Multivariable Advanced Calculus

156 MEASURES AND MEASURABLE FUNCTIONSand since ε is arbitrary, this establishes 2.).Next consider 3.). By definition, there exists a sequence of open intervals, {(a i , b i )} ∞ i=1whose union contains [a, b] such thatµ ([a, b]) + ε ≥∞∑(F (b i −) − F (a i +))i=1By Theorem 7.1.3, finitely many of these intervals also cover [a, b]. It follows there existsfinitely many of these intervals, {(a i , b i )} n i=1 which overlap such that a ∈ (a 1, b 1 ) , b 1 ∈(a 2 , b 2 ) , · · · , b ∈ (a n , b n ) . Therefore,It followsµ ([a, b]) ≤n∑(F (b i −) − F (a i +))i=1n∑(F (b i −) − F (a i +)) ≥ µ ([a, b])i=1Since ε is arbitrary, this shows≥≥n∑(F (b i −) − F (a i +)) − εi=1F (b+) − F (a−) − εµ ([a, b]) ≥ F (b+) − F (a−)but also, from the definition, the following inequality holds for all δ > 0.µ ([a, b]) ≤ F ((b + δ) −) − F ((a − δ) +) ≤ F (b + δ) − F (a − δ)Therefore, letting δ → 0 yieldsThis establishes 3.).Consider 4.). For small δ > 0,Therefore, from 3.) and the definition of µ,µ ([a, b]) ≤ F (b+) − F (a−)µ ([a + δ, b − δ]) ≤ µ ((a, b)) ≤ µ ([a, b]) .F ((b − δ)) − F ((a + δ)) ≤ F ((b − δ) +) − F ((a + δ) −)Now letting δ decrease to 0 it follows= µ ([a + δ, b − δ]) ≤ µ ((a, b)) ≤ F (b−) − F (a+)F (b−) − F (a+) ≤ µ ((a, b)) ≤ F (b−) − F (a+)This shows 4.)Consider 5.). From 3.) and 4.), for small δ > 0,F (b+) − F ((a + δ))≤ F (b+) − F ((a + δ) −)= µ ([a + δ, b]) ≤ µ ((a, b])≤µ ((a, b + δ)) = F ((b + δ) −) − F (a+)≤ F (b + δ) − F (a+) .