Why is $B_n = \{f \in L^1 : \int |f|^2 < n \}$, $n \in \mathbb{N}$ a nowhere dense subset of $L^1$? Please provide a proof without assuming that $L^2 \subsetneq L^1$.
Clarification:
$L^p$ here follows the common notation in mathematics, and refers to the functional space equipped with the $L^p$ norm. The underlying space is $X$, i.e., we are concerning ourselves with $L^p(X)$. The only restriction for $X$ is $\mu(X) < \infty$ where $\mu$ is the measure on $X$.
Edit:
Here is how I have approached the problem:
To show that $B_n$ is a nowhere dense set in $L^1$, we need to show that the closure of $B_n$ has an empty interior in $L^1$. Notice that the closure of $B_n$ is simply $\{f \in L^1 : \int |f|^2 \le n \}$, so we simply need to prove $\{f \in L^1 : \int |f|^2 \le n \}$ has an empty interior. My intuition would be proof by contradiction. Suppose the interior is not empty. Then $\exists \ f_0 \in L^1$ and $r_0 \in \mathbb{R}$ such that $B_{r_0}(f_0) = \{g \in L^1 \ | \ \|g - f_0\|_{L^1} < r_0\} \subset B_n$. But I do not know how to derive a contradiction from here.
Would be grateful for any helpful constructive comments.
Let $\varepsilon > 0,f \in B_n$. You need to show that the $L^1$-ball of radius $\varepsilon$ around $f$ is not contained in $B_n$. Here is a hint for how to do this: find $g \in L^1 \setminus L^2$ and consider $h=f+\frac{\varepsilon}{2} \frac{g}{\| g \|_1}$.
Note that the proof above requires that something is in $L^1 \setminus L^2$; this is not always the case, it depends on the underlying measure space. In particular, it requires that there exist sets of arbitrarily small positive measure. A sketch of the construction of something in $L^1 \setminus L^2$ under this assumption is as follows.
Take a sequence of disjoint sets $A_n$ with $0<\mu(A_n)\leq n^{-4}$. Take $g(x)=\frac{1}{n^2 \mu(A_n)}$ for $x$ in $A_n$ and zero elsewhere. Check that $\| g \|_1 < \infty$ and $\| g \|_2=\infty$.