12.11. COMPLETION OF MEASURES 327

If F ⊇ E and F ∈F , then µ (F)≥ µ (E) and so µ (E) is a lower bound for all such µ (F)which shows that

µ (E)≡ inf{µ (F) : F ∈F and F ⊇ E} ≥ µ (E) .

This verifies 2.Next consider 3. Let E ∈F and let S be a set. I must show

µ (S)≥ µ (S\E)+µ (S∩E) .

If µ (S) = ∞ there is nothing to show. Therefore, suppose µ (S)< ∞. Then from the defini-tion of µ there exists G⊇ S such that G ∈F and µ (G) = µ (S) . Then from the definitionof µ,

µ (S) ≤ µ (S\E)+µ (S∩E)

≤ µ (G\E)+µ (G∩E)

= µ (G) = µ (S)

This verifies 3.Claim 4 comes by the definition of µ as used above. The only other case is when

µ (S) = ∞. However, in this case, you can let G = Ω.It only remains to verify 5. Let the Ωn be as described above and let E ∈F such that

E ⊆Ωn. By 4 there exists H ∈F such that H ⊆Ωn, H ⊇Ωn \E, and

µ (H) = µ (Ωn \E) . (12.11.53)

Then let F ≡Ωn∩HC. It follows F ⊆ E and

E \F = E ∩FC = E ∩(H ∪Ω

Cn)

= E ∩H = H \ (Ωn \E)

Hence from 12.11.53µ (E \F) = µ (H \ (Ωn \E)) = 0.

It followsµ (E) = µ (F) = µ (F) .

In the case where E ∈F is arbitrary, not necessarily contained in some Ωn, it followsfrom what was just shown that there exists Fn ∈F such that Fn ⊆ E ∩Ωn and

µ (Fn) = µ (E ∩Ωn) .

Letting F ≡ ∪nFn

µ (E \F)≤ µ (∪n (E ∩Ωn \Fn))≤∑n

µ (E ∩Ωn \Fn) = 0.

Therefore, µ (E) = µ (F) and this proves 5. This proves the theorem.Now here is an interesting theorem about complete measure spaces.