Riemann-Stieltjes integral

  • 微積分主要與兩個函數的幾何問題有關:
    • 找出函數曲線上某一點的切線(微分)
    • 找出函數曲線下的面積(積分)

更細分割(Finer partition)

  • P={x0,x1,x2,,xn}P = \lbrace x_0, x_1, x_2, \cdots, x_n \rbrace with a=x0<x1<x2<<xn=ba=x_0 < x_1 < x_2 < \cdots < x_n = b為閉區間[a,b][a,b]的分割。
  • PPP \subseteq P^{'},則PP^{'}為是比PP更細的分割。(因為PP^{'}的分割點更多)。
  • 而閉區間[a,b][a,b]間所有可能的分割為P[a,b]\mathbb{P}[a,b]
  • 分割的範數(norm)為分割中最長的區間,P=sup1knΔxk=sup1kn[xkxk1]\Vert P \Vert = \sup_{ 1 \leq k \leq n} \Delta x_k = \sup_{ 1 \leq k \leq n} [x_k - x_{k-1}] .
  • PPPP P \subseteq P^{'} \Rightarrow \Vert P^{'} \Vert \leq \Vert P \Vert . (反之不一定成立)
    • E.g. P={0,1,3,6,7,10}P= \{0,1,3,6,7, 10\}為閉區間[0,10][0,10]的分割,若P1={0,1,2,3,4,6,7,10}P_1 = \{0,1,2,3,4,6,7,10 \} ,則PP1 P \subseteq P_1
    • E.g. PP同上,所以P=3 \Vert P \Vert = 3,若P2={0,1,3,5,7,9,10}P_2 = \{ 0,1,3,5,7,9,10 \},則P2=2\Vert P_2 \Vert = 2,但是P2P_2不是PP的更細分割。

Riemann-Stieltjes積分定義

  • 這邊先討論函數可積分的情形下之性質,之後的章節再討論函數在何種條件下可積分。

  • PP為閉區間[a,b][a,b]的分割,令點tk[xk1,xk], k=1,2,,nt_k \in [x_{k-1}, x_k],\ k=1,2,\cdots, n,稱S(P,f,α)=k=1nf(tk)Δαk S(P,f,\alpha) = \sum_{k=1}^n f(t_k) \Delta \alpha_k.為Riemann-Stieltjes sum of ff w.r.t. to α\alpha

  • S(P,f)=k=1nf(tk)Δxk S(P,f) = \sum_{k=1}^n f(t_k) \Delta x_k.為 Riemann sum of ff

  • Δαk=α(xk)α(xk1)\Delta \alpha_k = \alpha(x_k) - \alpha(x_{k-1})是將xx經函數α\alpha轉換後再將區間切細之差值,若α(x)=x\alpha(x)=x時,即為單純的Riemann積分。

  • 定義:函數ff 相對於α\alpha在區間[a,b][a,b] Riemann-Stieltjes可積分,或是簡寫為fRS(α)[a,b]f \in RS(\alpha)[a,b]ϵ>0 partition Pϵ[a,b]S(P,f,α)M<ϵ,PϵP and tk[xk1,xk]. \forall \epsilon >0 \exists \text{ partition } P_{\epsilon}[a,b] \ni |S(P,f,\alpha) - M| <\epsilon, \forall P_\epsilon \subseteq P \text{ and } \forall t_k \in [x_{k-1}, x_k].
    • 分割Pϵ P_{\epsilon}某一種滿足條件 S(P,f,α)M=ϵ|S(P,f,\alpha) - M| = \epsilon 的分割,當分割PPPϵP_{\epsilon}的更細分割時,即PϵP P_\epsilon \subseteq PS(P,f,α)S(P,f,\alpha)之值會逐漸接近MM

    • 可積分的條件之一是在定義域越切越細時,值域之加總值為有限值M<M < \infty ,即全變差之值為有界。並非所有連續函數都可積分。

    • 當函數可積分時,其積分值MM唯一決定,且積分可寫成abfdα or abf(x)dα(x)\int_a^b f d\alpha \text{ or } \int_a^b f(x) d \alpha(x)ff為被積分函數(integrand),而α\alpha為積分函數(integrator)。

