integrable function

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March 13th 2009, 12:53 PM
manjohn12

integrable function

Suppose $f$ is Riemann integrabe on $[a,b]$ . Let $\xi$ be any number in $[a,b]$ , and let $r$ be any real number different from $f(\xi)$ . Define a function $g$ as follows: $g(x) = \begin{cases} f(x) \ \ \ \ \text{if} \ x \neq \xi \\ r \ \ \ \ \ \ \ \ \ \text{if} \ x = \xi \end{cases}$ . Prove that $g$ is Riemann integrable on $[a,b]$ and that $\int_{a}^{b} g = \int_{a}^{b} f$ .

Proof. Let $\epsilon>0$ . Choose $\delta < \frac{\epsilon}{2|r|}$ . Let $P = \{x_{0}, x_{2}, \ldots, x_{n} \}$ be a partition of $[a,b]$ with $||P|| < \delta$ . Choose $x_{1}^{*}, x_{2}^{*}, \ldots, x_{n}^{*}$ arbitrarily such that $x_{i-1} \leq x_{i}^{*} \leq x_i$ . Then $\mathcal{R}(g,P) = \sum_{i=1}^{n} g(x_{i}^{*})(x_{i}-x_{i-1})$ .

So $\mathcal{R}(g,P) = \begin{cases} f(x_{i}^{*})(x_{i}-x_{i-1}) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{if} \ x_{i}^{*} \neq \xi \ \text{for any} \ i \\ r(x_{i}-x_{i-1}) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{if} \ x_{i}^{*} = \xi \ \text{for exactly one} \ i \\ r(x_{i+1}-x_{i})+r(x_{i}-x_{i-1}) \ \ \ \ \ \text{if} \ \xi = x_{i+1}^{*} = x_{i}^{*} \ \ \ \ \ \text{for some} \ i \end{cases}$

So $|\mathcal{R}(g,P)| \leq |\mathcal{R}(f,P)|+2|r|||P||$ . Thus $|\mathcal{R}(g,P)|-|\mathcal{R}(f,P)| \leq 2r||P||$ . Then $||\mathcal{R}(g,P)|-|\mathcal{R}(f,P)|| \leq |\mathcal{R}(g,P)-\mathcal{R}(f,P)| \leq 2|r|||P|| < \epsilon$ . Hence $g$ is Riemann integrable on $[a,b]$ and $\int_{a}^{b} g = \int_{a}^{b} f$ . $\blacksquare$

Is this correct?
March 13th 2009, 02:49 PM
Plato

It seem correct in gereral. (I have not carefully read it).
But I notice you may want $\delta < \frac{\varepsilon }{{2\left| r \right| + 1}}$ to avoid divison by zero.
March 13th 2009, 05:52 PM
ThePerfectHacker

Let $f,g$ be integrable on $[a,b]$ then $f\pm g$ is integrable on $[a,b]$ and $\smallint_a^b (f\pm g) = \smallint_a^b f \pm \smallint_a^b g$ .

If you know this fact then you can use it to prove your problem.

Lemma: Let $f$ be a zero function on $[a,b]$ except at finitely many points, then $f$ is integrable with $\smallint_a^b f = 0$ .

Proof: Let the non-zero points be $z_1,...,z_k$ so that $f(z_j)\not = 0$ and let $M = \max_{ 1\leq j\leq k} |f(z_j)|$ . For $\delta >0$ and a partition $P=\{x_0,x_1,...,x_n\}$ form the Riemann sum $R=\sum_{j=1}^n f(t_j) \Delta_j$ where $||P|| < \delta$ . Notice that $| R| \leq \sum_{j=1}^n |f(t_j)| \Delta_j \leq M\cdot k \delta$ . This can be made arbitrary small if $\delta$ is small enough. Thus, $\smallint_a^b f = 0$ .

To prove you problem, consider $f-g$ and use the lemma above. (Wink)