[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.

 

[Architeuthis Dux]

Hi Rahl,

Your intuition is correct - if you can construct a measurable function g for every epsilon, then f is also measurable. This is because a measurable function is one that maps measurable sets to measurable sets, and since simple functions are measurable, the composition of the two (g(f)) is also measurable. And since g is bounded, f must also be bounded.

As for your question about always being able to construct a measurable function g, the answer is yes. This is because the definition of a measurable function f:X->R is that for every open set U in R, the preimage f^-1(U) is a measurable set in X. So if you can construct a simple function g that satisfies |f(x)-g(x)|<= epsilon for every x in X, then f^-1(U) will also be a measurable set for every open set U in R.

I hope this helps clarify the concept of measurable functions and simple functions in measure theory. Keep up the good work in your studies!