Proof of separability of $L^2(\mathbb{R})$ without Stone-Weierstrass Theorem Is there a more elementary proof of the fact that $L^2(\mathbb{R})$ is a separable Hilbert Space? The one I am aware of uses the fact that the set of polynomials is dense in $C([0,1])$ (the Stone-Weierstrass Theorem), but I wonder if something simpler can be done.
 A: If $S$ is the set of all functions of the form $$f=\sum_1^n\alpha_j\chi_{(a_j,b_j)}$$where $\alpha_j$, $a_j$ and $b_j$ are all rational then $S$ is countable and it's not hard to show $S$ is dense in $L^2$.
A: You could use this argument: It's enough to show $L^2[-a,a]$ is separable for each $a>0.$ Each one of those is isomorphic to $L^2[0,2\pi].$ In the last space the exponentials $\{e^{int}:n\in \mathbb Z\}$ form an orthonormal basis. Hence the set of trigonmetric polynomials with rational coefficients, which is countable, is dense in $L^2[0,2\pi].$
A: David Ullrich's comment is the quickest proof I know.  Nonetheless, I'll expand on zhw's answer and show it's not necessary to use the Stone-Weierstrass Theorem to prove the exponentials densely span $L^{2}([0,1])$.  For each $n \in \mathbb{Z}$, let $e_{n} : [0,1] \to \mathbb{C}$ be given by $e_{n}(x) = e^{i 2 \pi n x}$.  I claim that $\text{span}\{e_{n} \, \mid \, n \in \mathbb{Z}\}$ is dense in $L^{2}([0,1])$.  
Recall the definition of the Fourier coefficients of a $L^{2}([0,2 \pi])$ function: if $f \in L^{2}([0,1])$ and $n \in \mathbb{Z}$, then the $n$th Fourier coefficient of $f$ is given by 
$$\hat{f}(n) = \int_{0}^{1} f(x) e^{- i 2 \pi n x} \, dx.$$
Given $N \in \mathbb{N}$, define the Fejer kernel $K_{N} : [0,1] \to \mathbb{C}$ by
$$K_{N}(x) = \sum_{|j| \leq N} \left(1 - \frac{|j|}{N} \right) e^{i 2 \pi j x}.$$
After some calculus, we obtain
$$K_{N}(x) = \frac{1}{N} \left(\frac{\sin(N \pi x)}{\sin(\pi x)}\right)^{2}.$$
In particular, $0 \leq K_{N}(x) \leq C N^{-1} \min\{N^{2},x^{-2}\}$.  Our original definition shows that $\int_{0}^{1} K_{N}(x) \, dx = 1$, and the inequality just stated shows that 
$$\forall \delta > 0 \quad \lim_{N \to \infty} \int_{\{x \in [-1/2,1/2] \, \mid \, |x| > \delta\}} K_{N}(x) \, dx = 0.$$   This proves the family $\{K_{N}\}_{N \in \mathbb{N}}$ is an approximate identity.  Specifically, the following fact is standard:
\begin{align*}
\forall f \in C(\mathbb{T}) \quad K_{N} * f \to f \quad \text{uniformly},
\end{align*}
where $C(\mathbb{T}) = \{f \in C([0,1]) \, \mid \, f(0) = f(1)\}$.  From this, density of $C(\mathbb{T})$ in $L^{2}([0,1])$, and the properties of $\{K_{N}\}_{N \in \mathbb{N}}$, we obtain
\begin{align*}
\forall f \in L^{2}([0,1]) \quad K_{N} * f \overset{L^{2}}\to f.
\end{align*}
However, note that
$$(K_{N} * f)(x) = \sum_{|j| \leq N} \left(1 - \frac{|j|}{N}\right) \hat{f}(j) e^{i 2 \pi j x}.$$
In particular, $K_{N} * f \in \text{span}\{e_{n} \, \mid \, n \in \mathbb{Z}\}$.  This proves $\text{span}\{e_{n} \, \mid \, n \in \mathbb{Z}\}$ is dense in $L^{2}([0,1])$.  
Given $a > 0$, our previous work extends to $L^{2}[-a,a]$ by working with the functions $e_{n}(x) = e^{\frac{i \pi n x}{a}}$ instead.  Now that we know that $L^{2}([-N,N])$ has a countable dense subset for each $N$, we can use
$$\forall f \in L^{2}(\mathbb{R}) \quad f \chi_{[-N,N]} \overset{L^{2}}\to f$$ to conclude.   
