Multivariable Advanced Calculus

9.4. LEBESGUE MEASURE ON R P 217If A k is Lebesgue measurable then ∏ pk=1 A k ∈ F p andm p( p∏k=1A k)=p∏m (A k ) .In addition to this, m p is translation invariant. This means that if E ∈ F p and x ∈ R p ,thenm p (x+E) = m p (E) .The expression x + E means {x + e : e ∈ E} .Proof: m p is regular on B (R p ) by Theorem 7.4.6 because it is finite on compactsets. Then from Theorem 7.5.7, it follows m p is regular on F p . This is because theregularity of m p on the Borel sets and the definition of the completion of a measuregiven there implies the uniqueness part of Theorem 7.5.6 can be used to conclude)(R p , F p , m p ) =(R p , B (R p ), m pNow for E ∈ F p , having finite measure, there exists F ∈ B (R p ) such that m p (F ) =m p (E) and F ⊇ E. Thus m p (F \ E) = 0. By regularity of m p on the Borel sets, thereexists G a countable intersection of open sets such that G ⊇ F and m p (G) = m p (F ) =m p (E) . Thus m p (G \ E) = 0. If E does not have finite measure, then lettingk=1E m ≡ (B (0, m) \ B (0, m − 1)) ∩ E,it follows there exists G m , an intersection of open sets such that m p (G m \ E m ) = 0.Hence if G = ∪ m G m , it follows m p (G \ E) ≤ ∑ m m p (G m \ E m ) = 0. Thus G is acountable intersection of open sets.To obtain F a countable union of compact sets contained in E such that m p (E \ F ) =0, consider the closed sets A m = B (0,m) \ B (0, m − 1) and let E m = A m ∩ E. Thenfrom what was just shown, there exists G m ⊇ (A m \ E m ) such thatm p (G m \ (A m \ E m )) = 0.and ( G m is) the countable intersection of open sets. The set on the inside equalsGm ∩ A C m ∪ (Gm ∩ E m ) . AlsoG C m ⊆ A C m ∪ E m so G C m ∩ A m ⊆ E mand G C m ∩ A m is the countable union of closed sets. Also⎛⎛ ⎞⎞=∅(m p Em \ ( G C )){ }} {⎜⎜m ∩ A m = m p ⎝(E m ∩ G m ) ∪ ⎝E m ∩ A C ⎟⎟m⎠⎠≤ m p((Gm ∩ A C m)∪ (Gm ∩ E m ) ) = 0.Denote this set G C m∩A m by F m . It is a countable union of closed sets and m p (E m \ F m ) =0. Let F = ∪ ∞ m=1F m . Then F is a countable union of compact sets andm p (E \ F ) ≤∞∑m p (E m \ F m ) = 0.m=1Consider the next assertion about the measure of a Cartesian product. By regularityof m there exists B k , C k ∈ B (R p ) such that B k ⊇ A k ⊇ C k and m (B k ) = m (A k ) =