# Appendix to Notes 8 (a)

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1 Appendix to Notes 8 () 13 Comprison of the Riemnn nd Lebesgue integrls. Recll Let f : [, b] R be bounded. Let D be prtition of [, b] such tht Let D = { = x 0 < x 1 <... < x n = b}. m i = inf{f(x) : x i 1 x x i } M i = sup{f(x) : x i 1 x x i }. Define the step functions (therefore, simple functions, since we hve ssumed tht f is bounded nd so M i < for ll i). nd So α D (x) = m i on [x i 1, x i ) for ll 1 i n, β D (x) = M i on [x i 1, x i ) for ll 1 i n. α D (x) f(x) β D (x) for ll x [, b]. Note tht if D D then α D (x) α D (x) nd β D (x) β D (x). Tht is, with finer prtition we get better pproximtions to f. With the nottion of integrls of simple functions we hve, with Lebesgue mesure on R, I(α D ) = n m i (x i x i 1 ) nd I(β D ) = i=1 n M i (x i x i 1 ), i=1 which re normlly known s L(D, f) nd U(D, f) in the theory of Riemnn integrtion. Then we obviously hve I(α D ) I(β D ) for ll D, nd if D D then I(α D ) I(α D ) nd I(β D ) I(β D ). Let f(x)dx = sup D I(α D ) nd Then f is Riemnn integrble if, nd only if, f(x)dx = inf D I(β D). 1

2 The common vlue is denoted by f(x)dx = R- f(x)dx. f(x)dx. Theorem 1 If f is Riemnn integrble on finite intervl [, b] then it is Lebesgue integrble with the sme vlue. Proof For ech n 1 we cn find, by the definition of supremum, prtition D α n such tht when, in prticulr, 0 f(x)dx I(α D α n ) < 1 n, I(α D α n ) f(x)dx s n. Similrly choose sequence of prtitions D β n such tht I(β D β n ) Set D n = Dn α Dn β then f(x)dx s n. nd Thus I(α D α n ) I(α Dn ) I(β D β n ) I(β Dn ) f(x)dx f(x)dx. I(α Dn ) f(x)dx nd I(β Dn ) f(x)dx s n. (1) 2

3 Replcing the sequence D 1, D 2, D 3,... by D 1, D 1 D 2, D 1 D 2 D 3,...nd relbeling we cn ssume tht D n D n+1 for ll n 1 while (1) still holds. Yet D n D n+1 mens tht α Dn (x) α Dn+1 (x) nd β Dn (x) β Dn+1 (x) for ll n nd x. In prticulr {α Dn } n 1 in n incresing sequence bounded bove by f. So lim n α Dn = g exists, nd stisfies g f. Similrly {β Dn } n 1 in n decresing sequence bounded below by f. So lim n β Dn = h exists, nd stisfies h f. Now {α Dn α D1 } n 1 is n incresing sequence of non-negtive simple F-mesurble functions tending to g α D1. So by Lebesgue s Monotone Convergence Theorem we hve L- (g α D1 )dµ = lim I(α Dn α D1 ) = lim I(α Dn ) I(α D1 ) = f(x)dx I(α D1 ). Since α D1 is simple function we hve L- α D 1 dµ = I(α D1 ) nd so L- gdµ = Similrly, by exmining β D1 β Dn L- hdµ = f(x)dx. (2) we find tht f(x)dx. So, if f is Riemnn integrble, tht is, f(x)dx = f(x)dx, then L (g h)dµ = 0. Yet h g 0, so h = g.e.(µ) on [, b]. But g f h nd so f = g.e.(µ) on [, b]. Hence L- fdµ = L- = = R- gdµ since f = g.e.(µ) on [, b] f(x)dx by (2) f(x)dx 3 since f is Riemnn integrble

4 Let (D) = mx 1 i n (x i x i 1 ). In Theorem 1 it is possible, by dding extr points to ech of the prtitions D n, to ssume tht (D n ) 0 s n. With the nottion nd ssumptions of Theorem 1 we cn prove Lemm 1 Assume tht (D n ) 0 s n. Then for ny x / k=1 D k we hve tht f is continuous t x if, nd only if, g(x) = f(x) = h(x). Proof Recll tht f is continuous t x if, nd only if, ε > 0 δ > 0 : y if y x < δ then f(y) f(x) < ε. (3) For ech k let I k be the subintervl of D k contining x. This is unique since x / k=1 D k. Write I k = [x i 1, x i ]. ( ) Let ε > 0 be given. From (D6) we find δ > 0. Since (D n ) 0 s n there exists N such (D n ) < δ for ll n N. Then l(i k ) < δ so if y I k we hve tht y x < δ. In which cse, from (D6) we get tht f(y) f(x) < ε. In turn this mens tht we hve both inf f(y) f(x) < ε nd sup f(y) f(x) < ε. I k Yet inf Ik f(y) nd sup Ik f(y) re the vlues of α k nd β k t x. Hence, combining the inequlities, β k (x) α k (x) < 2ε. Let k to deduce h(x) g(x) < 2ε. True for ll ε > 0 gives h(x) = g(x). ( ) Assume f is not continuous t x. So ε > 0 δ > 0 : y with y x < δ nd f(y) f(x) ε. (4) For ech k 1 choose δ k = min(x x i 1, x i x) so (x δ k, x + δ k ) I k. But then by (4) we cn find y k (x δ k, x+δ k ) such tht f(y k ) f(x) ε. In prticulr, I k sup f inf f ε, I k I k in which cse β k (x) α k (x) ε nd h(x) g(x) ε. Hence h(x) g(x). This leds to Theorem 2 Assume f : [, b] R is bounded. Then f is Riemnn integrble if, nd only if, f is continuous.e.(µ) on [, b]. Proof Choose sequence of prtitions, D k, s in Lemm 1. Then. 4

