I tried to prove the following. Please could somebody tell me if my proof is correct?
Let $f: [a,b]\to \mathbb R$ be Riemann integrable. Then changing one value of $f$ then $f$ is still integrable and it integrates to the same value.
My proof. Let $x \in [a,b]$ denote the point where $f$ is changed. Let $\widetilde{f}$ denote the new function with $\widetilde{f}(x) = z$ and the old function is $f(x) = y$. Let $M = |y-z|$. Let $\varepsilon > 0$. Let $U(f,P)$ denote the upper sum and $L(f,P)$ the lower sum for partition $P$. Since $f$ is integrable there exists a partition $P$ such that the upper sums minus the lower sums are less than epsilon:
$$ U(f,P) - L(f,P) < \varepsilon $$
Let $Q$ be the refinement of $P$ consisting of $P$ and $\{x-{\varepsilon \over 2M}, x + {\varepsilon \over 2M}\}$.
Then $U(f,P) \ge U(f,Q)$ and $L(f,P) \le L(f,Q)$.
Furthermore, $|U(f,Q)-U(\widetilde{f},Q)| \le {\varepsilon \over M}\cdot M = \varepsilon$. This is true because $f$ and $\widetilde{f}$ only differ at $x$ and at $x$ they can maximally differ by $M$. Since the partition $Q$ contains the interval $(x- {\varepsilon \over 2M}, x + {\varepsilon \over 2M})$ and this interval has lenght ${\varepsilon \over M}$ the maximal difference of these sums can be only $\varepsilon$. Similarly, $|L(f,Q)-L(\widetilde{f},Q)| \le \varepsilon$.
Hence
$$ |U(\widetilde{f},Q) - L(\widetilde{f},Q)| \le |U(\widetilde{f},Q) - U(f,Q)| + |U(f,Q) - L(f,Q)| + |L(f,Q) - L(\widetilde{f},Q)| \le \varepsilon $$