Are finite sets measurable?

Question

I know that countable sets are measurable. But is this set {1,2,3} measurable ? If so how can I think of this? I'm just struggling to get the concept behind it?

Answer 1

This is lacking quite a lot of of context, but assuming your are talking about Lebesgue algebra, then yes, since all finite sets are closed (union of finitely many closed sets are closed).

Answer 2

A finite set is a countable set. So you instantly get that the result carries over to finite sets.

To really answer your questions anyways, the way you show this (assuming you're working in the reals with the standard Euclidean metric), you can just observe that for any $\varepsilon > 0$, $$ I_n = \left(n - \frac{\varepsilon}{2*3}, n + \frac{\varepsilon}{2*3}\right) \qquad n = 1, 2, 3 $$ is a covering of $\{1,2,3\}$. Hence by countable subadditivity $$ m_*(\{1, 2, 3\}) \le \sum_{n = 1}^{3}\ell(I_n) = \epsilon $$ and since $\epsilon$ is arbitrary, we see that $m_*(\{1, 2, 3\}) = 0$. Since it has measure zero it is measurable. But you should know that finite sets are countable, and if you didn't, well now you know.