# Prove ....

• Mar 4th 2010, 01:02 AM
flower3
Prove ....
Let $\displaystyle c \in (a,b)$ .Define $\displaystyle g:[a,b] \to \mathbb{R}$ by:
$\displaystyle g(x)=\left\{\begin{array}{ll}0&\;\;if\,\,x \ \neq c\\1&\;\; \ x \ = c \end{array}\right.$

prove that if $\displaystyle f:[a,b] \to \mathbb{R}$ is continuous at c then $\displaystyle f \in \mathfrak{R} (g) \ on \ [a,b] \ with \ \int_a^bf \ dg =0$
• Mar 4th 2010, 09:05 AM
Drexel28
Quote:

Originally Posted by flower3
Let $\displaystyle c \in (a,b)$ .Define $\displaystyle g:[a,b] \to \mathbb{R}$ by:
$\displaystyle g(x)=\left\{\begin{array}{ll}0&\;\;if\,\,x \ \neq c\\1&\;\; \ x \ = c \end{array}\right.$

prove that if $\displaystyle f:[a,b] \to \mathbb{R}$ is continuous at c then $\displaystyle f \in \mathfrak{R} (g) \ on \ [a,b] \ with \ \int_a^bf \ dg =0$

Integrability is easy to prove. To prove that $\displaystyle \int_a^b\text{ }f=0$ merely notice that for every $\displaystyle \varepsilon>0$ choosing a partition $\displaystyle \mathcal{P}$ such that $\displaystyle x_{j-1}\leqslant c\leqslant x_j$ with $\displaystyle \Delta x_j=\frac{\varepsilon}{c}$ we have that $\displaystyle \int_a^b\text{ }f=\inf_{P\in\wp}U(P,f)\leqslant U\left(\mathcal{P},f\right)=c\cdot\frac{\varepsilo n}{c}=\varepsilon$. It follows that $\displaystyle \int_a^b\text{ }f=0$

Oh...I misread the question. Well...now you have an idea of where to go with it.
• Mar 4th 2010, 09:38 AM
Plato
Because $\displaystyle f$ is continuous, there are points $\displaystyle t~\&~s,~~a<t<c<s<b$ such that $\displaystyle t\le x<y\le s$ implies that $\displaystyle \left| {f(x) - f(y)} \right| < \frac{\varepsilon }{2}$.
For any partition of $\displaystyle [a,b]$ find a refinement that contains the connective intervals $\displaystyle [x_i,c],[c,x_j]$ such that $\displaystyle \left[ {x_i ,x_j } \right] \subseteq \left[ {t,s} \right]$.
A $\displaystyle g-$sum on that refinement is $\displaystyle \le \varepsilon$.