open and closed set in $\mathbb{(L_1(\mathbb{R^n})}$ In a test I had was written if $X=(f\in\mathbb{L_1}\mathbb{R^n}|m(f^{-1}((0,\infty))=0)$ is open or closed in $\mathbb{(L_1(\mathbb{R^n})}$.
I suspect that this set is none  but I am not sure. Is that true? Can you please help me?
 A: The condition $f\in X$ means that $f(x)\le 0$ (a.e.). In order to show that $X$ is closed, assume $f_n\in X$ and  $\|f_n-f\|_1\to 0.$ We have to show that $f(x)\le 0$ a.e., i.e. $f\in X.$ Although it seems obvious, the proof is given below.  We will base on the inequality $$|\min(a,0)-\min(b,0)|\le |a-b|\qquad (*)$$
As $f_n\in X,$ then $f_n(x)=\min(f_n(x),0).$ Denote $g(x)=\min(f(x),0).$ By $(*)$ we get
$$\|f_n-f\|_1=\int\limits_{\mathbb{R}^d}|f_n(x)-f(x)|\,dx\ge \int\limits_{\mathbb{R}^d}|f_n(x)-g(x)|\,dx=\|f_n-g\|_1$$
We obtain $f_n\to g$ in $L^1.$ Therefore $f=g\le 0, $ i.e. $f\in X.$
A: $X$ is indeed a closed subset in $L_1$. To wit, Let $f\in \overline{X}$ and $(f_n:n\in\mathbb{N})\subset X $ such that $\|f-f_n\|_1\xrightarrow{n\rightarrow\infty}0$. There is a subsequence $n_k$ such that $f_{n_k}(x)\xrightarrow{n\rightarrow\infty}f(x)$ for $m$ -almost all $x\in\mathbb{R}^d$. Without lose of generality assume $f_n\xrightarrow{n\rightarrow\infty}f$ pointwise $m$-almost surely. That is, $M=\{x:f_n(x)\not\rightarrow f(x)\}$ has $m(M)=0$.
Then
$$\{x:\in M^c:f(x)>0\}\subset\bigcup_n\bigcap_{k\geq n}\{x\in M^c: f_k(x)>0\}$$
Consequently
\begin{align}
m(f>0)&=m(M^c\cap\{f>0\})\leq m\Big(\bigcup_n\bigcap_{k\geq n}M^c\cap\{f_k>0\}\Big)\\
&=\lim_n m\Big(\bigcap_{k\geq n}M^c\cap\{f_k>0\}\Big)\leq\liminf_k m(M^c\cap\{f_k>0\})=0
\end{align}
That is $f\in X$.
