Inner regularity of the Haar measure Let $G$ be a locally compact Hausdorff topological group, and denote by $B$ the σ-algebra generated by the open subsets of $G$, an element of $B$ is called a Borel set.
A (left) Haar measure $m$ on $G$ is a measure $m$ on $B$ which is:


*

*outer regular on all Borel sets $B$

*inner regular on all open sets $U$

*finite on all compact sets $K$

*invariant under left translations: $m(gB)=m(B)$ for all Borel sets $B$ and all $g$ in $G$

*nontrivial: $m(B)> 0$ for all non-empty open sets $B$.


Wikipedia's Haar measure article says

The left Haar measure satisfies the inner regularity condition for all σ-finite Borel sets, but may not be inner regular for all Borel sets.

Can we conclude that Haar measure $m$ is inner regular on all finite Borel sets $B$, just making use of the conditions above?
Can we conclude that Haar measure $m$ is inner regular on all σ-finite Borel sets if  $m$ is inner regular on all finite Borel sets $B$?
 A: It's possible to imitate the proof of the Riesz-Markov representation theorem found in Rudin's Real and Complex Analyis (page 40). Only the first three conditions are necessary to prove inner regularity for $\sigma$-finite Borel sets.
Step 1: show that $E\cap K$ is inner regular for every Borel set $E$ and every compact set $K$.
$E^c \cap K$ has finite measure, so there exists (because of the outer regularity) an open set $U$ which contains $E^c \cap K$ so that 
$$m(U) \leq m(E^c\cap K) + \varepsilon$$
$m(U \cap K) \leq m(U)$ so we have that
$$ m(K) - m(U \cap K) \geq m(K) - m(E^c \cap K) - \varepsilon $$
Because $m$ is additive, this is equivalent to
$$m(U^c \cap K) \geq m(E \cap K) - \varepsilon$$
$U^c\cap K$ is contained in $E \cap K$ and is compact as a closed subset of $K$, so $E \cap K$ is inner regular.
Step 2: show that $E$ is inner regular if $E$ is a Borel set with finite measure. 
Use outer regularity of $E$ to find an open $U \supseteq E$ with finite measure, inner regularity of $U$ to find a compact $K \subseteq U$ so that $m(U) \leq m(K) + \varepsilon$ and step 1 to find a compact $L \subseteq E \cap K$ so that $m(E \cap K) \leq m(L) + \varepsilon$. No we have
$$m(E) = m(E \cap K) + m(E \setminus K) \leq m(L) + \varepsilon + m(U \setminus K) \leq m(L) + 2\varepsilon$$
so $E$ is inner regular.
Step 3: show that $E$ is inner regular if $E$ is a $\sigma$-finite Borel set
This can be found in this answer.
