(Rudin) $m(E) = 0$ for all Borel sets $E$ in $G$

Question

Let $G$ be a locally compact abelian group with Haar measure $m$.

Recall that $m$ is a Haar measure on $G$ if $m$ is a non-zero regular measure on $G$ which is translation invariant.

The following is extracted from 'Fourier Analysis on Groups' by Rudin, page $2:$

If $V$ is a non-empty open set of $G,$ then $m(V)>0.$

Proof: For if $m(V) = 0$ and $K$ is compact, finitely many translates of $V$ cover $K$ and hence $m(K) =0.$ The regularity of $m$ then implies that $m(E) = 0$ for all Borel sets $E$ in $G,$ a contradiction.

Question: How to conclude using regularity of $m$ that if $m(K) = 0$ for any compact set, then $m(E) = 0$ for all Borel sets $E$ in $G$?

Recall that $\mu$ is regular if if for every Borel subset $E$ of $G,$ \begin{align*} |\mu|(E) & = \sup\{|\mu|(K): K \subseteq E, K \text{ is compact} \} \\ & = \inf\{ |\mu|(O): O \supseteq E, O \text{ is open} \}. \end{align*}

In the definition, the compact set $K$ is a subset of $E,$ while in the proof $K$ contains $V.$

Answer 1

The proof works for arbitrary compact sets $K$. Let $x_0\in V$, then $$ K= \bigcup_{k\in K} (V+(k-x_0)).$$ Using compactness you can cover $K$ with finitely many translations of $V$ (by translation-invariance they have still measure zero). Therefore, $K$ has measure zero.