\(\star\): Show that for \(E\subseteq \RR^n\), TFAE:
\(E\) is measurable
\(E = H\union Z\) here \(H\) is \(F_\sigma\) and \(Z\) is null
\(E = V\setminus Z'\) where \(V\in G_\delta\) and \(Z'\) is null.
Solution
Let \(m\) denote Lebesgue measure on \(\RR^n\).
1. (1) \(\Rightarrow\) (2).
1.1. For each \(k \geq 1\) there is a closed \(F_k \subseteq E\) with \(m(E \setminus F_k) < 1/k\).
This is inner regularity of Lebesgue measure by closed sets, which holds for every measurable \(E \subseteq \RR^n\): apply outer regularity to \(E^c\) to get an open \(G \supseteq E^c\) with \(m(G \setminus E^c) < 1/k\), and put \(F_k = G^c\).
1.2. Q.E.D.
Put \(H \coloneqq \bigcup_k F_k\) and \(Z \coloneqq E \setminus H\). \(H\) is a countable union of closed sets, hence \(F_\sigma\), and \(E = H \cup Z\). Since \(Z \subseteq E \setminus F_k\) for every \(k\), step 1 gives \(m(Z) < 1/k\) for all \(k\), so \(m(Z) = 0\).
2. (1) \(\Rightarrow\) (3).
2.1. For each \(k \geq 1\) there is an open \(G_k \supseteq E\) with \(m(G_k \setminus E) < 1/k\).
This is outer regularity of Lebesgue measure.
2.2. Q.E.D.
Put \(V \coloneqq \bigcap_k G_k\) and \(Z' \coloneqq V \setminus E\). \(V\) is a countable intersection of open sets, hence \(G_\delta\). Since \(Z' \subseteq G_k \setminus E\) for every \(k\), step 1 gives \(m(Z') < 1/k\) for all \(k\), so \(m(Z') = 0\). Since \(E \subseteq V\), \(V \setminus Z' = V \setminus (V \setminus E) = E\).
3. (2) \(\Rightarrow\) (1).
\(H\) is a countable union of closed, hence measurable, sets, so \(H\) is measurable. \(Z\) is null, hence measurable because Lebesgue measure is complete. So \(E = H \cup Z\) is measurable.
4. (3) \(\Rightarrow\) (1).
\(V\) is a countable intersection of open, hence measurable, sets, so \(V\) is measurable. \(Z'\) is null, hence measurable. So \(E = V \cap (Z')^c\) is measurable.
5. Q.E.D.
Steps 1 and 2 give (1) \(\Rightarrow\) (2) and (1) \(\Rightarrow\) (3), and steps 3 and 4 give the converses.