Question 10.T.14: (First Approximation Theorem) For any E ⊆ R, the following s...
(First Approximation Theorem)
For any E ⊆ \mathbb{R}, the following statements are equivalent:
(i) E ∈\mathcal{M}.
(ii) For every ε > 0 there is an open set G ⊇ E such that
m^{∗}(G \setminus E) < ε.
(iii) For every ε > 0 there is a closed set F ⊆ E such that
m^{∗}(E \setminus F ) < ε.
(i)⇒(ii): Suppose m(E) < ∞, and let \left\{I_{i} : i ∈ \mathbb{N}\right\} be an \mathcal{I}-cover of E which satisfies
m(E) + \frac{ε}{2} > \sum\limits_{i=1}^{\infty}{l(I_{i})}.
Assuming I_{i} = [a_{i}, b_{i}), we can choose G = \cup_{i=1}^{∞} J_{i} with J_{i} = ( a_{i} − ε/2^{i+1}, b_{i}). G is clearly open and contains E. Furthermore,
m(G) ≤ \sum\limits_{i=1}^{\infty}{m(J_{i})} = \sum\limits_{i=1}^{\infty}{l(I_{i})} + \frac{ε}{2} < m(E) + ε.
Since both G and E are measurable
m^{∗}(G \setminus E) = m(G \setminus E) = m(G) − m(E) < ε.
If m(E) = ∞ then we set E_{n} = E ∩ (−n, n). Since m(E_{n}) ≤ 2n < ∞ for all n ∈ \mathbb{N}, there is an open set G_{n} such that G_{n} ⊇ E_{n} for each n and
m^{∗}(G_{n} \setminus E_{n}) < \frac{ε}{2^{n}}.
Clearly, the set G = \cup_{i=1}^{∞} G_{n} is open, contains E, and satisfies
G \setminus E = \cup_{i=1}^{∞} (G_{n} \setminus E) ⊆ \overset{\infty}{\underset{n=1}{\cup}} (G_{n} \setminus E_{n}).
Hence
m^{∗}(G \setminus E) ≤ \sum\limits_{n=1}^{\infty}{m^{∗}(G_{n} \setminus E_{n})} < ε.
(ii)⇒(i): For every n ∈ \mathbb{N}, let ε = 1/n and choose the open set G_{n} ⊇ E so that m^{∗}(G_{n}\setminus E) < 1/n. The intersection G = \cap_{n=1}^{∞} G_{n} is clearly measurable, contains E, and satisfies
G \setminus E = \overset{\infty}{\underset{n=1}{\cap}} (G_{n} \setminus E).
Consequently, m^{∗}(G \setminus E) < 1/n for every n ∈ \mathbb{N}, which implies m^{∗}(G \setminus E) = 0. G\E is therefore measurable, hence so is G\(G\E) = E.
(i)⇒(iii): Since E is measurable, so is E^{c}. Given ε > 0, from the equivalence of (i) and (ii), we can find an open set G ⊇ E^{c} such that m^{∗}(G \setminus E^{c}) < ε. Setting F = G^{c}, we see that F is closed, contained in E, and satisfies
E \setminus F = E ∩ G = G \setminus E^{c}
⇒ m^{∗}(E \setminus F) = m^{∗}(G \setminus E^{c}) < ε.
(iii)⇒(i): Since F^{c} \setminus E^{c} = E \setminus F , it is clear that that the set E^{c} satisfies statement (ii) with G = F^{c}. E^{c} is therefore measurable, so E is also measurable.