1264 CHAPTER 36. INTEGRATION ON MANIFOLDS

F,G such that G is the countable intersection of open sets each of which has finite measureand F is the countable union of compact sets such that µ (G\F) = 0 and F ⊆ E ⊆G. ThusH n (G\F) = 0,

H n (G) = µ (G) = µ (F) = H n (F)

By completeness of H n it follows E is H n measurable and H n (E) = µ (E) . If E is not offinite measure, consider Er ≡ E ∩B(0,r) . This is contained in the compact set Γ∩B(0,r)and so µ (Er) if finite. Thus from what was just shown, H n (Er) = µ (Er) and so, takingr→ ∞ H n (E) = µ (E) .

This shows you can simply use H n for the measure on Γ.