[measure theory] measurable function f and simple function g

In summary, The conversation discusses a function f:X->R and its measurability and boundedness. It is stated that if for each \epsilon > 0, there exists a simple function g satisfying |f(x)-g(x)| <= \epsilon for each x\in X, then f is measurable and bounded. The conversation also mentions the concept of pointwise limits of measurable functions and the measurability of simple functions. The poster also expresses difficulty understanding the topic and asks for any hints or clarification.
  • #1
rahl___
10
0
Hi everyone!

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.
 
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  • #2
Think about why pointwise limits of measurable functions ought to be measurable. Also, hopefully those simple functions are measurable.
 
  • #3


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!
 

FAQ: [measure theory] measurable function f and simple function g

1. What is a measurable function?

A measurable function is a function between two measurable spaces where the pre-image of any measurable set in the codomain is a measurable set in the domain. In simpler terms, it is a function that preserves the structure of measurable sets.

2. How do you determine if a function is measurable?

To determine if a function is measurable, you need to check the pre-image of every measurable set in the codomain and see if it is a measurable set in the domain. If this condition is satisfied, the function is considered measurable.

3. What is a simple function?

A simple function is a function that takes on a finite number of values over a measurable space. It can be written as a linear combination of indicator functions, where each indicator function takes on only two values (0 or 1).

4. How is a simple function related to a measurable function?

A measurable function can be approximated by a sequence of simple functions. This is known as the simple function approximation theorem. In other words, a measurable function can be broken down into simpler, finite-valued functions.

5. How is the Lebesgue integral of a measurable function calculated?

The Lebesgue integral of a measurable function is calculated by taking the supremum of the integrals of all simple functions that approximate the measurable function from below. This is known as the Lebesgue integral of the measurable function and is denoted by ∫f dμ.

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