Multivariable Advanced Calculus

14.3. SIMPLE CLOSED RECTIFIABLE CURVES 377is in S. Then S is open because if x ∈ S, then B (x, r) ⊆ Ω for small enough r and ify ∈ B (x, r) , you could go take γ x0x and from x follow the straight line segment joiningx to y. In addition to this, Ω \ S must also be open because if x ∈ Ω \ S, then choosingB (x, r) ⊆ Ω, no point of B (x, r) can be in S because then you could take the straightline segment from that point to x and conclude that x ∈ S after all. Therefore, since Ωis connected, it follows Ω \ S = ∅. Thus for every x ∈ S, there exists γ x0 x, a boundedvariation curve from x 0 to x.Define∫F (x) ≡ f · dγ x0xγ x0 xF is well defined by assumption. Now let l x(x+tek ) denote the linear segment from x tox + te k . Thus to get to x + te k you could first follow γ x0 x to x and from there followl x(x+tek ) to x + te k . HenceF (x+te k ) − F (x)t= 1 t∫ t0= 1 t∫l x(x+tek )f · dl x(x+tek )f (x + se k ) · e k ds → f k (x)by continuity of f. Thus ∇F = f and This proves the theorem. Corollary 14.2.15 Let Ω be a connected open set and f : Ω → R n . Then f has apotential if and only if every closed, γ (a) = γ (b) , bounded variation curve containedin Ω has the property that ∫f · dγ = 0γProof: Using Lemma 14.2.10, this condition about closed curves is equivalent tothe condition that the line integrals of the above theorem are path independent. Thisproves the corollary. Such a vector valued function is called conservative.14.3 Simple Closed Rectifiable CurvesThere are examples of space filling continuous curves. However, bounded variationcurves are not like this. In fact, one can even say the two dimensional Lebesgue measureof a bounded variation curve is 0.Theorem 14.3.1 Let γ : [a, b] → γ ∗ ⊆ R n where n ≥ 2 is a continuous boundedvariation curve. Thenm n (γ ∗ ) = 0where m n denotes n dimensional Lebesgue measure.Proof: Let ε > 0 be given. Let t 0 ≡ a and if t 0 , · · · , t k have been chosen, let t k+1be the first number larger than t k such that|γ (t k+1 ) − γ (t k )| = ε.If the set of t such that |γ (t) − γ (t k )| = ε is nonempty, then this set is clearly closedand so such a t k+1 exists until k is such thatγ ∗ ⊆ ∪ k j=0B (γ (t j ) , ε)