Show that $A_n$ has empty interior and is closed in a space Let $L_1([0,1],m)$ and $L_2([0,1],m)$ be the Lebesgue spaces on $[0,1]$. Define
$$
A_n=\left \{ f\in L_1([0,1],m):\int_{[0,1]}|f|^2dm\leq n \right \}
$$
for $n\geq 1$. Show that
1) $A_n$ has empty interior in $L_1([0,1],m)$ for all $n\geq 1$.
2) $A_n$ is closed in $L_1([0,1],m)$ for all $n\geq 1$.
What I tried: Note that
$$
L_p([0,1],m):=\left \{ f:[0,1]\to \mathbb{K}:\|f\|_p:=\left ( \int_{[0,1]}|f|^p \right )^{1/p} <\infty \right \}.
$$
for $p\geq 1$.
1) Want prove it by contradiction, and use the fact that $L_2([0,1],m)\subsetneq L_1([0,1],m)$. Suppose $(A_n)^\circ \neq \emptyset$ for some $n\geq 1$. Then there exists $f\in (A_n)^\circ$. So we have the open ball
$$
B(f,\epsilon):=\left \{ g\in L_1([0,1],m):\|f-g\|_1<\epsilon \right \}\subseteq A_n
$$
for some $\epsilon>0$. For $0\neq g\in L_1([0,1],m)$, we have $f+\frac{\epsilon}{2 \|g\|_1}g\in B(f,\epsilon)$, and so it belongs to $L_1([0,1],m)$. How do I conclude that it implies $g\in L_2([0,1],m)$, so that I got a contradiction?
2) I need a hand. I probably should use Fatou Lemma, but I do not know where to start.
 A: Hint:
1) if the expression is in $B(f, \varepsilon)$, it is in $A_n$, hence in
$L^2$. As $f\in L^2$, you get easily $g \in L^2$.
2) The unit ball of $L^2$ is weakly compact in $L^2$.
Lemma let $V$ be a subspace of a normed space $X$.
If $V$ contains a non empty open set $U$,
then $V = X$.
Proof Let $a \in U$ and $x \in  X$. We can choose $t > 0$
such that ${y}_{t} := \left(1-t\right) a+t x \in  U$. Indeed,
${y}_{t} \rightarrow  a$ when $t \rightarrow  {0}^{+}$ because
$\|y_t-a\| = |t| \|x-a\|$. It follows that
${y}_{t} \in  V$ and $a \in  V$, hence
$x = \frac{1}{t} \left({y}_{t}-\left(1-t\right) a\right) \in  V$.
Applying the lemma with $X = {L}^{1}$ and 
${A}_{n} \subset  V := {L}^{1} \cap  {L}^{2}$. We get that ${A}_{n}$
does not contain a non empty open set, hence its $L^1$-interior
is empty.
A: For $1).$ simply solve for $g.\ $  That is, let $h=f+\frac{\epsilon}{2 \|g\|_1}g.\ $ Then, $h\in B(f,\epsilon) \Rightarrow h\in L^2([0,1]).$ Therefore  $g=\frac{2\|g\|_1}{\epsilon }(h-f)\in L^2([0,1]),\ $ from which we conclude that $L^1([0,1])=L^2([0,1]),\ $ which is a contradiction. 
For $2).$ if you want to do it from scratch, take a sequence $A_n \ni(f_k)\to f\in L^1([0,1])$ and note that there is a subsequence $(f_{n_k})$ which converges pointwise a.e. So, we have $\int \liminf |f_{n_k}|^2=\int |f|^2=\|f\|^2_2\le \liminf \int |f_{n_k}|^2\le n\Rightarrow f\in L^2([0,1]).$
