Are Simple Functions Dense in Bounded Borel Functions on a Compact Space?

[Fermat1]

Let K be a compact space and let B be the space of bounded borel functions on K equipped with the supremum norm. Show that simple functions (i.e. functions attaining only a finite number of values) are dense in B.

Thanks

 

[Jameson]

Let K be a compact space and let B be the space of bounded borel functions on K equipped with the supremum norm. Show that simple functions (i.e. functions attaining only a finite number of values) are dense in B.

Thanks

Hi Fermat,

Please make an effort and show us what you've done.

Thanks.

 

[Fermat1]

I know I need to show $||f_{n}-f||->0$ where the $f_{n}$ are simple but I don't know where to start

 

[ThePerfectHacker]

I think this would do it,
Show simple functions are dense

 

[blue_raver22]

for bringing up this topic. Simple functions are indeed dense in the space of bounded Borel functions on a compact space. This means that any bounded Borel function on a compact space can be approximated by a sequence of simple functions.

To prove this, let f be a bounded Borel function on K. Since K is compact, f is also bounded. By the definition of boundedness, there exists a constant M such that |f(x)| ≤ M for all x∈K.

Now, consider the set of all possible values that f can take on K, denoted by V = {f(x) | x∈K}. Since f is bounded, V is a finite set. Let n be the cardinality of V.

Next, let ε > 0 be given. We will construct a simple function g such that ||f - g|| < ε.

Consider the partition P = {A1, A2, ..., An} of K, where Ai = {x∈K | f(x) = vi}, i=1,2,...,n. In other words, each Ai contains all the points in K where f takes on the value vi.

Now, define g as follows: g(x) = vi for all x∈Ai. In other words, g is a constant function on each Ai. Since f is bounded, g is also bounded and attains only a finite number of values, making it a simple function.

We claim that ||f - g|| < ε. To prove this, let x∈K be arbitrary. If x∈Ai for some i, then |f(x) - g(x)| = |f(x) - vi| = 0 < ε. If x∉Ai for all i, then x∈Aj for some j≠i. In this case, |f(x) - g(x)| = |f(x) - vj| ≤ M < ε.

Thus, for any x∈K, we have |f(x) - g(x)| < ε, which implies that ||f - g|| < ε. This shows that g is an ε-approximation of f and hence, simple functions are dense in B.

This result is important in many areas of mathematics, especially in functional analysis and measure theory. It allows us to approximate more complicated functions by simpler ones, making it easier to study and analyze