※ 引述《hau (小豪)》之銘言：
: Let λ denote Lebesgue measure and let f:[0,1]→[0,1] be a differentiable
: function such that for every Lebesgue measurable set A containing in [0,1] one
: has λ(f^(-1)(A))=λ(A). Prove that either f(x)=x or f(x)=1-x (the derivative
: f' is not assumed to be continuous).
這樣可以嗎？
Proof. First, the assumption says that for each closed subinterval [a,b]
of [0,1], we have
λ( f([a,b] ) ＝ λ( f^(-1) ( f([a,b]) ) )
≧ λ([a,b])
since [a,b] is a subset of f^(-1) ( f([a,b]) ). It means that the
measure of [a,b] is less than or equal to the measure of the image
f([a,b]) which is also a closed interval. (＊)
Next, if there exists a point c in [0,1] such that f'(c)＝ 0, then
by definition of differentiability, given ε＝1/3, there exists a
δ＞0 such that as x in (c－δ,c＋δ)∩[0,1]:＝(p,q), we have
|f(x) － f(c)| ＜ 1/3 |x － c|.
It implies
m ( f([p,q]) ) ≦ (2/3)|p－q|. (A)
Moreover, from (＊),
m ( f([p,q]) ) ≧ |p－q|. (B)
Hence, f'(x) ≠ 0 for all x in [0,1] via (A) and (B).
By the Darboux theorem for intermediate theorem of differentiability,
we know that f'(x)＞0 on [0,1] or f'(x)＜0 on [0,1]. That is, f is a
strictly monotone function on [0,1].
Consider two cases:
(1) f is strictly increasing on [0,1], and
(2) f is strictly decreasing on [0,1]
as follows.
(1) If f(x) ＞ x, then with help of (strictly) monotone definition,
the measure of [x,1] is bigger than the measure of [f(x),f(1)]
which is absurd by (＊).
If f(x) ＜ x, then with help of (strictly) monotone definition,
the measure of [0,x] is bigger than the measure of [f(0),f(x)]
which is absurd again by (＊).
So, in this case, we get f(x) ＝ x on [0,1].
(2) Consider g(x) ＝ 1 － f(x), and apply (1), g(x) ＝ x on [0,1].
that is, f(x) ＝ 1 － x on [0,1].
We have proved it. □
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