5 f is continuous.e.(µ) on [, b] iff f is continuous.e.(µ) outside D k on [, b] k iff g = h.e.(µ) on [, b] by Lemm 1, iff gdµ = hdµ iff iff f(x)dx = f(x)dx f is Riemnn integrble. Mesure preserving Trnsformtions These re specil cse of mesurble functions. Definition T : (, F, µ) (, F, µ) is mesure preserving trnsformtion if (i) T 1 A F for ll A F, (ii) µ(t 1 A) = µ(a) for ll A F. Definition Let A F. A point x A is sid to be recurrent with respect to A if there exists k 1 such tht T k x A. Theorem 3 Poincre s Recurrence Theorem Assume tht µ() <. Let F be the set of points of A which re not recurrent with respect to A. Then µ(f ) = 0. (So for every A F, lmost ll points of A re recurrent.) Proof Let x F. If there exists n 1 such tht T n x F then we hve both x F A nd T n x F A, i.e. x is recurrent point with respect to A, which contrdicts the definition of F. So T n x / F for ll n 1, tht is, T n F F = for ll n 1. Now, the preimge of n empty set is empty, so given ny k, n 1 we hve = T k n (T n F F ) = T k F T (n+k) F. Hence the sets F, T 1 F, T 2 F,... re pirwise disjoint. So 5

6 ( ) > µ() µ T k F = = µ(t k F ) k=0 µ(f ) k=0 k 0 by prt (ii) of definition. Hence µ(f ) = 0. We cn sk how long it tkes point x A to wnder bck into A. To this end define n A (x) = min{n 1 : T n x A}. Assume throughout the rest of this section tht µ() = 1. Definition A mesure preserving mp T : is Ergodic if either of the following hold. (i) Whenever A F is such tht µ(t 1 A A) = 0 then either µ(a) = 0 or 1. (ii) Whenever n integrble function f stisfies f(t x) = f(x) for.e. (µ) x in then f is constnt.e.(µ) on. The first definition here mens tht if T 1 A is lmost exctly A then either µ(a) = 0 or 1. So if 0 < µ(a) < 1 then T 1 A must differ quite lot from A. We sy tht T is mixing up the spce. We do not prove here tht (i) nd (ii) re equivlent. It cn be shown tht if T is ergodic then n A dµ = 1. A Since A dµ = µ(a) we hve, tht in some sense, n A is of size 1/µ(A). This is connected with the question of how often point x will wnder into the set A F. It will be shown below tht for B F nd T ergodic, 1 n 1 lim χ n n B (T k x) = µ(b).e. (µ), (5) k=0 where χ B is the chrcteristic function of the set B, i.e. χ B (x) = 1 if x B, 0 otherwise. So if µ(b) > 0 then lmost every point of wnders into B infinitely often. 6

7 Let S n (x) = #{1 i n : T i x B} nd A n (x) = S n (x)/n. It is not obvious tht the limit lim n A n (x) will exist. We will show tht it does by looking t the limsup nd liminf of the sequence {A n (x)}. So let A(x) = lim sup A n (x) which is trivilly 1. Lemm 2 Let { n } be sequence for which lim sup n <. Let {b n } be sequence for which lim b n = 0. Then lim sup( n + b n ) = lim sup n. Proof Write A = lim sup n. Let ε > 0 be given. There exists N 1 such tht ε < b n < ε for ll n N 1 nd there exists N 2 such tht A ε < sup r < A + ε r n for ll n N 2. Choose N = mx(n 1, N 2 ), so tht for ll n N we hve which gives the result. Lemm 3 A 2ε < sup( r + b r ) < A + 2ε r n Proof A(T x) = A(x). A n (T x) = 1 n 1 i n χ B (T i (T x)) = 1 n 2 i n+1 = A n (x) + χ B(x) χ B (T n+1 x) n nd n ppliction of Lemm 2 gives the result. Theorem 4 χ B (T i x) The limit lim A n (x) exists. n Proof For given x we follow the orbit of x, nmely x, T x, T 2 x, T 3 x,.... We cll the exponent n in T n x, the time. Let ε > 0 be given. It might be tht for ll sufficiently lrge n we hve A n (x) > A(x) ε which obviously shows tht the limit exists. Otherwise the sequence {m j } defined by 7

