積分函數定義域存在不連續點

  • 當積分函數α\alpha在區間[a,b][a,b]為常數時,積分abfdα=0\int_a^b f d\alpha = 0 必定存在 (因為在每一個子分割中的S(P,f,α)=0 S(P,f,\alpha) =0)。

  • 而當積分函數α \alpha在區間為常數,但只在某一點為跳躍值(jump)時,積分積分abfdα\int_a^b f d\alpha 不一定存在,且存在時也不一定為0。

  • Theorem: 給定 a<c<b a < c < b,令積分函數α(x)={α(a),ax<cα(b),c<xb \alpha(x) = \left \lbrace \begin{array}{rl} \alpha(a), & a \leq x < c \\ \alpha(b), & c < x \leq b \end{array} \right.,即 α\alpha 在點 cc 沒有定義。 令函數f:[a,b]Rf:[a,b] \rightarrow \mathbb{R},若f, αf,\ \alpha其中至少一個函數於點cc左側連續且至少一個函數於點cc右側連續時(左側與右側連續不必為同一函數),則函數ff在此區間可積分 (或者說當函數f, gf, \ g在點cc都不連續時,此時積分不存在),且可得出: abfdα=f(c)[α(c+)α(c)] \int_a^b f d \alpha = f(c) [ \alpha(c+) - \alpha(c-)] (因為α(c)=α(a)\alpha( c-) = \alpha(a)在點cc的左側都是定值,同理在點cc的右側都是定值,因此只有在跳躍處才有積分值).
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    Single point step function α \alpha .
    • Proof:
    • Let cPc \in P,因此函數α\alpha除了跳躍點α(c)\alpha(c)外的值均為0,即S(P,f,α)=f(tk1)[α(c)α(c)]+f(tk)[α(c+)α(c)] S(P, f, \alpha) = f(t_{k-1}) [\alpha(c) - \alpha(c-)] + f(t_k)[\alpha(c+) - \alpha(c)], tk1ctkt_{k-1} \leq c \leq t_k.

    • f(c)[α(c+)α(c)] f(c)[\alpha(c+) - \alpha(c-)] 為函數ff在點cc相對於函數α\alpha之值,當tk1t_{k-1}逼近tkt_k時,S(P,f,α)S(P,f, \alpha)加總應收斂到此值。

    • ΔS(P,f,α){f(c)[α(c+)α(c)]}=(f(tk1)f(c))(α(c)α(c))+(f(tk)f(c))(α(c+)α(c)). \begin{array}{rcl} \Delta & \equiv & S(P,f,\alpha)-\{ f(c)[\alpha(c+) - \alpha(c-)]\} \\ & = &( f(t_{k-1}) -f(c) ) (\alpha(c) - \alpha(c-)) + (f(t_k) - f(c))(\alpha(c+) - \alpha(c)). \end{array}

    • Δf(tk1)f(c)α(c)α(c)+f(tk)f(c)α(c+)α(c)\therefore | \Delta | \leq | f(t_{k-1}) -f(c) | |\alpha(c) - \alpha(c-) | + |f(t_k) - f(c)| |\alpha(c+) - \alpha(c)|.

    • 若函數ff在點cc連續時,ϵ>0 δ>0P<δf(tk1)f(c))<ϵ\forall \epsilon > 0 \ \exists \delta >0 \ni \Vert P \Vert < \delta \Rightarrow | f(t_{k-1}) - f(c)) | < \epsilon and f(tk)f(c)<ϵ | f(t_k) - f(c) | < \epsilon .

    • 所以可得不等式 Δϵα(c)α(c)+ϵα(c+)α(c) | \Delta| \leq \epsilon | \alpha(c) - \alpha(c-) | + \epsilon | \alpha(c+) - \alpha(c) |,但此不等式與函數ff是否在點cc是否連續無關。

    • Case 1 (α\alpha在點cc連續且函數ff在點cc兩側都不連續): 此時α(c)=α(c)\alpha(c) = \alpha(c-)α(c)=α(c+)\alpha(c) = \alpha(c+)時,可得到Δ=0\Delta = 0

    • Case 2 (α\alpha在點cc右側連續且函數ff在點cc左側連續): 必須滿足α(c)=α(c+)\alpha(c)=\alpha(c+),即α\alpha在點cc右側連續才可得出Δϵα(c)α(c) | \Delta| \leq \epsilon | \alpha(c) - \alpha(c-)| .

