Does this basic integral rule (property) hold true here? Consider a function $f(x)$ which is undefined / indeterminate at $x=a$.  Is it still true that, 

$$\int\limits_a^a f(a)\,\mathrm dx=0~?$$

 A: First note that the only possible partition of $[a,a]$ is $\{a\}$, which is again $a$. Therefore the Riemann sum is $$\sum_{n=1}^1f(a)(a-a)$$
This expression is not defined, so we really can't do anything with it, other than give it some prescribed value. Instead of going this far back though, we can directly assign a value to the original integral (also to not wind up with some inconsistency for discrete sums). The inspiration is that if $g$ is integrable over $[a,b]$ for some $b > a$, and $g$ is not defined at $a$, we have $$\int_{[a,b]} g(x)\,dx = \int_{(a,b]}g(x)\,dx$$
since we don't have to use $g(a)$ in taking the limit of Riemann sums (or in considering supremums/infimums). Therefore it's only sensible to define $\int_{[a,a]}g(x)\,dx$ as $0$, so that we have $$\int_{[a,b]} g(x)\,dx = \int_{[a,a]}g(x)\,dx + \int_{(a,b]}g(x)\,dx$$
and now the additivity of the integral holds in more generality. If such shameless and too-convenient-seeming definitions cause you discomfort (as they do me), note that you could always stop at every place in your theoretic deductions where you might have to integrate over degenerate intervals, and split into cases (where the case that the interval is degenerate most likely simplifies for some other trivial reason). No matter how much you want to avoid conventions, nobody in their right mind would take this approach to Riemann integration.
This mess is easily avoided with Lebesgue integration since we only care about equivalence classes of functions where a small amount of point-values don't matter.
A: I assume you intended to write $f(x)$ inside the integral; for instance
$$\int_0^0 \frac{1}{x}\ dx\ \stackrel{?}{=}\ 0.$$
This is a matter of convention and the specific definition of integration that you are using. I believe that most rigorous definitions of the Riemann integral using tagged partitions etc. will require $f$ to be defined on $[0,0]$, but it wouldn't surprise me if some theorem somewhere requires $\int_0^0 f(x)\,dx = 0$ regardless of $f$ (in analogy to the empty sum).
