18/01/01 17:11:49.95 dCRrvhl7.net
>>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
つづく