Question 10.T.15: (Second Approximation Theorem) Let E ⊆ R. If m^∗(E) 0, ther...
(Second Approximation Theorem)
Let E ⊆ \mathbb{R}. If m^{∗}(E) < ∞ then E ∈ \mathcal{M} if and only if, for every ε > 0, there is a set A ∈ \mathcal{E} such that m^{∗}(E Δ A) < ε.
Suppose E ∈ \mathcal{M} and ε > 0. Then, by Theorem 10.14, there is an open set G ⊇ E such that
m(G \setminus E) < \frac{ε}{2}. (10.14)
Since G is open, it can be represented by a union of disjoint open intervals \left\{I_{i} : i ∈ \mathbb{N}\right\}. Thus G = \cup_{i=1}^{∞} I_{i}, and, since m(E) < ∞, we can use countable additivity and (10.14) to write
\sum\limits_{i=1}^{\infty}{m(I_{i})} = m(G) < ∞.
Therefore there is a positive integer n such that
m(G) − m (\cup_{i=1}^{n} I_{i}) < \frac{ε}{2}. (10.15)
Setting I_{i} = (a_{i}, b_{i}), A_{0} = \cup_{i=1}^{n} I_{i}, and A = \cup_{i=1}^{n} [a_{i}, b_{i}), we note that A ∈ \mathcal{E} and differs from A_{0} by a finite number of points. Hence m(A) = m(A_{0}) < m(G) < ∞. Now the inclusion relation E \setminus A ⊆ G \setminus A_{0} and (10.15) imply
m(E \setminus A) ≤ m(G \setminus A_{0}) = m(G) − m(A_{0}) < \frac{ε}{2},
while the inclusion A_{0} \setminus E ⊆ G \setminus E and (10.14) imply
m(A \setminus E) = m(A_{0}\setminus E) ≤ m(G \setminus E) < \frac{ε}{2}.
Thus m(E Δ A) < ε.
Conversely, given m^{∗}(E) < ∞, let ε > 0 and choose A = \cup_{i=1}^{n} [a_{i}, b_{i}) ∈ \mathcal{E} so that m^{∗}(E Δ A) < ε. To prove that E is measurable, it is sufficient, by Theorem 10.14, to prove the existence of an open set G ⊇ E such that m^{∗}(G \setminus E) < 3ε. Let A_{1} = \cup_{i=1}^{n} (a_{i} − ε/2^{i}, b_{i}). Then A_{1} is an open set which satisfies
m^{∗}( E Δ A_{1}) ≤ m^{∗}(E \setminus A_{1}) + m^{∗}(A_{1} \setminus E)
< m^{∗}(E \setminus A) + m^{∗}(A \setminus E) + ε < 2ε. (10.16)
Furthermore, from the definition of m^{∗}(E \setminus A_{1}), there is a cover \left\{I_{i} ∈ \mathcal{I} : i ∈ \mathbb{N}\right\} of E \setminus A_{1} such that
m^{∗}(E \setminus A_{1}) + \frac{ε}{2} >\sum\limits_{i=1}^{\infty}{l(I_{i})}.
Following the procedure in the proof of Theorem 10.14, we can construct an open set A_{2} that contains E \setminus A_{1} and satisfies
m^{∗}(A_{2}) ≤ \sum\limits_{i=1}^{\infty}{l(I_{i})} + \frac{ε}{2} < m^{∗}(E \setminus A_{1}) + ε. (10.17)
By defining G = A_{1} ∪ A_{2}, we see that G ⊇ E and G \setminus E ⊆ A_{2} ∪ (A_{1} \setminus E).
Hence, using (10.16) and (10.17),
m^{∗}(G \setminus E) ≤ m^{∗}(A_{2}) + m^{∗}(A_{1} \setminus E).
< m^{∗}(E \setminus A_{1}) + m^{∗}(A_{1} \setminus E) + ε.
= m^{∗}(E Δ A_{1}) + ε = 3ε.