When dealing with a Radon measure, one can assert that the L^{p} spaces are separable. Recall this means that they have a countable dense subset.
Theorem 8.8.1 For p ≥ 1 and μ a Radon measure, L^{p}(ℝ^{n},μ) is separable. Recall this means there exists a countable set, D, such that if f ∈ L^{p}(ℝ^{n},μ) and ε > 0, there exists g ∈D such that f − g_{p} < ε.
Proof: Let Q be all functions of the form cX_{[a,b)}where

and both a_{i}, b_{i }are rational, while chas rational real and imaginary parts. Let Dbe the set of all finite sums of functions in Q.Thus, Dis countable. In fact Dis dense in L^{p}(ℝ^{n},μ).To prove this it is necessary to show that for every f ∈ L^{p}(ℝ^{n},μ), there exists an element of D, s such that s − f_{p} < ε.If it can be shown that for every g ∈ C_{c}

Therefore, assume at the outset that f ∈ C_{c}
Let P_{m} consist of all sets of the form [a,b) ≡∏ _{i=1}^{n}[a_{i},b_{i})where a_{i} = j2^{−m}and b_{i} = (j + 1)2^{−m} for jan integer. Thus P_{m} consists of a tiling of ℝ^{n} into half open rectangles having diameters 2^{−m}n^{}

 (8.15) 
Let

Since f(a_{i}) = 0except for finitely many values of i, the above is a finite sum. Then 8.15 implies s_{m} ∈D.If s_{m} converges uniformly to f then it follows
Since f ∈ C_{c}

and

Therefore, using the triangle inequality, it follows that

and since x is arbitrary, this establishes uniform convergence. ■
Here is an easier proof if you know the Weierstrass approximation theorem.
Theorem 8.8.2 For p ≥ 1 and μ a Radon measure, L^{p}(ℝ^{n},μ) is separable. Recall this means there exists a countable set, D, such that if f ∈ L^{p}(ℝ^{n},μ) and ε > 0, there exists g ∈D such that f − g_{p} < ε.
Proof: Let P denote the set of all polynomials which have rational coefficients. Then P is countable. Let τ_{k} ∈ C_{c}

Corollary 8.8.3 Let Ω be any μ measurable subset of ℝ^{n} and let μ be a Radon measure. Then L^{p}(Ω,μ) is separable. Here the σ algebra of measurable sets will consist of all intersections of measurable sets with Ω and the measure will be μ restricted to these sets.
Proof: Let

and so the countable set