I am reading this question
Let $f:[a,b]→R$ be Riemann integrable. Then changing one value of $f$ then $f$ is still integrable and it integrates to the same value.
At the link that I pasted is the answer for this quesiton. But as I understand proof of only first part.
How proof that it integrates to same value?
I tried this.
As we know $lim_{\lambda\to0}\sigma=I$(if $I$ if finite) is Riemann integral.
Now $\sigma$ = $\sum_{i=0}^{n-1}f(\phi)\triangle x_i$
So for $f$
$\sigma$ = $\sum_{i=0}^{n-2}f(\epsilon_i)\triangle x_i + f(\epsilon_j)(x-x_0)$
for $\widetilde f $ $\sigma= \sum_{i=0}^{n-2}f(\epsilon_i)\triangle x_i + (f(\epsilon_j) + f(\epsilon_k)) \frac{\epsilon}{2M}$
Will be glad if you can say what I did is right and help with why it integrates to same value.
Riemann-integrable functions are bounded, so $f$ won't be Riemann-integrable anymore if you change its value at a single point to $\pm \infty$.
If $f=g$ everywhere except when $x=x_0$, then $f-g = 0$ unless $x = x_0$, and, for sufficiently small $\varepsilon > 0$, $$ \int_a^b (f-g)\, dx= \int_a^{\max(a,x_0 - \varepsilon)} + \int_{\max(a,x_0 - \varepsilon)}^{\min(x_0 + \varepsilon,b)} + \int^b_{\min(x_0 + \varepsilon,b)}. $$ We have $$ \min(x_0 + \varepsilon,b) - \max(a,x_0 - \varepsilon) = \begin{cases} \varepsilon & \text{if $x_0=a$ or $x_0=b$}\\ 2\varepsilon & \text{if $a < x_0 < b$.} \end{cases} $$ The second integral does not exceed $$ \int\limits_{\max(a,x_0 - \varepsilon)}^{\min(x_0 + \varepsilon,b)} \!\!\!\!\!|f-g|\,dx \leq \int\limits_{\max(a,x_0 - \varepsilon)}^{\min(x_0 + \varepsilon,b)} \!\!\!\!\!\sup_{[a,b]}|f-g|\,dx \leq 2\varepsilon\cdot |f(x_0) - g(x_0)| $$ by absolute value. The remaining two integrals are both 0.