Let $f$ be a real-valued function on $[a,b]$ such that $f(x) = 0$ for all $x \neq c_1,...,c_n$. Prove that $f \in R[a,b]$ with $\int_a^b f = 0$ 
Let $f$ be a real-valued function on $[a,b]$ such that $f(x) = 0$ for all $x \neq c_1,...,c_n$. Prove that $f \in R[a,b]$ with $\int_a^b f = 0$.

Case 1: m > 0
Choose partition $P = \{a, c_1 - \frac{\epsilon}{4m}, c_1 + \frac{\epsilon}{4m}, b\}$ where $c_1 \in (a,b)$.
$L(P,f) = 0$ since $f(x) = 0$ for all $x \neq c_1,...,c_n$
$U(P,f) = m_i \Delta x_i = m(c_1 + \frac{\epsilon}{4m} - (c_1 - \frac{\epsilon}{4m})) = m(\frac{\epsilon}{2m}) = \frac{\epsilon}{2} < \epsilon.$
Thus $f \in R[a,b]$
$\int_a^\overline b f\le U(P,f) < \epsilon$ for all $\epsilon > 0$ as $\epsilon \to 0$
$0 = L(P,f) \le \int_\underline a^b f.$
Therefore $$0 = \int_\underline a^b f\le \int_a^\overline b f\le 0$$
Thus $$\int_a^b f = 0$$
Case 2: m = 0
$U(P,f) - L(P,f) = 0 < \epsilon$
Thus $f \in R[a,b]$
$\int_a^\overline b f\le U(P,f) = 0$
$0 = L(P,f) \le \int_\underline a^b f.$
Therefore $$0 = \int_\underline a^b f\le \int_a^\overline b f = 0$$
Thus $$\int_a^b f = 0$$
Is my proof correct and sufficient or did I overdo it?
 A: You don't specify what $m$ is in your problem statement.
In any event,
$$\int_{a}^{b}f=\lim_{\delta\to0}\delta\sum_{i=1}^{n}f(c_{i})=0$$
where the partition is uniform, i.e. $\delta=(b-a)/N$ wherer $N$ is the number of sub-intervals in the partition.
To use the Darboux definition, let $C=\max\;\{f(c_{i})\}$ and $c=\min\;\{f(c_{i})\}$.  Choose a partition $P$ such that $||P||\leq\epsilon$, where $\epsilon$ and the arrangement of sub-intervals in $P$ are both chosen so that each of the $c_{i}$ lie in unique sub-intervals of $P$.  Then for any refinement $P_{\epsilon}$ of $P$ we have
$$U(P_{\epsilon},f)=\sum_{i=1}^{N}\sup_{x\in[x_{i-1},x_{i}]}f(x)(x_{i}-x_{i-1})\leq nC\epsilon\to0$$
and
$$L(P_{\epsilon},f)=\sum_{i=1}^{N}\inf_{x\in[x_{i-1},x_{i}]}f(x)(x_{i}-x_{i-1})\geq nc\epsilon\to0.$$
(Observe that $n$, the number of $c_{i}$, is independent of the partition, i.e. does not grow as $\epsilon\to0$ like $N$ does, the number of actual sub-intervals in the partition.)
The claim now follows from $L(P,f)\leq\int_{a}^{b} f\leq U(P,f)$.
