Here is the link of the question:
$L^2([0,1])$ is a set of first category in $L^1([0,1])$?
Where the answer is given there as follows:
For $n \in \mathbb N$ set $B_n = \{f \in L^2[0,1]\mid \|f\|_2 \le n \}$. We will show that $B_n$ is nowhere dense in $L^1$. Let $g \in L^1[0,1]\setminus L^2[0,1]$ and $f \in B_n$, then $f + \frac 1k g \to f$ in $L^1$ but $f+\frac 1k g \not\in B_n$ for all $k$. Hence $f \not\in \mathring{B_n}$ and $B_n$ doesn't have inner points. On the other hand, $B_n$ is closed in $L^1$: Let $g \in L^1$ and $g_k \in B_n$, $g_k \to g$. Then $g_{k_\ell} \to g$ almost everywhere for some subsequence, it follows by Fatou's Lemma \[ \int_0^1 |g|^2\, dx \le \liminf_{l \to \infty} \int_0^1 |g_{k_\ell}|^2 \, dx \le n^2 \] so $g \in B_n$.
As $L^2[0,1] = \bigcup_n B_n$ is a countable union of nowhere dense sets, it is of first category.
My question is:
why we take $g \in L^1[0,1]\setminus L^2[0,1]$ and $f \in B_n$, and why these leads to that then $f + \frac 1k g \to f$ in $L^1$ and why $f+\frac 1k g \not\in B_n$ for all $k$ ? and why these leads to that $f \not\in \mathring{B_n}$?
Could anyone explain these for me please? I was having hard times trying to understand the solution of this problem.