    • 積分值MM為函數f,αf, \alpha與積分區間a,ba,b的函數。

  • 定義:若 a<b abfdα=bafdα a < b \ \int_a^b f d \alpha = - \int_b^a f d \alpha if bafdα<\int_b^a f d \alpha < \infty
  • 定義:aafdα=0 \int_a^a f d \alpha = 0 .
  • 另一個常用的Riemann-Stieltjes積分定義如下:

  • 定義: ϵ>0 δ>0 PP[a,b],P<δ and tk[xk1,xk],S(P,f,α)M<ϵ\forall \epsilon > 0 \ \exists \delta > 0 \ \ni \forall P \in \mathbb{P}[a,b], \Vert P \Vert < \delta \text { and } \forall t_k \in [x_{k-1}, x_k], \vert S(P,f,\alpha) - M \vert < \epsilon .
    • P<δ \Vert P \Vert < \delta 即分割PP越切越細,子區間最大值小於δ\delta值時分割。

    • 不同定義的區別在於第一種定義的更細分割PP是分割PϵP_{\epsilon}的子集合,而第二種分割只要更細分割的最大子分割小於ϵ\epsilon即可,不必為原分割的子集合。

    • 兩種定義的想法都是當積分函數區間切的越細時,函數加總值會逐漸逼近積分值MM

積分線性性質

  • f, gRS(α)[a,b]f,\ g \in \mathbf{RS}(\alpha)[a,b],則c1f+c2gRS(α)[a,b], c1,c2Rab(c1f+c2g)dα=c1abfdα+c2abgdαc_1 f+ c_2 g \in \mathbf{RS}(\alpha)[a,b], \ \forall c_1, c_2 \in \mathbb{R} \Leftrightarrow \int_a^b (c_1 f + c_2 g) d \alpha = c_1 \int_a^b f d\alpha + c_2 \int_a^b g d \alpha
    .

    • Proof: let h=c1f+c2gh = c_1 f + c_2 g, and given PP[a,b]P \in \mathbb{P}[a,b] then

    • S(P,h,α)=k=1nh(tk)Δαk=c1k=1nf(tk)Δαk+c2k=1ng(tk)Δαk=c1S(P,f,α)+c2S(P,g,α). \begin{array}{rcl} S(P, h, \alpha) & = & \sum_{k=1}^n h(t_k) \Delta \alpha_k \\ & = & c_1 \sum_{k=1}^n f(t_k) \Delta \alpha_k + c2 \sum_{k=1}^n g(t_k) \Delta \alpha_k \\ & = & c_1 S(P,f,\alpha) + c_2 S(P,g,\alpha). \end{array}

      • fRS[a,b]ϵ>0 \because f \in \mathbf{RS}[a,b] \therefore \forall \epsilon > 0 choose PϵPS(P,f,α)abfdα<ϵP_{\epsilon}^{'} \subseteq P \ni \vert S(P,f,\alpha) - \int_a^b f d\alpha \vert < \epsilon, and
    • gRS[a,b]PϵPS(P,g,α)abgdα<ϵ \because g \in \mathbf{RS}[a,b] \therefore P_{\epsilon}^{''} \subseteq P \ni \vert S(P,g,\alpha) - \int_a^b g d\alpha \vert < \epsilon.

    • 此時PP是切的最細的分割,令Pϵ=PϵPϵP_{\epsilon} = P_{\epsilon}^{'} \cup P_{\epsilon}^{''}PϵP\because P_{\epsilon} \subseteq P,則

    • S(P,h,α)c1abfdαc2abgdαc1ϵ+c2ϵ \left \vert S(P, h, \alpha) - c_1 \int_a^b f d \alpha - c_2 \int_a^b g d \alpha \right \vert \leq | c_1 | \epsilon + | c_2 | \epsilon (QED).

  • fRS(α)[a,b]f \in \mathbf{RS}(\alpha)[a,b] and fRS(β)[a,b]f \in \mathbf{RS}(\beta)[a,b] then fRS(c1α+c2β)[a,b]abfd(c1α+c2β)=c1abfdα+c2abgdβ.f \in \mathbf{RS}(c_1 \alpha + c_2 \beta)[a,b] \Leftrightarrow \int_a^b f d (c_1 \alpha + c_2 \beta) = c_1 \int_a^b f d\alpha + c_2 \int_a^b g d \beta.
  • c[a,b]c \in [a,b], 當右式三個積分有兩個積分值存在時,則第三個積分值也存在 acfdα+cbfdα=abfdα<\Leftrightarrow \int_a^c f d \alpha + \int_c^b f d \alpha = \int_a^b f d \alpha < \infty
    • 或寫成 acfdα+cbfdα+cafdα=0\int_a^c f d \alpha + \int_c^b f d \alpha + \int_c^a f d \alpha = 0 .

