- #1

rahl___

- 10

- 0

my problem:

If [tex]f:X->R[/tex] is a function that for each [tex]\epsilon > 0[/tex] exists such simple function g satisfying [tex]|f(x)-g(x)| <= \epsilon[/tex] for each [tex]x\in X[/tex], then f is measurable and bounded.

since every simple function is bounded, we at once know, that either is our function f, cause:

[tex] - \epsilon + g(x) <= f(x) <= \epsilon + g(x)[/tex], so that's obviously not the problem here. this whole measure stuff doesn't get into my intuition and I don't have any idea how to try to solve this task. if i knew, that for each [tex]\epsilon[/tex] i could get a measurable function g, it would be obvious, that f is measurable too [wouldn't it?], but can i really always have a measurable function g?

i would be very grateful for any hints. hope my english isn't terrible enough to disturb the sense of this post.

rahl.