Riemann integrable function proof? Let a < b and f a Riemann integrable function on [a,b]. For a function g such that f = g on [a,b] except for finitely many points, it seems intuitive that g is also  Riemann integrable on [a,b], but is there a proof for this? If so, how would you go about proving that the integrals across [a,b] are equal for both functions?
 A: An outline for you to complete:
Suppose $f \neq g$ only at the points $x_1 < x_2 < ... < x_n$. Then, construct a partition of the following kind : Let $\epsilon >0$ be small enough so that for each $x_i$, $[x_i - \epsilon,x_i+\epsilon] \subset [a,b]$, and these are disjoint intervals for $i \neq j$.
Now create a partition which contains at least all of these intervals $[x_i - \epsilon, x_i + \epsilon]$. For this partition $P = (y_i)$(where $y_i$ are the endpoints of the intervals) , we can consider the upper and lower sums for $f$ and $g$:
$$
\mathcal U(f,P) := \sum_{P} (y_i - y_{i+1})(\sup_{x \in [y_i,y_{i+1}]} f(x)) \quad \mathcal U(g,P) := \sum_{P} (y_i - y_{i+1})(\sup_{x \in [y_i,y_{i+1}]} g(x)) \\
\mathcal L(f,P) :=\sum_{P} (y_i - y_{i+1})(\inf_{x \in [y_i,y_{i+1}]} f(x)) \quad \mathcal L(g,P) := \sum_{P} (y_i - y_{i+1})(\inf_{x \in [y_i,y_{i+1}]} g(x)) \\
$$
Now, we can disregard intervals where $f=g$, this leaves us to only care about the intervals $[x_i-\epsilon,x_i+\epsilon]$, so $\mathcal U(f,P) - \mathcal U(g,P) = \sum_i (2\epsilon)(f(x_i)-g(x_i))$, and similarly for $\mathcal L$.If we let $\epsilon \to 0$ now, we get the result, namely that $g$ is integrable and $\int g = \int f$, since their upper and lower sums are the same (and equal to each other due to integrability).
