Approximating a continuous function with compact support by simple functions from above?

Question

Let $f\in C_c(X), f:X \to \mathbb{R}$ be a continuous function with compact support on a measure space $X$ with $f \ge 0$.

Allegedly it is possiple to approximate that function by a monotonic decreasing sequence of simple functions $s_n$ of the form $s_n(x) = \sum_j c_j \chi_{K_j}(x)$, where the $K_j$ are compact.

Well, if I divide the co-domain into intervals of the form $[\frac{j-1}{2^n}, \frac{j}{2^n}]$, I get compact sets within the support of $f$ by $f^{-1}([\frac{j-1}{2^n}, \frac{j}{2^n}])$, but then these sets are not disjoint and I can't define a simple function of the form described above.

How can I remedy that?

An idea I have is dividing the co-domain into intervals of the form $(\frac{j-1}{2^n}, \frac{j}{2^n}]$ and then taking $f^{-1}([a, \frac{j}{2^n}])$, where $[a, \frac{j}{2^n}]\subset (\frac{j-1}{2^n}, \frac{j}{2^n}]$ is the largest compact interval that is contained in $(\frac{j-1}{2^n}, \frac{j}{2^n}]$.

Does this work?

Answer 1

There seems to be some issues with the requirement that the partitioned compact sets in the support must be disjoint. For example, take some function whose domain is R. It would be impossible to partition a compact set of R into an arbitrary number disjoint compact sets (as each partition must itself be closed and bounded there is at least some intersection at the boundaries).

Answer 2

After some research, I think I can answer my own question.

The problem I ran into was definition-wise. A simple function is defined as a function that assumes only a finite number of distinct values. The requirement that the $K_j$'s in the representation $s_n(x) = \sum_{j=1}^{n} c_j \chi_{K_j}(x)$ be disjoint is however not part of the definition. Often this is convenient, since the representation of $s_n(x) $ is unique for distinct sets $K_j$. However, if you don't need uniqueness you can just have the sets $K_j$ not necessarily disjoint and find a sequence of those functions that converges to an $f$ pointwise.

So, the big misunderstanding here, for me, was that I thought that for a function to be a simple function the "defining sets" had to be disjoint which is not necessary at all.