Showing two functions have equal Riemann integrals Let $f:[a,b] \to \mathbb{R}$ and $g:[a,b] \to \mathbb{R}$ be bounded and Riemann integrable on $[a,b]$.  If $(x_n)$ is a sequence converging to $x$ with $(x_n) \subseteq [a,b]$ and $f(w)=g(w)$ except when $w=x_i$ for $i \in \mathbb{N}$, how can I show $\int_{a}^{b} f = \int_{a}^{b} g$?
I'm only sure about how to work from the definition of Riemann integrable functions. Let $\epsilon > 0$.  Since $f,g \in \mathcal{R}[a,b]$ there is a $\delta > 0$ so that if $\dot{P}$ is a tagged partition of $[a,b]$ satisfying $||\dot{P}|| < \delta$ then we have $$\left | S(f,\dot{P}) - \int_{a}^{b} f \right | < \epsilon$$ and $$\left | S(g,\dot{P}) - \int_{a}^{b} g \right | < \epsilon$$
I think to work from here I need to be able to say something about $$\left | S(f,\dot{P}) - S(g,\dot{P}) \right |$$
I know that $$\left | S(f,\dot{P}) - S(g,\dot{P}) \right | =\left | \sum_{i=1}^{n}(f(t_i)-g(t_i))(x_i-x_{i-1})\right |$$
Now I think $f(t_i)-g(t_i)=0$ except it $t_i \in (x_n)$ so I only need to worry about the times that the $t_i \in (x_n)$.  Since $f,g$ are bounded, the worst value this difference can have is the case when $f(x_i)$ is at its maximum value and $g(x_i)$ is at its minimum value.  Also we have $(x_i-x_{i-1})< \delta $.  Assuming the preceding is correct, my confusion is about how many times $f(t_i)-g(t_i)$ can take its maximum value.  Is there a way to count these?  Do I even need to be able to count them? I think if I knew a maximum number of times this could happen, I could define a new $\delta$ allowing result to hold.
 A: In this answer we want to use the fact that if two functions differ only at finitely many points, then the Riemann integral are the same. Using this we can more or less get rid of all the analysis to deal with partial sums. 
Also you have to use the essentially fact that $x_n \to x$. First of all, for all $\epsilon >0$, we have
$$\int_a^b f = \int_a ^{x-\epsilon} f + \int_{x-\epsilon}^{x + \epsilon} f + \int_{x+\epsilon} ^b f$$
and similar for $g$ (we assume $a <x-\epsilon < x+\epsilon <b$). Now as $x_n \to x$, there is $N_\epsilon \in \mathbb N$ so that $|x- x_n| < \epsilon$ for all $n \ge N_\epsilon$. Thus away from $[x-\epsilon, x+ \epsilon]$, $f$ and $g$ differs only at finitely many points. In particular, 
$$\int_a ^{x-\epsilon} f = \int_a ^{x-\epsilon} g,\ \ \ \int_{x+\epsilon} ^b f = \int_{x+\epsilon} ^b g. $$
This implies 
$$\left| \int_a^b f - \int_a^b g \right| \le \left|\int_{x-\epsilon}^{x+ \epsilon} (f-g)\right| \le 4M \epsilon,$$
where $M$ is the upper bound of $f$, $g$. As $\epsilon>0$ is arbitrary, take $\epsilon \to 0$ to get 
$$\int_a ^b f = \int_a^b g.$$
A: It seems like if you prove that $|S(g, P) - S(f, P)| < \varepsilon$, then it means that $\int_a^b g(x) dx = S(g, P)$. Also, $|S(f, P) - S(f, P)| = 0 < \varepsilon$, so $\int_a^b f(x) dx = S(g, P)$. Hence $\int_a^b f(x) dx = \int_a^b g(x) dx$.
