# Prove ....

• Mar 4th 2010, 01:02 AM
flower3
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 $f \in \mathfrak{R} (g) \ on \ [a,b] \ with \ \int_a^bf \ dg =0$
• Mar 4th 2010, 09:05 AM
Drexel28
Quote:

Originally Posted by flower3
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.
• Mar 4th 2010, 09:38 AM
Plato
Because $f$ is continuous, there are points $t~\&~s,~~a such that $t\le x 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 $\le \varepsilon$.