Riemann integral on a single point

Question

Let $f$ be a continuous function over $\mathbb{R}$. Let $c\in(-\infty,\infty)$. Then, it is believed that $\int_{c}^{c}f(\xi)d\xi=0$ in the sense of Riemann integration. Is this just a definition? But the definition of the Riemann integral on wikipedia or on text book always start with a closed interval [a,b] with $b>a$ and then partition the interval. So, the definition of Riemann integral does not cover the case $a=b$. So, $\int_{c}^{c}f(\xi)d\xi=0$ seems to be a property of a Riemann integral inferred by its definition. Can anyone tell me if $\int_{c}^{c}f(\xi)d\xi=0$ is a definition or not and show me some proof?

Answer 1

According to Apostol's Calculus Vol. 1 (second edition) page 74, it is a definition that $\int_{c}^{c}f(x)dx=0$.

If the integral on a single point hasn't been defined, one cannot prove what its value is since it's meaningless.

Note:

It is reasonable to define $\int_{c}^{c}f(x)dx=0$ because it is clear that the area of the line segment between the point $(c, 0)$ and $(c, f(c))$ is zero. Moreover, if we want to define it so that the theorem of additivity $$ \int_{a}^{b}f(x)dx+\int_{b}^{c}f(x)dx=\int_{a}^{c}f(x)dx$$ remains valid, then we must have $$ \int_{a}^{c}f(x)dx+\int_{c}^{c}f(x)dx=\int_{a}^{c}f(x)dx,$$ so $\int_{c}^{c}f(x)dx=0$.

Answer 2

Yes, $\int_c^cf(\xi)d\xi=0$ is a definition. As you pointed out, the standard definition of the Riemann integral does not cover this case. However, using the Riemann-Stieltjes integral we can define integration on a single point, in which case the equality $\int_c^cf(\xi)d\xi=0$ becomes a theorem.

Answer 3

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.