8 m j = min{m > m j 1 : A m (x) > A(x) ε} hs infinitely mny gps. Note tht this sequence depends on x. The question must be how lrge cn these gps be? Define τ(x) = min{n : A n (x) > A(x) ε}. We first ssume tht there exists M such tht τ(x) < M.e. (µ). Let S be the exceptionl set here. Assume there is gp fter m j so A mj (x) > A(x) ε but A mj +1(x) A(x) ε. Then if T m j x / S we know there exists n < M such tht A n (T m j x) > A(T m j x) ε = A(x) ε by the lemm bove. So we hve both nd m j +1 i m j +n 1 i m j χ B (T i x) > m j (A(x) ε) χ B (T i x) = 1 i n Adding these two inequlities gives tht is 1 i m j +n χ B (T i (T m j x)) > n(a(x) ε). χ B (T i x) > (m j + n)(a(x) ε), A mj +n(x) > A(x) ε. Thus m j+1 m j + n < m j + M. So if T m j x / S, the gp m j+1 m j is less thn M. So if x / T k S k=1 set of mesure zero, ll the gps re less thn M. Thus given x / k=1 T k S nd given N choose j (which will depend on x s well N since the sequence of m j depends on x) such tht m j N < m j+1, then for lmost ll x we hve 8

9 S N (x) S mj (x) > m j (A(x) ε) > (N M)(A(x) ε). Since this inequlity is true for lmost ll x we cn integrte to get (A(x) ε)dµ = 1 N M 1 N M = Nµ(B) N M, n N χ T n Bdµ µ(t n B) since T is mesure preserving. Let N nd then ε 0 to deduce A(x)dµ µ(b). (6) The ssumption bove concerning M my not hold. Define S M = {x : τ(x) > M}, the collection of which is nested sequence of sets, S 1 S 2 S 3... From the definition of lim sup A n (x) we know tht given ε > 0 nd x there exists (infinitely mny) N such tht A N (x) > A(x) ε in which cse τ(x) N nd so x / S N. Let S = M 1 S M, then we hve seen tht x / S for ll x, tht is, S =. Since the S N re nested we cn thus find n M such tht µ(s M ) < ε. Let B = B S M. We pply the rguments bove with B replced by B, so S n(x) = #{1 i n : T i x B }, A n(x) = S n(x)/n nd A (x) = lim sup A n(x). We follow the method bove looking t the gps in the sequence of {m j }. If fter some m j we hve gp then A m j (x) > A (x) ε but A m j +1(x) A (x) ε, in prticulr A m j +1(x) < A m j (x). If it were the cse tht T m j+1 x B then n N A m j +1(x) = S m j (x) + χ B (T mj+1 x) m j + 1 > S m j (x) = A m m j (x), j = S m j (x) + 1 m j + 1 hving used the observtion tht D > C > 0 implies C+1 > C. Hence D+1 D we must hve T mj+1 x / B. In prticulr T mj+1 x / S M, in which cse 9

10 τ(t mj+1 x) M nd so, s in the rgument bove, the gp fter m j is bounded by M. Thus following the rgument tht led to (6) will led to A (x)dµ µ(b ). But A n(x) A n (x) for ll n in which cse A (x) A(x), while µ(b ) < µ(b) + ε. Let ε 0 to deduce A(x)dµ µ(b). Hence this inequlity holds whtever we ssume bout M. A similr rgument gives µ(b) A(x)dµ where A(x) = lim inf A n (x). Hence A(x)dµ = A(x)dµ = µ(b) (7) nd so A(x) = A(x) for.e. (µ) x. Hence lim n A n (x) exists for lmost every (µ) x (nd is integrble). The remining question must be wht is the vlue of this limit? From the definition of ergodic bove the pproprite one for the present sitution sttes tht whenever n integrble function f stisfies f(t x) = f(x) for.e. (µ) x in then f is constnt.e.(µ) on. From the lemm bove we hve tht A(T x) = A(x) for ll x nd so for the points t which the limit exists we hve lim n A n (T x) = lim n A n (x), i.e. this holds.e.(µ) on. Hence if T is ergodic we hve tht lim n A n (x) = c, constnt,.e.(µ). For the vlue of c use (7) tht shows tht µ(b) = lim A n(x)dµ = cdµ = cµ() = c, n since µ() = 1. Hence.e.(µ). #{1 i n : T i x B} lim n n = µ(b) 10

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