    • 使用數學歸納法,可得出將閉區間切成有限個子區間時,均為可積

    • Proof: 令分割PP[a,b] and cPP \in \mathbb{P}[a,b] \text{ and } c \in P,令P=P[a,c], P=P[c,b]P^{'} = P \cap [a,c], \ P^{''} = P \cap [c,b],分別為區間[a,c], [c,b[a,c],\ [c,b上的分割.

    • S(P,f,α)=S(P,f,α)+S(P,f,α) \therefore S(P, f, \alpha) = S(P^{'}, f, \alpha) + S(P^{''}, f, \alpha) .

    • 假設acfdα and cbfdα\int_a^c f d \alpha \text{ and } \int_c^b f d \alpha 存在,則:

    • Given ϵ>0,Pϵ[a,c]S(P,f,α)acfdα<ϵ/2, PϵP \epsilon > 0, \exists P_{\epsilon}^{'}[a,c] \ni \left| S(P^{'}, f, \alpha) - \int_a^c f d\alpha \right| < \epsilon/2,\ P_{\epsilon}^{'} \subseteq P^{'}

    • Given ϵ>0,Pϵ[c,b]S(P,f,α)cbfdα<ϵ/2, PϵP \epsilon > 0, \exists P_{\epsilon}^{''}[c,b] \ni \left| S(P^{''}, f, \alpha) - \int_c^b f d\alpha \right| < \epsilon/2,\ P_{\epsilon}^{''} \subseteq P^{''}

    • Pϵ=PϵPϵP_{\epsilon} = P_{\epsilon}^{'} \cup P_{\epsilon}^{''} [a,b][a,b]的分割,則PϵPPϵP and PϵPP_{\epsilon} \subseteq P \Rightarrow P_{\epsilon}^{'} \subseteq P \text{ and } P_{\epsilon}^{''} \subseteq P

      +。S(P,f,α)acfdαcbfdα<ϵ\therefore \left| S(P,f,\alpha) - \int_a^c f d \alpha - \int_c^b f d \alpha \right| < \epsilon (QED).

分部積分(Integration by parts)

  • If fRS(α)[a,b] f \in \mathbf{RS}(\alpha)[a,b] then αRS(f)[a,b]\alpha \in \mathbf{RS}(f)[a,b], 即積分 abfdα\int_a^b f d\alpha 存在時,abαdf \int_a^b \alpha df也存在,反之亦然。 abf(x)dα(x)+abαdf(x)=f(b)α(b)f(a)α(a) \Leftrightarrow \int_a^b f(x) d \alpha(x) + \int_a^b \alpha d f(x) = f(b) \alpha(b) - f(a) \alpha(a ).
    • Proof:
    • 給定ϵ>0 \epsilon > 0 ,因為積分 abfdα\int_a^b f d \alpha存在,即存在分割PϵP[a,b]PϵP, S(P,f,α)abfdα<ϵP_{\epsilon} \in \mathbb{P}[a,b] \ni \forall P_{\epsilon} \subseteq P^{'}, \ \left| S(P^{'}, f, \alpha) - \int_a^b f d \alpha \right| < \epsilon .

    • 考慮積分abαdf\int_a^b \alpha df的Riemann-Stieltjes sum,給定PϵPP_{\epsilon} \subseteq P,

    • S(P,α,f)=k=1nα(tk)Δfk=k=1nα(tk)f(xk)k=1nα(tk)f(xk1)S(P, \alpha, f) = \sum_{k=1}^n \alpha(t_k) \Delta f_k = \sum_{k=1}^n \alpha(t_k) f(x_k) - \sum_{k=1}^n \alpha(t_k) f(x_{k-1}).

    • A=f(b)α(b)f(a)α(a)A = f(b)\alpha(b) - f(a)\alpha(a),可得等式A=k=1nf(xk)α(xk)k=1nf(xk1)α(xk1)A = \sum_{k=1}^n f(x_k) \alpha(x_k) - \sum_{k=1}^n f(x_{k-1}) \alpha(x_{k-1}).

