Is diameter of a set a measure?

Question

Suppose the diameter of a nonempty set $A$ is defined as

$$\sigma(A) := \sup_{x,y \in A} d(x,y)$$

where $d(x,y)$ is a metric.

Is $\sigma(.)$ a 'measurement'? I.e., how do I prove the countable additivity for this particular case?

Answer 1

Observe that the diameter of singletons is $0$ and the diameter of set $\{x,y\}$ is $d(x,y)>0$ if $x\neq y$. So there is no additivity.

Answer 2

It's not even finitely additive. If $X$ and $Y$ are two disjoint closed intervals on the real line then the diameter of their union is not the sum of their diameters.

Answer 3

... not to mention

$\sigma( \text{rational numbers between A and B}) + \sigma( \text {irrational numbers between A and B}) \ne \sigma( \text{ real numbers between A and B})$.

This is pretty much the perfect example of something that absolutely can not be a measure and illustrates why we need a concept of measure.

Answer 4

Let $A$ be a circle of diameter 1, and let $B$ a circle of diameter 2, having the same center of $A$. Note that $A \subseteq B$.

Now, $\sigma(B) = 2$, yet $\sigma(B \setminus A) + \sigma(A) = 2 + 1 = 3$.