Riemann integral on a single point

This is not simply a definition. It follows from the basic definition of the Riemann-Darboux integral.

If $f$ is bounded and by a partition of $\{c\}$ we mean a single degenerate interval $[c,c]$ with length $c-c = 0$, then (vacuously) for any lower and upper Darboux sum we have

$$L(P,f) = \inf_{x \in \{c\}}f(x) (c-c) = 0,$$ $$U(P,f)= \sup_{x \in \{c\}}f(x) (c-c) =0,$$

and, thus,

$$\sup_P L(P,f) = \inf_PU(P,f) =\int_c^c f(x) \,dx = 0$$

Similarly, the Lebesgue integral $\int_E f = 0$ for $E = \{c\}$ or any other zero-measure set. Again, this is not simply an arbitrary definition for the Lebesgue integral. It can be proved from the definition of the Lebesgue integral in terms of a supremum of integrals of simple functions.

Alternate Proof

We can also "prove" this using the change-of variable theorem for Riemann integrals. Take $g:x \in [a,b] \mapsto c$. With $f$ continuous on $g([a,b]) = \{c\}$ and $g' = 0$ we have

$$\int_c ^c f(x) \,dx = \int_{g(a)}^{g(b)} f(x) \, dx = \int_a^b f(g(t))g'(t) \, dt = \int_a^b f(c) \cdot 0 \, dt = 0$$

This is less of a proof and more of a consistency check that defining the integral over $[c,c]$ to be $0$ is not arbitrary.