Sur les residus des integrales doubles by Poincare H.

By Poincare H.

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By Poincare H.

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1. P ∗ ⊂ B. Proof. It is easy to see that any G ∈ P contains an open set H ∈ TO such that H ⊂ G ⊂ cl (H). Thus, G = H ∪(G\H) ∈ B, as G\H ⊂ cl (H)\H ∈ I. 2. The following conditions are equivalent: (i): the set A is superporous at x; and, (ii): given s ∈ (0, 1) there exists Ds > 0 and Rs ∈ (0, 1) such that whenever 0 < D < Ds and (y − δ, y + δ) ⊂ (x − D, x + D) \ {x} with 2δ/D > s, then there is an interval J ⊂ (y − δ, y + δ) ∩ Ac with m(J)/2δ > Rs . I-DENSITY CONTINUOUS FUNCTIONS 31 Proof. Since porosity is translation invariant, it may be assumed that x = 0.

It suffices to show that {x : f (x) ≥ 0} is a Gδ set. To do this, for each p ∈ N, let Up = {x : f (x) > −1/p} and, for p, q, r, k ∈ N, define (17) A(p, q, r, k) = x ∈ R: k−1 k , q q ∩ r (Up − x) = ∅ and q A(p, q, r) = (18) A(p, q, r, k). k=1 It is easy to see that each A(p, q, r) is an open set. Next, define U= (19) A(p, q, r). p∈N q∈N r≥q It is clear that U is a Gδ set. We will show that U = {x : f (x) ≥ 0}. To show that U ⊂ {x : f (x) ≥ 0}, fix p ∈ N and let Vp = A(p, q, r). q∈N r≥q Suppose that 0 ∈ Vp .

If B = ber c such that n∈N (an , bn ) ⊂ [−1, 1] and there exists a positive num- bn − an > c, max{|an |, |bn |} for every n ∈ N, then 0 is not an I-dispersion point of B. Proof. Without loss of generality we may assume that B = n∈N (an , bn ) ⊂ [0, 1]. Put tn = 1/bn for n ∈ N. Then, for every subsequence {tnk }k∈N of {tn }n∈N , (1 − c, 1) ⊂ lim sup (tnk B) . 2(iii), 0 is not an I-dispersion point of B. 28 K. CIESIELSKI, L. LARSON AND K. 3. 3 it was shown that TI ∩ B = TI ∩ B. The purpose of this section is to prove that the family TI = TI ∩ B = TI ∩ B forms a topology on R.

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