Here is my proof.
Theorem: If $f$ is Riemann integrable on $[a,b],c\in[a,b]$, and $g(x)=f(x)$ except for finitely many points $c_1,\cdots,c_k$ in $[a,b]$, then $g$ is Riemann integrable on $[a,b]$, and $\int_{a}^{b}f(x)dx=\int_{a}^{b}g(x)dx.$
Suppose another function $g$ such that $g(x)=f(x)$ except for finitely many points $c_1,\cdots,c_k$. Since $f$ is Riemann integrable, for any $\epsilon>0$, there exists $\delta=\frac{\epsilon}{4n\sum_{i=1}^{k}|g(c_i)-f(c_i)|}$ such that for any partition $P=(a=x_0,\cdots,x_n=b)$ with $\left\lVert P\right\rVert<\delta$, we have $\left|R(f,P)-\int_{a}^{b}f\right|<\frac{\epsilon}{2}.$ Now,
\begin{align*}
\left|R(g,P)-\int_{a}^{b}f\right|&=\left|R(g,P)-R(f,P)+R(f,P)-\int_{a}^{b}f\right|\\
&\leq\left|R(g,P)-R(f,P)\right|+\left|R(f,P)-\int_{a}^{b}f\right|\\
&<\left|R(g,P)-R(f,P)\right|+\frac{\epsilon}{2}\\
&<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.\\
\end{align*}where \begin{align*}
|R(g,P)-R(f,P)|&=\left|\sum_{i=1}^{n}g(x_i*)(x_i-x_{i-1})-\sum_{i=1}^{n}f(x_i*)(x_i-x_{i-1})\right|\\
&=\left|\sum_{i=1}^{n}(g(x_i^*)-f(x_i^*))(x_i-x_{i-1})\right|\\
&<\sum_{i=1}^{k}|g(c_i)-f(c_i)|(2n\delta)\\
&=2n\sum_{i=1}^{k}|g(c_i)-f(c_i)|\left(\frac{\epsilon}{4n\sum_{i=1}^{k}|g(c_i)-f(c_i)|}\right)\\
&<\frac{\epsilon}{2}
\end{align*}
Is my proof of this theorem correct?
Best Answer
Your approach via Riemann sums can be simplified with lesser use of symbolism. Let $M$ be a positive upper bound for both $|f|, |g|$ on $[a, b] $. Since $f$ is integrable, say with integral $I$, then for every $\epsilon>0$ we have a $\delta'>0$ such that for any partition $P$ of $[a, b] $ with norm $||P||<\delta'$ we have $$|S(f, P) - I|<\frac{\epsilon} {2}$$ where $$P=\{x_0,x_1,x_2,\dots,x_n\}$$ and $$S(f, P) =\sum_{i=1}^{n}f(t_i)(x_i-x_{i-1}),\, t_i\in[x_{i-1},x_i]$$ Since $g(x) =f(x) $ except for a finite number $k$ of exceptional points we have $$|S(g, P) - S(f, P) |<2Mk||P||$$ and thus if we choose $\delta$ such that $0<\delta <\min(\delta',\epsilon/(4Mk))$ we have $$|S(g, P) - S(f, P) |<\frac{\epsilon} {2}$$ whenever $||P||<\delta$ and thus $$|S(g, P) - I|<\epsilon$$ whenever $||P||<\delta$. This means that $g$ is also Riemann integrable on $[a, b] $ with the same integral $I$.
In short, changing the values of a function at a finite number of points affects neither its Riemann integrability nor the value of its integral (if it exists).
The word finite in above statement can't be replaced by infinite.