    • Case 3 (α\alpha在點cc左側連續且函數ff在點cc右側連續):則必須滿足α(c)=α(c)\alpha(c)=\alpha(c-),即α\alpha在點cc左側連續才可得出Δϵα(c+)α(c) | \Delta| \leq \epsilon | \alpha(c+) - \alpha(c)| .

    • 由以上三個case可知除了函數在兩側同時都不連續時才會無法積分(QED).

  • E.g. α(x)={0,x0,1,x=0.\alpha(x) = \left\lbrace \begin{array}{ll} 0, & x \neq 0, \\ -1, & x=0. \end{array} \right., f(x)=1,1x1f(x) = 1, -1 \leq x \leq 1.

    • 因為在x=0x=0時,函數f(0)=1f(0)=1,所以11fdα=0\int_{-1}^1 f d \alpha = 0 .
  • E.g. α(x)\alpha(x)定義同上,但函數f(x)=1,x0f(x)=1, \forall x \neq 0,因為f,αf, \alphax=0x=0時均無定義,所以11fdα\int_{-1}^1 f d \alpha不存在。

    • S(P,f,α)=f(tk)(α(xk)α(0))+f(tk1)(α(0)α(xk2))=f(tk)f(tk1) \because S(P,f,\alpha) = f(t_k)(\alpha (x_k) - \alpha(0)) + f(t_{k-1})(\alpha(0) - \alpha(x_{k-2})) = f(t_k) - f(t_{k-1}) , xk2tk10tkxkx_{k-2} \leq t_{k-1} \leq 0 \leq t_k \leq x_k,上述之值依tk,tk1t_k, t_{k-1}之值,可能為0, 1, 或-1(無法唯一決定),因此積分不存在。

以有限和表示Riemann-Stieltjes積分

階梯函數(Step function)

  • 定義:函數α:[a,b]R\alpha: [a,b] \rightarrow \mathbb{R}被稱為階梯函數:存在一分割 a=x1<x2<<xn=ba = x_1 < x_2 < \cdots < x_n = b 使得函數α\alpha在每一個開區間(xk1,xk), k=1,2,,n(x_{k-1}, x_k), \ k=1,2,\cdots,n均為常數值ckc_k
    • α(xk+)α(xk)\alpha(x_k+) - \alpha(x_k-)稱為在點xk, 1<k<nx_k, \ 1 < k < n的跳躍(jump),而在端點的跳躍為α(x1+)α(x1) and α(xn)α(xn)\alpha(x_1 + ) - \alpha(x_1) \text{ and } \alpha(x_n ) - \alpha(x_n -).

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      (跳躍點)右連續的階梯函數。
  • E.g. 最簡單的階梯函數是greatest-integer function,即f(x)=[x], [x]x<[x]+1f(x) = [x], \ [x] \leq x < [x]+1.

  • Theorem: 令α:[a,b]R\alpha: [a,b] \rightarrow \mathbb{R}為階梯函數,且在xkx_k之跳躍值為αk\alpha_k。函數f:[a,b]Rf: [a,b] \rightarrow \mathbb{R}x1,x2,,xnx_1,x_2,\cdots, x_n每一點不會與函數α\alpha同時在左側或右側不連續,則abfdα\int_a^b f d\alpha存在且 abf(x)dα(x)=k=1nf(xk)αk\int_a^b f(x) d \alpha(x) = \sum_{k=1}^n f(x_k) \alpha_k
    .

    • c(a,b), acfdα+cbfdα=abfdα \because \forall c \in (a,b),\ \int_a^c f d \alpha + \int_c^b f d \alpha = \int_a^b f d\alpha [1].
    • 而且在每一個跳躍點,只要f,αf, \alpha不要同時不連續即可積分 [2]。
    • 由[1][2]可得 abf(x)dα(x)=k=1nf(xk)αk\int_a^b f(x) d \alpha(x) = \sum_{k=1}^n f(x_k) \alpha_k (QED).

