推 znmkhxrw :看到了! 謝謝 09/12 14:32
造個反例：
Let x_n = Σ 1/k^{3/2} k=1 to n.
y_n = Σ 1/k^{3/2} + 1/2n^{3/2} k=1 to (n-1).
where n\in N.
Now define
h(x) = + n^{-1/2} for any x \in [x_{n-1},y_n]
- n^{-1/2} for any x \in [y_n,x_n]
h is L^{\infty} => L^1. So
x
g(x):=∫ h(t) dt
0
is absolutely continuous, g(y_n)= n^{-1/2}1/2n^{3/2}=1/2n^2 and g(x_n)=0.
This implies that \sqrt(g)(y_n)= 1/\sqrt(2)n and \sqrt(g)(x_n)=0. Therefore
\sqrt(g) is not absolutely cont.
Let f(x)=\sqrt(x) and g(x) defined as above. Here we have a counter example.
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