A function that is equal to zero at all points except one is Riemann integrable Let $c$ be inside $(a,b)$ and let $d$ be inside the set of all real numbers. Define $f:[a,b]\rightarrow R$ as
$$f(x):=\cases{d\quad& if $x$ is equal to c\cr 0& if $x$ is not equal to c\cr}\ $$ Prove that $f$ is Riemann integrable and compute  $\displaystyle \int \limits_{a}^{b}f$ using the definition of the integral.

To show it is Reimann integrable I know I have to show that   $\sup  L(p,f) =\inf U(p,f) $  (notation-wise this mean the $\sup$ (lower Darboux Sum) = $\inf$ (upper Darboux Sum)).
I am running into confusion determining these values though. I know $L(p,f)=0$ and I have $U(p,f)=d(b-a)$. Is this correct? If so how do I determine the $\sup$ and $\inf$?
 A: Hint: Details depend on precisely how the Riemann integral is presented, so this can only be a guide. 
Take a partition $\Pi$ of the interval $[a,b]$. Then any Riemann sum $S$ based on $\Pi$ satisfies the inequality 
$$0\le S\le \epsilon d,$$
where $\epsilon$ is the mesh of the partition, that is, the maximum length of the subintervals of $\Pi$.
As $\epsilon\to 0$, $S\to 0$.
A: Some useful things to note:
This also works if a function disagrees with $0$ at finitely many points, just repeat Andre's argument $n$ times, surrounding each point of issue by an interval 
$\frac{\epsilon}{nM}$ where $n$ is the number of points and $M$ is the max of these points (this may be done as it is finite list).
Notice that this stems from a more general issue that if $f \in \mathcal{R}$ and we define a set $ \{x_1, \ldots, x_k \}$, then we may define
$$g(x) : = \begin{cases}
f(x) \qquad x \notin J \\
d_i \qquad x = x_i
\end{cases}$$  
Then we can prove that $g \in \mathcal{R}$ and even that $$\int_{[a,b]} f \, dx = \int_{[a,b]} g \, dx.$$  
In fact, this even works if, say, $f$ and $g$ disagree on all of $\mathbb{Q} \cap [a,b]$!
