I'm not sure if the following prove is correct. It seems incomplete. Hope you can help me:
Theorem: Suppose $f:[a,b] \to \mathbb R$ is Riemann integrable. Suppose $g: [a,b] \to \mathbb R$ is a function such that $g(x) = f(x)$ for all except finite many $x \in [a,b]$. Prove that $g$ is Riemann integrable in $[a,b]$. and
$$ \int_{a}^{b} f = \int_{a}^{b} g $$
Prove:
If f is integrable then for each partition $P$ of $[a,b]$ there exist an $\epsilon\gt 0$ such that
$$
\\ U(f,[a,b])-L(f,[a,b]) < \epsilon \\\
\implies U(f,[a,b])-L(f,[a,b]) \le U(f,P,[a,b])-L(f,P,[a,b]) \\\
\\\ = \sum_{j=1}^{n} (x_{j}-x_{j-1}) \sup f - \sum_{j=1}^{n} (x_{j}-x_{j-1}) \inf f \\\
$$
since $f(x)=g(x)$ for $x \in [a,b]$
$$ \sum_{j=1}^{n} (x_{j}-x_{j-1}) \sup g - \sum_{j=1}^{n} (x_{j}-x_{j-1}) \inf g \\\ = U(g,[a,b])-L(g,[a,b]) \le U(g,P,[a,b])-L(g,P,[a,b]) \\\ $$
Therefore $g$ is also Riemann integrable and
$$ \int_{a}^{b} f = \int_{a}^{b} g $$
Edit:
Next try:
If e.g. the value of the last point of an interval $[a,b]$ is $g(x)=f(x)+1$ and for the rest of the points $g(x)=f(x)$ then with any partition P
$$ U(h,P,[a,b])=\sum_{i=1}^{n-1}(x_{j}-x_{j-1})\sup_{[x_{j-1},x_{j}]} h + (x_{n}-x_{n-1})\sup_{[x_{n-1},x_{n}]} h \\\ \lim_{n\to\infty} \sum_{i=1}^{n-1}(x_{j}-x_{j-1})\sup_{[x_{j-1},x_{j}]} h + (x_{n}-x_{n-1})\sup_{[x_{n-1},x_{n}]} h = 0 $$
with
$$ h(x) = f(x)-g(x) $$
This can be applied to each subinterval of [a,b].
Does this work?