Prove r is an outer measure $r(A)=\frac{1}{d(A,Y)}$

Question

Prove r is an outer measure $r(A)=\frac{1}{d(A,Y)}$ where Y is y-axis and $d$ is usual metric on $R^2$. How to prove the condition countable subadditive? I have proved that it satisfied others conditions.

Answer 1

I preassume that in this context $r(A)=+\infty\iff d(A,Y)=0$.

Let $A_n\subseteq\mathbb R^2$ for $n=1,2,\dots$ and $A=\bigcup_{n=1}^{\infty}A_n$.

Now let us assume that $\sum_{n=1}^{\infty}r(A_n)<r(A)\leq+\infty$.

It is enough now to prove that this assumption leads to a contradiction.

If $r(A)<\infty$ then we can find some $a\in A$ with $\sum_{n=1}^{\infty}r(A_n)<r(\{a\})\leq r(A)$.

But from $a\in A$ it follows that $a\in A_n$ for some $n$ and consequently $r(\{a\})\leq r(A_n)$.

On base of this contradiction we conclude that $\sum_{n=1}^{\infty}r(A_n)<r(A)=+\infty$.

$A\cap Y\neq\varnothing$ implies that $A_n\cap Y\neq\varnothing$ for some $n$ hence cannot be true because $r(A_n)<+\infty$.

Combining this with $r(A)=+\infty$ we conclude that a sequence $(a_k)_k$ must exist with $a_k\in A$ for every $k$ and $r(\{a_k\})\to+\infty$.

Let it that $a_k\in A_{n_k}$, so that $r(A_{n_k})\geq r(\{a_k\})$.

Then also $r(A_{n_k})\to+\infty$ contradicting that $\sum_{n=1}^{\infty}r(A_n)<\infty$.

A contradiction is found!