Convex cone of nonnegative functions in L2 has empty interior 
Convex cone $S:=\{f\in L^2(\mathbb{R},\mu):f\geq 0\}$ has empty
  interior in $L^2(\mathbb{R},\mu)$ when $\mu$ is Lebesgue measure.

I wanted to prove it but i have major holes in my knowledge of measure and integral theory.  
So, I want to prove that for every $f\in S$ and $\varepsilon >0$ exists some $g\in L^2(\mathbb{R},\mu)$ such that $||f-g||_2<\varepsilon$ and $g$ has at least one negative value. I am a little bit confused with this also, is one negative value enough or i need a set of negative values with positive measure? (there is no a.e. in definition of $S$) 
So, if only one negative value is enough i can 
define function $g$ to be same as $f$ everywhere except in some point $x_0$ and there I put $g(x_0)$ has negative value. Then $g$ would not be nonnegative and $||f-g||_2=\sqrt{\int_{\mathbb{R}}^{}|f-g|^2d\mu}=0$ since one point doesnt effect the integral. Is this ok?
What conclusions can i make of $f\in S$ in general? I know it must be bounded. Does $f$ goes to zero when $x\rightarrow\pm\infty$? Is there a type of statement that says something like: $\forall\varepsilon >0$ there exists $E\in\mathcal{B}(\mathbb{R}), \mu(E)>0$ so that $\int_{E}^{}f<\varepsilon$. And if so must it be $\mu(E)=\infty$?
I would really appreciate your comments and helpful tips. This is not any type of homework, I graduated years ago, I just never really grasped whole Lebesgue integration and everything that goes with it and i am looking for helpful tips. I apologize in front for my language mistakes since I'm not a native speaker.
 A: As pointed out in comments, you can't just set one value of $g$ to be negative and set $g = f$ everywhere else, because in order to define $L^p$ you make equivalence classes of functions that agree everywhere except a measure-zero set. So then you would essentially have $g = f$ under this paradigm, so this would not show the interior of your set is empty. Try the following: For a given $\epsilon > 0$, take one value of your function $f$ and make it some negative value to start to define $g$, and then make the function $g$ return to the original values of $f$ nearby on both sides in a continuous fashion, say with piecewise linear interpolators. Then as long as you get $g$ to return back to $f$ "fast enough" the $L^2$ norm of $|f - g|$ will be less than $\epsilon$.
A: Take $f \in L^2$, $f \geq 0$. Define $A_\varepsilon = \{ x \in \Bbb{R} : f(x) \geq \varepsilon\}$ for every $\varepsilon >0$. Now
$$
\mu(A_\varepsilon) < \infty\,,
$$
hence $\mu( \Bbb{R} \setminus A_\varepsilon) = \infty$. Now
$$
\mu( \Bbb{R} \setminus A_\varepsilon) = \mu \left( \bigcup_{q \in \Bbb{Q}} B(q, \varepsilon) \cap (\Bbb{R} \setminus A_\varepsilon) \right) \leq \sum_{q \in \Bbb{Q}} \mu (B(q, \varepsilon) \cap (\Bbb{R} \setminus A_\varepsilon)) = \infty\,,
$$
so $\mu (B(q', \varepsilon) \cap (\Bbb{R} \setminus A_\varepsilon)) > 0$ for some $q' \in \Bbb{Q}$.
Define $g$ to be such that $g = f$ outside of $B(q', \varepsilon) \cap (\Bbb{R} \setminus A_\varepsilon)$ and $g(x) = f(x) - \varepsilon$ in $B(q', \varepsilon) \cap (\Bbb{R} \setminus A_\varepsilon)$. Now
$$
\|f-g\|_2^2 = \int_{B(q', \varepsilon) \cap (\Bbb{R} \setminus A_\varepsilon)} |f(x) - f(x) + \varepsilon|^2 \leq \int_{B(q', \varepsilon)} \varepsilon^2 = \varepsilon^3\,,
$$
and $g$ is negative on a set $B(q', \varepsilon) \cap (\Bbb{R} \setminus A_\varepsilon)$ of positive measure.
