Showing that g is integrable and $\int^b_a{f}$ = $\int^b_a{g}$ Let $f$ be integrable on $[a,b]$, and suppose g is a function on $[a,b]$ such that $g$($x$) = $f$($x$) except for finitely many $x$ in $[a,b]$. Show $g$ is
integrable and $\int^b_a{f}$ = $\int^b_a{g}$.
I started out solving this problem by letting a function $h = f - g$ such that $h(x) = 0$ expect at one point in [a,b]. Since $f$ is integrable, then there exists a partition P such that $U(f,P)-L(f,P)$ < $\epsilon$. I'm not really sure how to proceed from here. Any hints and help is much appreciated.   
 A: First of all you may prove the result for the case where there is only one point  say $c$ where the f and g have different values.
Note that  in that case for any partition of [a,b] the difference between Riemman Sum of $f$ and the Riemman Sum for $g$ is at most $$|f(c)-g(c)|\times \delta$$ where $\delta$ is the maximum length of your subpartition intervals.
Now you pick a partition such that $$ \delta <\frac {\epsilon}{|f(c)-g(c)|}$$
The rest is easy to complete.    
A: Put $h (x)=f (x)-g (x) $ and let us prove that $h $ is integrable at $[a,b] $
Assume that $$\forall x\in (a,b] \; h (x)=0$$ and $$h (a)=c>0.$$
Let $\epsilon>0$ enough small given.
$h $ is continuous at $[a+\frac {\epsilon}{2c},b]$ and then integrable.
there exist a partition $\sigma_1$ of $[a+\frac {\epsilon}{2c},b] $ such that
$$U (h,\sigma_1)-L (h,\sigma_1)<\frac {\epsilon}{2}$$
put now $$\sigma=\sigma_1\cup \{0\}$$
$\sigma $ is a partition of $[a,b] $
now
$$U (h,\sigma)-L (h,\sigma)=$$
$$U (h,\sigma_1)-L (h,\sigma_1)+c\frac {\epsilon}{2c} <\epsilon.$$
This proves that $h $ is integrable at $[a,b] $.
$g=f-h $ is integrable since $f $ and $h $  are integrables at $[a,b] $.
A: Assuming $f\ne g$ so that $S=\{x: f(x)\ne g(x)\}$ is finite and not empty. Let $M=\max \{|f(x)-g(x)|:x\in S\}.$ Let $|S|$ be the number of members of $S.$
For a partition $P$ let $\|P\|$ be the length of the largest interval belonging to $P$. For  any $\epsilon >0$ there exists $\delta_{\epsilon}>0$ such that $\forall P\;(\|P\|<\delta_{\epsilon} \implies U(f,P)-L(f,P)<\epsilon/2).$
So if $\|P\|< \delta_{\epsilon}\cdot \min (1, \epsilon(M|S|)^{-1}/2)$ then $|U(f,P)-U(g,P)|<\epsilon$ and $|L(f,P)-L(g,P)|<\epsilon.$
The idea is simply that, since $S$ is finite, the contribution,  ( to a finite sum that approximates the integral of $f$ or of $g$ ), of  intervals containing members of $S, $ will be negligible if $\|P\|$ is small enough.
