By Krzysztof Ciesielski, Lee Larson, Krzysztof Ostaszewski
The classical method of exhibiting the parallel among theorems referring to Lebesgue degree and theorems bearing on Baire class at the actual line is specific to units of degree 0 and units of first class. the reason for this is that classical Baire type idea doesn't have an analogue for the Lebesgue density theorem. by utilizing ${\mathcal I}$-density, this deficiency is got rid of, and masses of the constitution of measurable units and features might be proven to exist within the experience of type in addition. This monograph explores classification analogues to things like the density topology, approximate continuity, and density continuity. moreover, a few questions on topological semigroups of genuine services are replied.
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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.