    • AS(P,α,f)=k=1nf(xk)[α(xk)α(tk)]+k=1nf(xk1)[α(tk)α(xk1)]=S(P,f,α)\therefore A - S(P,\alpha, f) = \sum_{k=1}^n f(x_k)[\alpha(x_k) - \alpha(t_k)] + \sum_{k=1}^n f(x_{k-1})[\alpha(t_k) - \alpha(x_{k-1})] = S(P^{'}, f, \alpha),這邊的分割PP^{'}由是xk, tkx_k,\ t_k得來。

    • PPPϵP\therefore P \subseteq P^{'} \Rightarrow P_{\epsilon} \subseteq P^{'}.

    • 所以可得PϵP, AS(P,α,f)abfdα<ϵ \forall P_{\epsilon} \subseteq P, \ \left| A - S(P,\alpha, f) - \int_a^b f d \alpha \right| < \epsilon (QED).

變數變換(Change of variable)

  • 函數 fRS(α)[a,b]f \in \mathbf{RS}(\alpha)[a,b] 且 函數 g:[c,d]Rg:[c,d] \rightarrow \mathbb{R} 為嚴格單調連續函數 (也可不必為單調函數),定義在端點為 c,dc,d 的區間 SS

  • a=g(c), b=g(d)a = g(c), \ b = g(d) 且函數 h(x)=f(g(x))β(x)=α(g(x)), xS h(x) = f(g(x)) \beta(x) = \alpha(g(x)), \ \forall x \in S 為複合函數。

  • then hRS(β)[c,d] h \in \mathbf{RS}(\beta)[c,d] and abfdα=cdhdβ \int_a^b f d \alpha = \int_c^d h d \beta. i.e. g(c)g(d)f(t)dα(t)=cdf(g(x))d(α(g(x)))\int_{g(c)}^{g(d)} f(t) d\alpha(t) = \int_c^d f(g(x)) d (\alpha(g(x))) .
    • Proof:
    • 假設ggSS嚴格遞增,因此gg為1-1,所以g1:[a,b]Rg^{-1}: [a,b] \rightarrow \mathbb{R}為連續嚴格遞增函數。
    • 即對於所有的分割P={y0,y1,,yn}[c,d]P=\lbrace y_0, y_1, \cdots, y_n \rbrace \in [c,d],都存在對應的分割P={x0,x1,,xn}[a,b]P^{'}=\lbrace x_0, x_1,\cdots, x_n \rbrace \in [a,b]xk=g(yk)x_k = g(y_k)
    • P=g(P)P^{'} = g(P) P=g1(P)P = g^{-1}(P^{'}),且更細的分割都存在此種關係。

    • fR[a,b]f \in \mathbf{R}[a,b],給定ϵ>0Pϵ[a,b]S(P,f,α)abfdα<ϵ, PϵP\epsilon > 0 \exists P_{\epsilon}^{'}[a,b] \ni \left| S(P^{'}, f, \alpha) - \int_a^b f d \alpha \right| < \epsilon,\ \forall P_{\epsilon}^{'} \subseteq P^{'}.

    • Pϵ=g1(Pϵ)P_{\epsilon} = g^{-1}(P_{\epsilon}^{'})[c,d][c,d]的分割,且PϵP={y0,y1,,yn}P_{\epsilon} \subseteq P = \lbrace y_0, y_1, \cdots, y_n \rbrace
    • S(P,h,β)=k=1nh(uk)Δβk, uk[yk1,yk], Δβk=β(yk)β(yk1)S(P, h, \beta ) = \sum_{k=1}^n h(u_k)\Delta \beta _k,\ u_k \in [y_{k-1}, y_k],\ \Delta \beta_k = \beta(y_k) - \beta(y_{k-1}) .
    • tk=g(uk), xk=g(yk)t_k = g(u_k),\ x_k = g(y_k),則PϵP={x0,x1,,xn}P_{\epsilon} \subseteq P^{'} = \lbrace x_0, x_1, \cdots, x_n \rbrace [a,b][a,b]更細的分割。
    • 可得 S(P,h,β)=k=1nf[g(uk)](α[g(yk)]α[g(yk1])=k=1nf(tk)[α(xk)α(xk1)]=S(P,f,α). \begin{array}{rcl} S(P,h,\beta ) & = & \sum{k=1}^n f[g(u_k)](\alpha[g(y_k)] - \alpha[g(y_{k-1}]) \\ & = & \sum_{k=1}^n f(t_k)[\alpha(x_k) - \alpha(x_{k-1})] \\ & = & S(P^{'}, f, \alpha). \end{array}

    • tk[xk1,xk], S(P,h,β)abfdα<ϵ\because t_k \in [x_{k-1}, x_k],\ \therefore \left| S(P,h,\beta) - \int_a^b f d \alpha \right| < \epsilon (QED).

