2012
DOI: 10.1016/j.apal.2012.05.010
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Algorithmic randomness, reverse mathematics, and the dominated convergence theorem

Abstract: We analyze the pointwise convergence of a sequence of computable elements of L 1 (2 ω) in terms of algorithmic randomness. We consider two ways of expressing the dominated convergence theorem and show that, over the base theory RCA 0 , each is equivalent to the assertion that every G δ subset of Cantor space with positive measure has an element. This last statement is, in turn, equivalent to weak weak König's lemma relativized to the Turing jump of any set. It is also equivalent to the conjunction of the state… Show more

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Cited by 28 publications
(73 citation statements)
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References 16 publications
(44 reference statements)
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“…The standard proof of the Lemma of Kučera relativizes directly to the formulation given above (see for instance the proof of [19, Lemma 6.10.1]). This leads to the following observation (see also [1,Theorem 3.7] for an account of the situation for 2-randomness in reverse mathematics). We recall that a point p ∈ 2 N is called (n + 1)-random for n ∈ N, if p ∈ MLR(∅ (n) ).…”
Section: Probabilistic Properties Of the Baire Category Theoremmentioning
confidence: 80%
See 2 more Smart Citations
“…The standard proof of the Lemma of Kučera relativizes directly to the formulation given above (see for instance the proof of [19, Lemma 6.10.1]). This leads to the following observation (see also [1,Theorem 3.7] for an account of the situation for 2-randomness in reverse mathematics). We recall that a point p ∈ 2 N is called (n + 1)-random for n ∈ N, if p ∈ MLR(∅ (n) ).…”
Section: Probabilistic Properties Of the Baire Category Theoremmentioning
confidence: 80%
“…In particular, y is not a cluster point of (x n ) n and hence there is some s > 0 such that B(y, s) ⊆ B(x, r) and B(y, s) only contains finitely many x n . Since X is perfect, there is some z ∈ B(y, s 2 ) that is different from all these finitely many x n and hence positively bounded away from all x n , i.e., there is some t > 0 such that d(z, x n ) > t for all n ∈ N. This implies z ∈ B(x, r) \ A and since B(x, r) ⊆ B, this is a contradiction to (1). Hence, B(x, r) ⊆ A and hence x ∈ A • .…”
Section: Cluster Points and Boundary Approximationmentioning
confidence: 94%
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“…Proof. It is obvious that over RCA 0 the instance of (3.1) for Π 0 2 sets implies 2−POS, which is the statement introduced by Avigad et al [1] that every Π 0 2 set with positive outer measure is non-empty. By [1, Theorem 3.7 ], RCA 0 + 2−POS ⊢ BΣ 0 2 .…”
Section: Perfect Sets and Reverse Mathematicsmentioning
confidence: 88%
“…A Π 0,X ′ 1 set is also a Π 0,X 2 set. By [1,Proposition 3.4], over RCA 0 + BΣ 0 2 every positive Π 0,X 2 set contains a positive Π 0,X ′ 1 set. Moreover, that every positive Π 0,X ′ 1 set contains a perfect subset where X ranges over all second order elements, trivially implies 2 − WWKL and thus also implies BΣ 0 2 by [1, Theorem 3.7].…”
Section: Perfect Sets and Reverse Mathematicsmentioning
confidence: 99%