1. ## 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?

2. 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.

3. 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.