) + (µ−(X) + ν−(X)) = (µ+(X) + µ−(X)) + (ν+(X) + ν−(X)) = jµj(X) + jνj(X) = kµkmeas + kνkmeas From here it is easy to see that (Meas(X; A); k · kmeas) is a normed linear space. Theorem 5.1.2. Let (X; A) be a measurable space. Then (Meas(X; A); k · kmeas) is Banach space. Proof. Let fµng be a Cauchy sequence in (Meas(X; A); k · kmeas). Let E 2 A. Since jµn(E) − µm(E)j ≤ jµn − µmj(E) ≤ kµn − µmkmeas we see that fµn(E)g is also Cauchy in R. We define for E 2 A, µ(E) = lim µn(E): n!1 72 Note that the convergence above is actually uniform on A. It is easy to see that µ satisfies the first two conditions on a signed measure. To see that µ is actually a 1 S signed measure, let fEng ⊆ A be a sequence of pairwise disjoint measurable sets and let E = En. First n=1 observe that k k [ [ µ( En) = lim µj( En) j!1 n=1 n=1 k X = lim µj(En) j!1 n=1 k X = lim µj(En) j!1 n=1 k X = µ(En) n=1 for every k 2 N. In particular, µ is finitely additive. Let > 0. Choose an N 2 N so that if n ≥ N, then jµ(A) − µ (A)j < n 3 for every A 2 A. We can choose a K0 such that if k ≥ K0, then k k [ X jµ (E) − µ ( E )j = jµ (E) − µ (E )j < : N N n N N n 3 n=1 n=1 Now assume that k ≥ K0. Then k k X [ jµ(E) − µ(En)j = jµ(E) − µ( En)j n=1 n=1 k k k [ [ [ ≤ jµ(E) − µN (E)j + jµN (E) − µN ( En)j + jµN ( En) − µ( En)j n=1 n=1 n=1 < + + 3 3 3 = This shows that 1 X µ(E) = µ(En): n=1 5.2 Functions of Bounded Variation In this section, we introduct the space of functions of bounded variation. Definition 5.2.1. Let f :[a; b] ! R. For a partion Π = fa = x0 < x1 < x2 < ··· < xn = bg of [a; b], define n b X Va (f; Π) = jf(xi) − f(xi−1)j: i=1 73 b b Note that if Π1 ⊆ Π2, then Va (f; Π1) ≤ Va (f; Π2). The variation of f on [a; b] is b b Va (f) = sup Va (f; Π): Π b We say that f is of bounded variation on [a; b] if Va (f) < 1. Let BV [a; b] = ff :[a; b] ! R j f is of bounded variation on [a; b]g: Example 5.2.2. 1) Let f :[a; b] ! R be increasing. Then f 2 BV [a; b] with b Va (f) = f(b) − f(a): If f = f1 − f2 where both f1 and f2 are increasing then f 2 BV [a; b] with b Va (f) ≤ (f1(b) − f1(a)) + (f2(b) − f2(a)): 2) Let f :[a; b] ! R be differentiable with jf 0(x)j ≤ M for all x 2 [a; b]. Then f 2 BV ([a; b]) and since by the Mean Value Theorem given Π = fa = x0 < x1 < x2 < ··· < xn = bg, n n X X jf(xi) − f(xi−1)j ≤ M(xi − xi−1) = M(b − a) i=1 i=1 we have b Va (f) ≤ M(b − a): Remark 5.2.3. BV ([a; b]) is a vector space and it is easy to see that b kfkBV = Va (f) defines a semi-norm on BV ([a; b]). Moreover, kfkBV = 0 if and only if f is constant on [a; b]. If we let NBV ([a; b]) = ff 2 BV ([a; b])j f is right continuous on [a; b]:g; then b kfkNBV = Va (f) + jf(a)j makes NBV ([a; b]) into a normed linear space. Definition 5.2.4. Let f : R ! R. Define x Vf (x) = lim Va (f): a→−∞ Vf is called the variation function of f. We say that f is of bounded variation on if V ar(f) = lim Vf (x) < 1. R x!1 Let BV (R) = ff : R ! R j f is of bounded variation on Rg: If f 2 BV ( ), then lim Vf (x) is called the total variation of f. R