化簡為Riemann積分

  • 當函數α\alpha的微分函數 α \alpha^{'}連續時,可將積分abf(x)dα(x)\int_a^b f(x) d \alpha(x) 的積分函數dα(x)d \alpha(x) 轉換成 α(x)dx \alpha^{'}(x) dx .

  • Theorem: 函數 fRS(α)[a,b] f \in \mathbf{RS}(\alpha)[a,b],且函數 α\alpha 在閉區間 [a,b][a,b] 存在連續的微分函數 α\alpha^{'} ,則 abf(x)dα(x)=abf(x)α(x)dx \int_a^b f(x) d \alpha(x) = \int_a^b f(x) \alpha^{'}(x) dx
    .

  • 之後可證明微分函數 α\alpha^{'} 不須連續上式也成立。

    • Proof:
    • g(x)=f(x)α(x)g(x) = f(x)\alpha^{'}(x) 且考慮Riemann sum S(P,g)=k=1ng(tk)Δxk=k=1nf(tk)α(tk)Δxk S(P,g) = \sum_{k=1}^n g(t_k)\Delta x_k = \sum_{k=1}^n f(t_k) \alpha^{'}(t_k) \Delta x_k

    • 相同的分割PP 與相同的切割點tkt_k可得到Riemann-Stieltjes sum S(P,f,α)=k=1nf(tk)Δαk S(P,f,\alpha) = \sum_{k=1}^n f(t_k) \Delta \alpha_k.

    • 由均值定理(MVT),可得到Δαk=α(vk)Δxk, vk(xk1,xk)\Delta \alpha_k = \alpha^{'}(v_k) \Delta x_k, \ v_k \in (x_{k-1}, x_k),

    • S(P,f,α)S(P,g)=k=1nf(tk)[α(vk)α(tk)]Δxk \therefore S(P,f,\alpha) - S(P,g) = \sum_{k=1}^n f(t_k)[\alpha^{'}(v_k) - \alpha^{'}(t_k)]\Delta x_k .

    • 因為函數ff有界,令0<f(x)M, x[a,b] 0 < \vert f(x) \vert \leq M, \ \forall x \in [a,b].

    • α\because \alpha^{'}[a,b][a,b] 連續,所以給定ϵ>0, δ>0α(x)α(y)<ϵ2M(ba), 0xy<ϵ\epsilon > 0,\ \exists \delta > 0 \ni |\alpha^{'}(x) - \alpha^{'}(y)| < \frac{\epsilon}{2M(b-a)}, \ 0 \leq | x- y| < \epsilon.

    • 取分割PϵP^{'}_{\epsilon},且其範數Pϵ<δ \Vert P_{\epsilon}^{'} \Vert < \delta,則PϵP,α(vk)α(tk)<ϵ/(2M(ba))\forall P_{\epsilon}^{'} \subseteq P, |\alpha^{'}(v_k) - \alpha^{'}(t_k) | < \epsilon/(2M(b-a))必定成立。

    • S(P,f,α)S(P,g)<ϵ2\therefore | S(P,f,\alpha) - S(P,g) | < \frac{\epsilon}{2} [1].

    • fRS[a,b],PϵPS(P,f,α)abfdα<ϵ2\because f \in \mathbf{RS}[a,b], \therefore \exists P_{\epsilon}^{''} \subseteq P \ni \left| S(P,f,\alpha) - \int_a^b f d \alpha \right| < \frac{\epsilon}{2} [2].

    • 由[1][2],取PϵPϵPP_{\epsilon}^{'} \cup P_{\epsilon}^{''} \subseteq P,可得S(P,g)abfdα<ϵ \left| S(P,g) - \int_a^b f d \alpha \right| < \epsilon (QED).

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