現代数学の系譜工学物理雑談古典ガロア理論も読む49

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70: 現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE  [sage] 2018/01/01(月) 17:10:50.85 ID:dCRrvhl7

>>69 つづき

http://www.math.wvu.edu/~kcies/prepF/BookIdensity/BookIdensity.pdf
I-DENSITY CONTINUOUS FUNCTIONS Krzysztof Ciesielski他 1994- 被引用数: 84
(抜粋)
CHAPTER 1
The Ordinary Density Topology
1.1. A Simple Category Topology

To gain some insight into what is happening with limits like this, it is useful
to generalize this idea to a topological setting.
A nonempty family J ⊂P(X) of subsets of X is an ideal on X if A ⊂ B and
B ∈ J imply that A ∈ J and if A∪B ∈ J provided A,B ∈ J. An ideal J on X
is said to be a σ-ideal on X if ∪n∈N An ∈ J for every family {An : n ∈ N} ⊂ J.
Let J be an ideal on R and Ｔo be the ordinary topology on R. The set
T (J) = {G ＼ J : G ∈ Ｔo, J ∈ J}
is a topology on R which is finer than Ｔo. The following proposition is evident
from the definitions.

Proposition 1.1.1. Let J be a σ-ideal on R and T (J ) be as above. For
f : (R, T (J )) → (R, Ｔo) and x0 ∈ R the following statements are equivalent to
each other.
(i): f is continuous at x0.
(ii): Given ε > 0 there is a δ > 0 such that
{x ∈ (x0 − δ, x0 + δ) : |f(x) − f(x0)| ≧ ε} ∈ J.

つづく

http://rio2016.5ch.net/test/read.cgi/math/1514376850/70

71: 現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE  [sage] 2018/01/01(月) 17:11:20.87 ID:dCRrvhl7

>>70 つづき

Theorem 1.1.4. Let J be a σ-ideal and f : R → R. The following statements
are equivalent.
(i): The function f is J -continuous J -a.e.
(ii): There exists a set K ∈ J such that the restricted function f|Kc is continuous.

Furthermore, if the ideal J contains no interval, then the following statement is
equivalent to (i) and (ii)
(iii): There exists a function g : R → R such that f = g J -a.e. and g is continuous in the ordinary sense J -a.e.

Proof. The fact that (ii) implies (i) is obvious. Suppose (i) is true and let
f be J -continuous on a set M = Jc for J ∈ J. For each n ∈ N and x ∈ M,
by Proposition 1.1.1(ii) there is an open interval I(n, x) and a J(n, x) ∈ J such
that
x ∈ I(n, x) \ J(n, x) ⊂ f−1((f(x) − 1/n, f(x) + 1/n)).
For each fixed n, there must be a countable sequence xn,m ∈ M such that
M ⊂∪m∈N I(n, xn,m).
Let
K = J ∪ ∪ n,m∈N J(n, xn,m) ∈ J.
If x ∈ Kc and ε > 0, then there must exist natural numbers n and m such that
2/n < ε and x ∈ I(n, xn,m). Then |f(x) − f(xn,m)| < 1/n so that
f(x) ∈ (f(xn,m) − 1/n, f(xn,m) + 1/n) ⊂ (f(x) − ε, f(x) + ε)

つづく

http://rio2016.5ch.net/test/read.cgi/math/1514376850/71

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