[過去ﾛｸﾞ] 現代数学の系譜 工学物理雑談 古典ガロア理論も読む49 (658ﾚｽ)
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71(1): 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(7/27) AAS
>>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

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72(1): 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) 17:11:49.95 ID:dCRrvhl7(8/27) AAS
>>71 つづき

and
I(n, xn,m) ∩ Kc ⊂ f−1((f(x) − ε, f(x) + ε)).
Hence, f|Kc is continuous at x.

To prove the last part of the theorem, note first that (iii) implies (ii) even
without the restriction that J contains no interval. Now suppose that J contains
no interval and that f,K are as in (ii). Define
(1) G(x) = lim sup t→x,t∈Kc f(t)
and
(2) g(x) = G(x) when G(x) is finite,
　 or = f(x) otherwise.
In particular, it follows from (ii) that f|Kc = g|Kc . Let x ∈ Kc and ε > 0.
According to (ii) there is a δ > 0 such that
(3) |g(y) − g(x)| = |f(y) − f(x)| < ε/2
whenever y ∈ (x − δ, x + δ) ∩Kc. If z ∈ (x − δ, x + δ) ∩K, then the assumption
that K can contain no nonempty open set implies the existence of a sequence
{zn : n ∈ N} ⊂ (x − δ, x + δ) ∩ Kc
such that f(zn) → G(z). Hence, by (3), G(z) is finite, so g(z) = G(z) and
|g(z) − g(x)| ? ε/2 < ε. Therefore, g is continuous at x. QED

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