For any subset $A ⊂ R^d$ (not necessarily measurable), show that there exists a $G_δ$ set $G$ with $A ⊆ G$ and $m∗(A) = m(G).$
Here, $G_δ$ is a countable intersection of open sets. The outer measure associated to $m$ is defined for any set $A⊂ R^d$ by
$m_*(A)=inf \sum|Qj|$, with the infimum taken over all countable coverings, $A⊂\cup Q_j$ by closed cubes.
How does one go about proving this statement, given that $A$ is not necessarily measurable?
We can replace in the definitions closes cubes by open cubes. This is easy to see: Replace a covering of closed cubes $Q_j$ by an open covering $O_j$ such that $O_j$ is the interior of $Q_j$ scaled by a factor $f>1$ around the center of the cube (so you make the cube slightly larger by a factor $f$ to accomodate that border). Then the sum of the volumes of the cubes differs by a factor of $f^d$. By letting $f\to 1$ we thus get the same infimum.
Thus now take a sequence of open coverings $O_j^1,O_j^2,\ldots$ of $A$ so that $$ \sum_j |O_j^i| \xrightarrow{i\to\infty} m_\ast(A) $$
Then we choose $G := \bigcap_i\bigcup_j O_j^i$. Then we have $A\subset G$ and $G$ is a $G_\delta$-set. Also by monotonity of $m_\ast$ we get $$ m_\ast(A) \leq m_\ast(G) \leq m_\ast(\bigcup_j O_j^i) \leq \sum_j |O_j^i| $$ for all $i$, thus also for the infimum over all $i$. But as $$ \inf_i\sum_j|O_j^i| = m_\ast(A)$$ we get $$ m_\ast(A) = m_\ast(G)$$