單調遞增的積分函數

    定義:分割PP[a,b]P \in \mathbb{P}[a,b]
  • Mk(f)=sup{f(x)x[xk1,xk]}M_k(f) = \sup \lbrace f(x) \vert x \in [x_{k-1}, x_k] \rbrace ,即第kk個子分割中,函數的最大值。
  • mk(f)=inf{f(x)x[xk1,xk]}m_k(f) = \inf \lbrace f(x) \vert x \in [x_{k-1}, x_k] \rbrace ,即第kk個子分割中,函數的最小值。
  • Upper Stieltjes sum of ff w.r.t. α\alpha for partition PP, U(P,f,α)=k=1nMk(f)Δαk U(P, f, \alpha) = \sum_{k=1}^n M_k(f) \Delta \alpha_k .
  • Lower Stieltjes sum of ff w.r.t. α\alpha for partition PP, L(P,f,α)=k=1nmk(f)Δαk L(P, f, \alpha) = \sum_{k=1}^n m_k(f) \Delta \alpha_k .
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黃色部份為upper sum,灰色部份為lower sum。
  • 由定義可得出 mk(f)Mk(f)m_k(f) \leq M_k(f),若函數α\alpha在區間[a,b][a,b]為遞增函數,則Δαk0\Delta \alpha_k \geq 0,因此可得出
    • tk[xk1,xk], mk(f)f(tk)Mk(f)\forall t_k \in [x_{k-1}, x_k],\ m_k(f) \leq f(t_k) \leq M_k(f).
    • 所以在α\alpha為遞增函數時,可得L(P,f,α)S(P,f,α)U(P,f,α)L(P, f, \alpha) \leq S(P, f, \alpha) \leq U(P, f, \alpha).

<<<<<<< HEAD

    • Theorem: 若函數α:[a,b]R\alpha: [a,b] \rightarrow \mathbb{R}為遞增函數,若PPP \subseteq P^{'}為更細的分割,則
    • U(P,f,α)U(P,f,α)U(P^{'}, f, \alpha) \leq U(P, f, \alpha) (切細時,上總和之值會變小)
    • L(P,f,α)L(P,f,α)L(P^{'}, f, \alpha) \geq L(P, f, \alpha) (切細時,下總和之值會變大)
    • =======
      Theorem: 若函數α:[a,b]R\alpha: [a,b] \rightarrow \mathbb{R}為遞增函數,若PPP \subseteq P^{'}為更細的分割,則
      • U(P,f,α)U(P,f,α)U(P^{'}, f, \alpha) \leq U(P, f, \alpha) (分割切細時,上總和之值會變小)
      • L(P,f,α)L(P,f,α)L(P^{'}, f, \alpha) \geq L(P, f, \alpha) (分割切細時,下總和之值會變大)
      • >>>>>>> 1bd8ee3f8f2acbe338a56a1029e34331a6aec367
      • 對於任意兩種分割P1, P2P_1, \ P_2,可得L(P1,f,α)U(P2,f,α)L(P_1, f, \alpha) \leq U(P_2, f, \alpha)
  • Proof (1)(2):

    • PP and P{c}=P, c[xi1,xi]P \subseteq P^{'} \text{ and } P \cap \{ c \} = P^{'}, \ c \in [x_{i-1}, x_i] , then
    • U(P,f,α)=k=1, kinMk(f)Δαk+M[α(c)α(xi1)]+M[α(xi)α(c)]U(P^{'},f ,\alpha) = \sum_{k=1, \ k\neq i}^n M_k(f) \Delta \alpha_k + M^{'}[\alpha(c) - \alpha(x_{i-1})] + M^{''}[\alpha(x_i) - \alpha(c)],
    • M=sup{f(x)x[xi1,xi]}, M=sup{f(x)x[xi1,c]}, M=sup{f(x)x[c,xi]} M = \sup \{ f(x) | x \in [x_{i-1}, x_i] \},\ M^{'} = \sup \{ f(x) | x \in [x_{i-1}, c] \}, \ M^{''} = \sup \{f(x)| x \in [c, x_i] \}.
    • MMi(f) and MMi(f)U(P,f,α)U(P,f,α) \because M^{'} \leq M_i(f) \text{ and } M^{''} \leq M_i(f) \Rightarrow U(P^{'}, f, \alpha) \leq U(P, f, \alpha) (QED).
  • Proof (3):

    • P=P1P2P = P_1 \cup P_2, 可得 L(P1,f,α)L(P,f,α)U(P,f,α)U(P2,f,α)L(P_1,f,\alpha) \leq L(P, f, \alpha) \leq U(P,f,\alpha) \leq U(P_2,f,\alpha). (QED).

