Density of $\mathcal{L}^2(\mathbb{R})^+\cap\mathcal{L}^{\infty}(\mathbb{R})^+$ in $\mathcal{L}^{\infty}(\mathbb{R})^+$ I am getting in a world of confusion here. One of my problems is the nomenclature for topologies on $\mathcal{L}^{\infty}(\mathbb{R})$ (or rather $B(H)$ for $H$ a Hilbert space) so straight away I want to talk about the initial topology $\tau$ on $\mathcal{L}^{\infty}(\mathbb{R})$ that makes the family of functions
$$\left\{\langle g|:\mathcal{L}^{\infty}(\mathbb{R})\rightarrow \mathbb{C},\quad f\mapsto \int_\mathbb{R}\overline{g}f\,d\lambda\quad\right|\left.\quad g\in\mathcal{L}^1(\mathbb{R})\right\},$$
continuous.
Here $\mathcal{L}^\infty(\mathbb{R})$ is the set of essentially bounded, Lesbesgue measurable functions, $\mathbb{R}\rightarrow \mathbb{C}$, and $\mathcal{L}^p(\mathbb{R})$ functions such that $\displaystyle \int_{\mathbb{R}}|f|^p\,d\lambda<\infty$.
My understanding is that with respect to the topology above, that $\mathcal{L}^2(\mathbb{R})^+\cap \mathcal{L}^{\infty}(\mathbb{R})^+$ is dense in $\mathcal{L}^{\infty}(\mathbb{R})^+$.
My usual approach to such a problem is given a set $A$ in which $B$ is claimed to be dense, I take an arbitrary $a\in A$ and $\varepsilon>0$, and construct a $b\in B$ such that:
$$d(a,b)<\varepsilon,$$
for $d:A^2\rightarrow \mathbb{R}^+$ a metric. My understanding is that $\mathcal{L}^{\infty}(\mathbb{R})\subset B(\mathcal{L^2}(G))$ (via multiplication) and my understanding is that topologies on bounded operators on Hilbert spaces are usually not metrisable so I am already out of my comfort zone.
Initially my intuition told me this was impossible. Sure how could something square integrable be 'close' to the constant function $\mathbb{1}_{\mathbb{R}}$?
But then as I played around for a little and noted that $\langle e^{-x^2}|\mathbf{1}_\mathbb{R}=\sqrt{\pi}$. From this I made a little open set in $\mathbb{C}$, $B_{\varepsilon}=\{z\in\mathbb{\mathbb{C}}:|z-\sqrt{\pi}|<\varepsilon\}$, and we know that the preimage of $B_{\varepsilon}$ under $\langle e^{-x^2}|$ is open.
Furthermore, for a fixed $a>0$, the indicator function $\mathbb{1}_{[-a,a]}\in\mathcal{L}^2(\mathbb{R})\cap \mathcal{L}^{\infty}(\mathbb{R})^+$ and as $\langle e^{-x^2}|\mathbb{1}_{[-a,a]}=\sqrt{\pi}\operatorname{erf}(a)$, and $\operatorname{erf}(a)\rightarrow 1$ as $a\rightarrow \infty$, we can get for finite $a$,
$$\mathbf{1}_{[-a,a]}\in \left(\langle e^{-x^2}|\right)^{-1}(B_{\varepsilon}),$$
that is in a neighbourhood of $\mathbb{1}_{\mathbb{R}}$.
However, this feels like a cheat, surely I have to find an $\mathcal{L}^2(\mathbb{R})\cap \mathcal{L}^{\infty}(\mathbb{R})^+$ function in every neighbourhood of $\mathbb{1}_{\mathbb{R}}$? 
I don't really know how close this is to a full proof. I am happy that my initial intuition is incorrect and this density is a lot "looser" (trying not to prejudice one of the questions below!), and the result true, but fear I might have major gaps in my understanding. I am afraid to use something like the density of $C_0(\mathbb{R})$ in $\mathcal{L}^1(\mathbb{R})$ as I am not sure what topology that is in...
So, the questions:

  
*
  
*What is the topology above commonly referred to as? See the tags to see the kind of contexts I am happy to hear answers within.
  
*Is this anywhere close to a proof of the full result (as given by the title of the question). Do I need to construct the $\mathcal{L}^{2}(\mathbb{R})\cap \mathcal{L}^{\infty}(\mathbb{R})^+$ function for an arbitrary $\mathcal{L}^1(\mathbb{R})$ function? Or for every possible $\mathcal{L}^1$ function simultaneously... do I need to find a base of the topology $\tau$ and can such a base be inferred from the "initialness" of the definition?

The slightly longer game is to understand how $\displaystyle \int_{\mathbb{R}}$ is what is called a semi-finite Haar weight on $\mathcal{L}^{\infty}(\mathbb{R})$.
 A: This topology is called the ultraweak or $\sigma$-weak or weak$^\ast$ topology on $L^\infty$. A net $(f_i)_{i\in I}$ converges to $f$ in this topology if and only if $\int f_i g\,d\lambda\to\int fg\,d\lambda$ for all $g\in L^1$. Some people will not like this, but I find this characterization much more useful in practice than the definition as initial topology.
If you don't feel comfortable with nets, that doesn't matter to much, because this topology is metrizable on norm bounded subsets of $L^\infty$ (as long as $L^1$ is separable, which is the case for $L^1(\mathbb{R})$. In any case, to check that $L^2_+\cap L^\infty_+$ is dense in $L^\infty_+$, it suffices to find for every $f\in L^\infty_+$ a sequence $(f_n)$ in $L^2_+\cap L^\infty_+$ such that $\int f_n g\to \int f g$ for all $g\in L^1$. In particular, the sequence $(f_n)$ must be simultaneously "good" for all $g\in L^1$.
This is not a problem in your example (and the function $x\mapsto \exp(-x^2)$ does not play a special role). Whenever $g\in L^1$, then $\int 1_{[-n,n]}g\to\int 1_{\mathbb{R}} g$ by the dominated convergence theorem.
In fact, you are almost done with the proof. The positive linear combinations of characteristic functions are uniformly dense in $L^\infty_+$ (this is standard measure theory). Hence it is sufficient to show that you can approximate arbitrary characteristic functions by characteristic functions of sets with finite measure. That works pretty similar to the example above.
