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Math Help - Prove ....

  1. #1
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    Prove ....

    Let  c \in (a,b) .Define  g:[a,b] \to \mathbb{R} by:
    g(x)=\left\{\begin{array}{ll}0&\;\;if\,\,x \ \neq c\\1&\;\; \ x \ = c \end{array}\right.

    prove that if  f:[a,b] \to \mathbb{R} is continuous at c then  f \in \mathfrak{R} (g) \ on \ [a,b] \ with \ \int_a^bf \ dg =0
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by flower3 View Post
    Let  c \in (a,b) .Define  g:[a,b] \to \mathbb{R} by:
    g(x)=\left\{\begin{array}{ll}0&\;\;if\,\,x \ \neq c\\1&\;\; \ x \ = c \end{array}\right.

    prove that if  f:[a,b] \to \mathbb{R} is continuous at c then  f \in \mathfrak{R} (g) \ on \ [a,b] \ with \ \int_a^bf \ dg =0
    Integrability is easy to prove. To prove that \int_a^b\text{ }f=0 merely notice that for every \varepsilon>0 choosing a partition \mathcal{P} such that x_{j-1}\leqslant c\leqslant x_j with \Delta x_j=\frac{\varepsilon}{c} we have that \int_a^b\text{ }f=\inf_{P\in\wp}U(P,f)\leqslant U\left(\mathcal{P},f\right)=c\cdot\frac{\varepsilo  n}{c}=\varepsilon. It follows that \int_a^b\text{ }f=0

    Oh...I misread the question. Well...now you have an idea of where to go with it.
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  3. #3
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    Because f is continuous, there are points t~\&~s,~~a<t<c<s<b such that t\le x<y\le s implies that \left| {f(x) - f(y)} \right| < \frac{\varepsilon }{2}.
    For any partition of [a,b] find a refinement that contains the connective intervals [x_i,c],[c,x_j] such that \left[ {x_i ,x_j } \right] \subseteq \left[ {t,s} \right].
    A g-sum on that refinement is \le \varepsilon.
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