Lebesgue outer measure of a finite set

I was trying to work out the Lebesgue outer measure of a finite set.

I took a simple example, considered the finite set $$\{a,b\}$$.

Then, I wrote $$\{a,b\}$$ as $$\{a\} \cup \{b\}$$.

Since the Lebesgue outer measure of singleton sets is zero, therefore, it would probably turn out that the Lebesgue outer measure of the set $$\{a,b\}$$ that I chose is also zero as it is the union of two singleton sets.

To show this i assumed that if $$\{a,b\}$$ = $$\{a\} \cup \{b\}$$ then $$m^*\{a,b\} \le m^*\{a\} + m^*\{b\}$$, is there any property of lebesgue outer measure that says this ?

Also, I think that the same concept (of proving outer measure zero) can be extended to any countably finite or even infinite set?

Am I correct? I have just started learning this so I am just trying to get the basics right.

• That comes frome the definition of outer measure – Lolman Aug 20 '14 at 15:25
• About "...any finite or even infinite set", only to countable setes. – Martín-Blas Pérez Pinilla Aug 20 '14 at 15:30

Notice that if $A$ is finite (or at most denumerable), we have $A = \{ a_1, \ldots, a_n, \ldots \}$, and hence $$0 \leq \mathfrak{m}^*A = {\frak m}^*\left( \bigcup_{n \geq 1} \{a_n\} \right) \leq \sum_{n \geq 1} {\frak m}^*\{a_n\} = 0$$ by subadditivity of $\frak m^*$, so ${\frak m}^*A = 0$.