Darboux上積分與下積分

  • 定義:函數α:[a,b]R\alpha: [a,b] \rightarrow \mathbb{R}為遞增函數,則
    • Upper Stieltjes integral: I(f,α)abfdα=inf{U(P,f,α)PP[a,b]} \overline{I}(f, \alpha) \equiv \overline{\int}_a^b f d\alpha = \inf \lbrace U(P,f,\alpha) \vert P \in \mathbb{P}[a,b] \rbrace .
    • Lower Stieltjes integral: I(f,α)abfdα=sup{L(P,f,α)PP[a,b]} \underline{I}(f, \alpha) \equiv \underline{\int}_a^bf d\alpha = \sup \lbrace L(P,f,\alpha) \vert P \in \mathbb{P}[a,b] \rbrace .
  • Theorem: 若積分函數α:[a,b]R\alpha: [a,b] \rightarrow \mathbb{R}為遞增函數,則下積分之值必定小於等於上積分之值。I(f,α)I(f,α)\Leftrightarrow \underline{I}(f, \alpha) \leq \overline{I}(f, \alpha).
    • proof:
    • Given ϵ>0,P1P[a,b]U(P1,f,α)<I(f,α)+ϵ\epsilon > 0, \exists P_1 \in \mathbb{P}[a,b] \ni U(P_1, f, \alpha) < \overline{I}(f, \alpha) + \epsilon.
    • 因為I(f,α) \overline{I}(f, \alpha)L(P,f,α)L(P,f,\alpha)的上界,所以I(f,α)I(f,α)\underline{I}(f, \alpha) \leq \overline{I}(f, \alpha) (QED).
  • 任意函數均可拆成兩個遞增函數的差值,而積分滿足線性相加性質。

  • E.g. α(x)=x\alpha(x) = x and f(x)={1,x is rational,0,x is irrationalf(x) = \left\lbrace \begin{array}{ll} 1, & x \text{ is rational,} \\ 0, & x \text{ is irrational} \end{array} \right. , then I(f,α)<I(f,α) \underline{I}(f, \alpha) < \overline{I}(f, \alpha).

    • PP[0,1], Mk(f)=1 and mk(f)=0\because \forall P \in \mathbb{P}[0,1], \ M_k(f) =1 \text{ and } m_k(f) = 0 U(P,f)=1, L(P,f)=0\therefore U(P,f)=1, \ L(P,f) = 0.

上、下積分的線性性質

  • a<c<b,abfdα=acfdα+cbfdα a < c < b, \overline{\int}_a^b f d \alpha = \overline{\int}_a^c f d \alpha + \overline{\int}_c^b f d \alpha .

  • 但下式只滿足不等式

    • ab(f+g)dαabfdα+abgdα \overline{\int}_a^b (f+g) d \alpha \leq \overline{\int}_a^b f d \alpha + \overline{\int}_a^b g d \alpha .
    • ab(f+g)dαabfdα+abgdα \underline{\int}_a^b (f+g) d \alpha \geq \underline{\int}_a^b f d \alpha + \underline{\int}_a^b g d \alpha .

Riemann條件

  • 由於上積分之值在定義域切更細的分割時會下降,而下積分之值在定義域切更的分割時會上升,兩者互為上、下界,而在上積分與下積分非常接近時,或是上積分與下積分的差值趨近於零時,可視為等號成立,即積分存在。

  • 定義:函數ff 在閉區間[a,b][a,b] 滿足 Riemann條件 w.r.t. 函數αϵ>0 PϵP 0U(P,f,α)L(P,f,α)<ϵ\alpha \Leftrightarrow \forall \epsilon > 0 \ \exists P_{\epsilon} \subseteq P \ \ni 0 \leq U(P, f, \alpha) - L(P,f,\alpha) < \epsilon.
  • 此定義即當分割PP切細時,若函數ff相對於函數α\alpha的上、下總和逐漸收斂時,則可積分。

    1. Theorem: 若函數α:[a,b]R\alpha: [a,b] \rightarrow \mathbb{R}為遞增函數,則以下三個敘述等價
    2. 函數fRS(α)[a,b]f \in \mathbf{RS}(\alpha)[a,b]
    3. 函數ff在閉區間[a,b][a,b]相對於函數 α\alpha 滿足Riemann條件。
    4. 函數ff Darboux上積分等於下積分,I(f,α)=I(f,α) \overline{I}(f, \alpha) = \underline{I}(f, \alpha).
  • Proof from (1) to (2):

