Multivariable Advanced Calculus

240 THE LEBESGUE INTEGRAL FOR FUNCTIONS OF P VARIABLESTo see this, let x ∈ QK ε . Then there exists k ∈ QK such that |k − x| < ε. Therefore,|(x 1 , · · · , x p−1 ) − (k 1 , · · · , k p−1 )| < ε and |x p − k p | < ε and so x is contained in the seton the right in the above inclusion because k p = 0. However, the measure of the set onthe right is smaller than[2 (diam (QK) + ε)] p−1 (2ε) = 2 p [(diam (K) + ε)] p−1 ε.This proves the lemma. Note this is a very sloppy estimate. You can certainly do much better but thisestimate is sufficient to prove Sard’s lemma which follows.Definition 9.9.7 If T, S are two nonempty sets in a normed vector space,dist (S, T ) ≡ inf {||s − t|| : s ∈ S, t ∈ T } .Lemma 9.9.8 Let h be a C 1 function defined on an open set, U ⊆ R p and let K bea compact subset of U. Then if ε > 0 is given, there exists r 1 > 0 such that if |v| ≤ r 1 ,then for all x ∈ K,|h (x + v) − h (x) − Dh (x) v| < ε |v| .Proof: Let 0 < δ < dist ( K, U C) . Such a positive number exists because if thereexists a sequence of points in K, {k k } and points in U C , {s k } such that |k k − s k | → 0,then you could take a subsequence, still denoted by k such that k k → k ∈ K and thens k → k also. But U C is closed so k ∈ K ∩ U C , a contradiction. Then for |v| < δ itfollows that for every x ∈ K,x+tv ∈ Uand|h (x + v) − h (x) − Dh (x) v||v|≤≤∣ ∫ 10 Dh (x + tv) vdt − Dh (x) v ∣ ∣∣∫ 10|v||Dh (x + tv) v − Dh (x) v| dt.|v|The integral in the above involves integrating componentwise. Thus t → Dh (x + tv) vis a function having values in R p ⎛Dh 1 (x+tv) v⎞⎜⎝.Dh p (x+tv) v⎟⎠and the integral is defined by⎛⎜⎝∫ 10 Dh 1 (x+tv) vdt.∫ 10 Dh p (x+tv) vdt⎞⎟⎠Now from uniform continuity of Dh on the compact set, {x : dist (x,K) ≤ δ} it followsthere exists r 1 < δ such that if |v| ≤ r 1 , then ||Dh (x + tv) − Dh (x)|| < ε for everyx ∈ K. From the above formula, it follows that if |v| ≤ r 1 ,|h (x + v) − h (x) − Dh (x) v||v|≤∫ 10|Dh (x + tv) v − Dh (x) v| dt|v|