[measure theory] measurable function f and simple function g

[rahl___]

Hi everyone!

my problem:

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

since every simple function is bounded, we at once know, that either is our function f, cause:
$$- \epsilon + g(x) <= f(x) <= \epsilon + g(x)$$, 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 $$\epsilon$$ 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.

 

[rochfor1]

Think about why pointwise limits of measurable functions ought to be measurable. Also, hopefully those simple functions are measurable.