x!1 Lemma 5.2.5. Let f 2 BV (R). Then n X Vf (x) = supf jf(xi) − f(xi−1)j j − 1 < x0 < x1 < x2 < ··· < xn = xg: i=1 74 Proof. Observe that if a < a ≤ x, then V x (f) ≤ V x (f). Moreover, if Π = fx < x < x < ··· < x = xg; 1 2 a2 a1 0 1 2 n then n X jf(x ) − f(x )j ≤ V x (f) i i−1 x0 i=1 hence n X supf jf(xi) − f(xi−1)j j − 1 < x0 < x1 < x2 < ··· < xn = xg ≤ Vf (x): i=1 Now let > 0 and choose x0 < x such that x Vf (x) ≤ V (f) + : x0 2 We can choose Π = fx0 < x1 < x2 < ··· < xn = xg so that n x X V (f) ≤ jf(xi) − f(xi−1)j + : x0 2 i=1 It follows that n X Vf (x) ≤ jf(xi) − f(xi−1)j + i=1 which establishes the lemma. The following statements are all easy to establish. In each case, the proof is left to the reader. Remark 5.2.6. 1) If F : R ! R is bounded and increasing, then F 2 BV (R) and VF (x) = F (x) − lim F (y): y→−∞ 2) If F; G 2 BV (R), then for each α; β 2 R, we have that αF + βG 2 BV (R) and V ar(αF + βG) ≤ jαjV ar(F ) + jβjV ar(G): In particular, kF kBV = V ar(F ) defines a seminorm on BV (R) with kF − GkBV = 0 If and only if F = G + c for some c 2 R. 3) If F 2 BV (R), then Fj[a;b] 2 BV ([a; b]) and b Va (f) = VF (b) − VF (a): Conversely, if F 2 BV ([a; b]) and 8 F (x) if x 2 [a; b]; <> G(x) = F (a) if x < a; :>F (b) if x > b; b then G 2 BV (R) and V ar(G) = Va (F ). As a consequence of the previous observation, much of what we say about BV (R) will carry over to BV ([a; b]). 4) Assume that F 2 BV (R), then F is bounded. 75 The following theorem characterizes functions of bounded variation. Theorem 5.2.7. Let F 2 BV (R), then both VF + F and VF − F are bounded and increasing. In particular, V + F V − F F = F − F 2 2 so F is the difference of two increasing functions. Conversely, assume that F : R ! R and G : R ! R are both increasing an bounded. Then if H = F − G we have that H 2 BV (R) and V ar(H) ≤ V ar(F ) + V ar(G) = ( lim F (x) − lim F (x)) + ( lim G(x) − lim G(x)): x!1 x→−∞ x!1 x→−∞ Proof. Let F 2 BV (R). Then the first statement follows immediately from the observation that if x < y, then y VF (y) − VF (x) = Vx (F ) ≥ jF (y) − F (x)j: The second statement follows from the fact that H = F − G 2 BV ([a; b]) for all a < b and b Va (H) ≤ (F (b) − F (a)) + (G(a) − G(b)): Definition 5.2.8. Let F 2 BV (R). The decomposition V + F V − F F = F − F 2 2 VF +F VF −F is called the Jordan Decomposition of F . 2 and 2 are called the positive and negative variation functions of F . Remark 5.2.9. Let F 2 BV (R), Then since F is the difference of two monotonic functions the following statements hold. 1) The set of discontinuities of F is countable. 2) We have that lim F (x) exists for each a 2 R. x!a+ 5.3 The Spaces Meas(R; B(R)) and Meas([a; b]; B([a; b])) In this section we will use what we know about functions of bounded variation to charaterize both Meas(R; B(R)) and Meas([a; b]; B([a; b])). We will need the following technical proposition. Proposition 5.3.1. If F 2 BV (R), then lim VF (x) = 0: x→−∞ Moreover, if F (x) is right continuous, then so is VF (x). 76 Proof.