Proving the Riemann integrability of a function

Question

Let $f : [a, b] \to \mathbb{R}$ be a Riemann integrable function. If $g : [a, b] \to \mathbb{R}$ is another function and $S = \lbrace x : f(x) \neq g(x)\rbrace$ contains exactly $n$ points, show that $g$ is also Riemann integrable from any of the equivalent definitions of Riemann integrability. Can anyone give me some hints?

Answer 1

Let $S:=\{x_{1},x_{2},\cdots,x_{n}\}$. I will suppose below that $a<x_{1}$ and $x_{n}<b$. Clearly, $S$ is isolated. Thus, we can find $\delta>0$ such that $x_{k+1}-x_{k}>2\delta$. We may also suppose that $x_{1}-a>\delta$ and $b-x_{n}>\delta$. Now, consider the following. \begin{align} \int_{a}^{b}f(u)\mathrm{d}u ={}&\int_{a}^{x_{1}-\delta}f(u)\mathrm{d}u+\int_{x_{1}+\delta}^{x_{2}-\delta}f(u)\mathrm{d}u+\cdots+\int_{x_{n-1}+\delta}^{x_{n}-\delta}f(u)\mathrm{d}u+\int_{x_{n}+\delta}^{b}f(u)\mathrm{d}u\\ {}&+\int_{x_{1}-\delta}^{x_{1}+\delta}f(u)\mathrm{d}u+\cdots+\int_{x_{n}-\delta}^{x_{n}+\delta}f(u)\mathrm{d}u\\ ={}&\int_{a}^{x_{1}-\delta}g(u)\mathrm{d}u+\int_{x_{1}+\delta}^{x_{2}-\delta}g(u)\mathrm{d}u+\cdots+\int_{x_{n-1}+\delta}^{x_{n}-\delta}g(u)\mathrm{d}u+\int_{x_{n}+\delta}^{b}g(u)\mathrm{d}u\tag{1}\label{eq1}\\ {}&+\int_{x_{1}-\delta}^{x_{1}+\delta}f(u)\mathrm{d}u+\cdots+\int_{x_{n}-\delta}^{x_{n}+\delta}f(u)\mathrm{d}u.\tag{2}\label{eq2} \end{align} Note that as $\delta\to0$, each term in \eqref{eq2} tends to $0$, and \eqref{eq1} becomes $\int_{a}^{b}g(u)\mathrm{d}u$.