1525

showing that all the inequalities must be equal signs. Hence, referring to the top and fourthlines above,

µ (Ω) = µ∗ (Ω\F)+µ

∗ (F \E)+µ∗ (E ∩F) .

Subtracting µ∗ (Ω\F) = µ (Ω\F) from both sides gives

µ∗ (S) = µ (F) = µ

∗ (F \E)+µ∗ (E ∩F)≥ µ

∗ (S\E)+µ∗ (E ∩S) ,

This proves the lemma.The next theorem is a major result. It states that the Suslin subsets are measurable under

appropriate conditions. This is sort of interesting because something being a Suslin subsethas to do with topology and this topological condition implies that the set is measurable.

Theorem 47.0.9 Let Ω be a metric space and let (Ω,F ,µ) be a complete Borel measurespace with µ (Ω) < ∞. Denote by µ∗ the outer measure generated by µ. Then if A is aSuslin subset of Ω, it follows that A is µ∗ measurable. Since the original measure space iscomplete, it follows that the completion produces nothing new and so in fact A is in F . SeeProposition 12.1.5.

Proof: We need to verify that

µ∗ (Ω)≥ µ

∗ (A)+µ∗ (Ω\A) .

We know from Corollary 47.0.6, there exists a continuous map, f : NN→ A which is onto.Let

E (k)≡{

n ∈ NN : n1 ≤ k}.

Then E (k) ↑ NN and so from Lemma 47.0.8 we know µ∗ ( f (E (k))) ↑ µ∗ (A) . Therefore,there exists m1 such that

µ∗ ( f (E (m1)))> µ

∗ (A)− ε

2.

Now E (k) is clearly not compact but it is trying to be as far as the first component isconcerned. Now we let

E (m1,k)≡{

n ∈ NN : n1 ≤ m1 and n2 ≤ k}.

Thus E (m1,k) ↑ E (m1) and so we can pick m2 such that

µ∗ ( f (E (m1,m2)))> µ

∗ ( f (E (m1)))−ε

22 .

We continue in this way obtaining a decreasing list of sets, f (E (m1,m2, · · · ,mk−1,mk)) ,such that

µ∗ ( f (E (m1,m2, · · · ,mk−1,mk)))> µ

∗ ( f (E (m1,m2, · · · ,mk−1)))−ε

2k .

Therefore,

µ∗ ( f (E (m1,m2, · · · ,mk−1,mk)))−µ

∗ (A)>k

∑l=1−(

ε

2l

)>−ε.