Consider, $L^{1}(\mathbb R)$= The space of Lebesgue integrable functions on $\mathbb R$; for $f\in L^{1}(\mathbb R),$ we define its norm, by $\|f\|_{L^{1}}=\int_{\mathbb R}|f(x)| dx$; It is well-known that, $(L^{1}(\mathbb R), \|\cdot\|_{L^{1}})$ is a a complete norm linear space(Banach space).
Put, $A=\{f\in L^{1}(\mathbb R): f^{2}\in L^{1}(\mathbb R)\}$(=$L^{1}\cap L^{2}$).
My Questions:(1) Is $A$ is closed in $L^{1}(\mathbb R)$ ? (2) Is any thing known about the closed sets(characterization) in $L^{1}(\mathbb R)$ ?
( Trivial Examples of closed sets in $L^{1}$: one may consider, (a) closed balls; then (i)its finite union; and (ii) its inter sections)
Thanks,
$A$ is not closed. Take any function $f\in L^1$ such that $f$ is not in $L^2$, then approximate it by functions in $A$.
I.e. $f(x) = \chi_{(0,1)}(x) x^{-1/2}$, then $f\in L^1$, $f\not\in L^2$. Define $f_n(x) = \min(n,f(x))$. Then $f_n\to f$ in $L^1$, $\|f_n\|_{L^2}\to\infty$.
As to your second question: sets like $\{f\in L^1: \ f\ge 0 \}$ are closed as well.