    • α(a)=α(b)\alpha(a) = \alpha(b)時,則(2)一定滿足,因此假設α(a)<α(b)\alpha(a) < \alpha(b).
    • fRS(α)[a,b]\because f \in \mathbf{RS}(\alpha)[a,b], given ϵ>0PPϵ,tk1,tk2[xk1,xk]\epsilon > 0 \exists P \supseteq P_{\epsilon}, t_{k1}, t_{k2} \in [x_{k-1}, x_k] \ni . +k=1nf(tk1)ΔαkA<ϵ3, and k=1nf(tk2)ΔαkA<ϵ3 ,where A=abfdα \left| \sum_{k=1}^n f(t_{k1}) \Delta \alpha_k - A \right| < \frac{\epsilon}{3}, \text{ and } \left| \sum_{k=1}^n f(t_{k2}) \Delta \alpha_k - A \right| < \frac{\epsilon}{3} \text{ ,where } A = \int_a^b f d \alpha .
    • k=1n[f(tk1)f(tk2)]Δαk<23ϵ \therefore \left| \sum_{k=1}^n [f(t_{k1}) - f(t_{k2})]\Delta \alpha_k \right| < \frac{2}{3}\epsilon .
    • Given k,Mk(f)mk(f)=sup{f(x)f(y)x,y[xk1,xk]}k, \because M_k(f) - m_k(f) = \sup \{f(x) - f(y) \vert x, y \in [x_{k-1}, x_k] \} .
    • h>0,f(tk1)f(tk2)>Mk(f)mk(f)h \therefore \forall h >0, f(t_{k1}) - f(t_{k2}) > M_k(f) - m_k(f) - h.
    • choose h=1/3ϵ[α(b)α(a)]h = \frac{1/3 \epsilon }{[\alpha(b) - \alpha(a)]}
    • U(P,f,α)L(P,f,α)=k=1n[Mk(f)mk(f)]Δαk<k=1n[f(tk1)f(tk2)]Δαk+hk=1nΔαk<ϵ. \begin{array}{rcl} U(P,f,\alpha) - L(P,f, \alpha) & = & \sum_{k=1}^n [M_k(f) - m_k(f)]\Delta \alpha_k \\ & < & \sum_{k=1}^n [f(t_{k1}) - f(t_{k2})]\Delta \alpha_k + h \sum_{k=1}^n \Delta \alpha_k < \epsilon. \end{array} (QED).
  • Proof from (2) to (3):

    • 若Riemann條件成立,即給定ϵ>0 \epsilon > 0 ,存在分割PPϵU(P,f,α)<L(P,f,α)+ϵ P \supseteq P_{\epsilon} \ni U(P,f,\alpha) < L(P,f,\alpha) + \epsilon .
    • I(f,α)U(P,f,α)L(P,f,α)+ϵI(f,α)+ϵ \therefore \overline{I}(f, \alpha) \leq U(P,f,\alpha) \leq L(P,f,\alpha) + \epsilon \leq \underline{I}(f, \alpha) + \epsilon .
    • ϵ>0, I(f,α)I(f,α)+ϵ \therefore \forall \epsilon > 0, \ \overline{I}(f,\alpha) \leq \underline{I}(f, \alpha) + \epsilon [1].
    • α \because \alpha 在閉區間[a,b][a,b]為遞增函數,所以I(f,α)I(f,α) \underline{I}(f,\alpha) \leq \overline{I}(f, \alpha) [2].
    • By[1][2], 可得I(f,α)=I(f,α) \underline{I}(f,\alpha) = \overline{I}(f, \alpha) . (QED).
  • Proof from (3) to (1):

    • I(f,α)=I(f,α)=A \underline{I}(f,\alpha) = \overline{I}(f, \alpha) = A,要證明abfdα\int_a^b f d \alpha 存在且等於AA
    • 給定ϵ>0\epsilon > 0,選定分割Pϵ,1U(P,f,α)<I(f,α)+ϵ, PPϵ,1.P_{\epsilon,1} \ni U(P,f,\alpha) < \overline{I}(f, \alpha) + \epsilon, \ \forall P \supseteq P_{\epsilon,1}.
    • 另外選定分割Pϵ,2L(P,f,α)>I(f,α)ϵ, PPϵ,2. P_{\epsilon,2} \ni L(P, f, \alpha) > \underline{I}(f,\alpha) - \epsilon, \ \forall P \supseteq P_{\epsilon, 2}.
    • Pϵ=Pϵ,1Pϵ,2P_{\epsilon} = P_{\epsilon,1} \cup P_{\epsilon,2} ,可得下式
    • I(f,α)ϵ<L(P,f,α)S(P,f,α)U(P,f,α)<I(f,α)+ϵ, PPϵ. \underline{I}(f, \alpha) - \epsilon < L(P,f,\alpha) \leq S(P,f,\alpha) \leq U(P,f,\alpha) < \overline{I}(f,\alpha) + \epsilon , \ \forall P \supseteq P_{\epsilon}.
    • I(f,α)=I(f,α)=A, S(P,f,α)A<ϵ PPϵ \because \overline{I}(f,\alpha) = \underline{I}(f,\alpha) = A, \ \therefore | S(P,f,\alpha) - A | < \epsilon \ \forall P \supseteq P_{\epsilon} (QED).

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