Do we have a name for this space? Let $\Omega$ be a bounded domain in $\mathbb{R}^{n}$. Consider the class 
$$
\mathcal{F}=\{f\in L^{1}(\Omega):\exists C>0 \text{ s.t. } \int_{U}|f|\leq C\sqrt{|U|},\text{ for any }U\subset \Omega.\}
$$
Here $U$ runs over all open subsets of $\Omega$. 
It is clear that $L^{2}(\Omega)\subset \mathcal{F}$.
My question: do we have a name for this class $\mathcal{F}$? Did anyone study it before?
Thanks. 
 A: $\newcommand{\Om}{\Omega}
\newcommand{\F}{\mathcal{F}}$Prompted by a comment by Nate Eldredge, I found this in Wikipedia : 

For any $0<r<p$ the expression 
  $$\||f|\|_{L^{p,\infty}}=\sup_{0<\mu(E)<\infty}\mu(E)^{-1/r+1/p}\Big(\int_E
|f|^r\,d\mu\Big)^{1/r}$$
  is comparable to the $L^{p,w}$-norm. 

Taking here $r=1$ and $p=2$, we conclude that 

$\mathcal F$ is the weak $L^2$ space $L^{2,w}$. 


Let us also provide a direct proof of the latter statement. Recall that 
\begin{equation*}
 L^{2,w}=\Big\{f\in L^0(\Om)\colon\;\exists C\in(0,\infty)\ \forall t\in(0,\infty)\  P_f(t)\le C/t^2\Big\},
\end{equation*}
where $$P_f(t):=\big|[|f|>t]\big|$$ and $[|f|>t]:=\{x\in\Om\colon|f(x)|>t\}$. 
On the other hand, as was shown in my other answer on this web page, 
\begin{equation*}
\F=\Big\{f\in L^1(\Om)\colon\;\exists C\in(0,\infty)\ \forall t\in(0,\infty)\,\int_{[|f|>t]}|f|\le C\sqrt{P_f(t)}\Big\}. \tag{0}
\end{equation*}
Take any $f\in L^0(\Om)$ and any $t\in(0,\infty)$. Then 
\begin{align*}
 \int_{[|f|>t]}|f|&=\int_\Om 1_{[|f|>t]}\,|f| \\ 
 &=\int_\Om 1_{[|f|>t]}\int_0^\infty ds\,1_{[|f|>s]} \\ 
 &=\int_0^\infty ds\,\int_\Om 1_{[|f|>t]}1_{[|f|>s]} \\ 
 &=\int_0^t ds\,\int_\Om 1_{[|f|>t]}
 + \int_t^\infty ds\,\int_\Om 1_{[|f|>s]} \\ 
 &=tP_f(t) + \int_t^\infty ds\,P_f(s). \tag{1}
\end{align*}
If now $f\in\F$ then, by (1),
\begin{equation*}
 tP_f(t)\le \int_{[|f|>t]}|f|\le C\sqrt{P_f(t)},
\end{equation*}
so that $P_f(t)\le C^2/t^2$ and hence $f\in L^{2,w}$. Thus,
\begin{equation*}
 \F\subseteq L^{2,w}. \tag{2}
\end{equation*}
To prove that $\F\supseteq L^{2,w}$, take any $f\in L^{2,w}$, so that 
\begin{equation*}
 P(t):=P_f(t)\le C/t^2 \tag{3}
\end{equation*}
for some real $C>0$ and all real $t>0$. The function $P$ is nonincreasing and right-continuous on $(0,\infty)$, with $P(\infty-)=0$. Consider the generalized inverse $P^{-1}$ of $P$ given by the formula 
\begin{equation*}
 t_u:=P^{-1}(u):=\inf\{s\ge0\colon P(s)\le u\}=\min\{s\ge0\colon P(s)\le u\}
\end{equation*}
for $u\in(0,P(0))$. Then for all real $s\ge0$ and all $u\in(0,P(0))$ we have 
\begin{equation*}
 s<t_u\iff u<P(s). \tag{4}
\end{equation*}
So, for any real $t>0$
\begin{align*}
 \int_t^\infty ds\,P_f(s)&=\int_t^\infty ds\,\int_0^{P(s)}du \\ 
 &=\int_0^{P(t)}du \int_t^{t_u} ds\,\\ 
 &=\int_0^{P(t)}du\, (t_u-t)\,\\ 
 &\le\int_0^{P(t)}du\,t_u.   
\end{align*}
By (3), for any real $u>0$ we have $P(\sqrt{C/u})\le u$, whence, by (4), $t_u\le\sqrt{C/u}$. Now the latter multi-line display yields 
\begin{equation*}
 \int_t^\infty ds\,P_f(s)\le2\sqrt C\,\sqrt{P_f(t)}. 
\end{equation*}
Also, (3) is obviously equivalent to $tP_f(t)\le\sqrt C\,\sqrt{P_f(t)}$. 
So, by (1), 
\begin{align*}
 \int_{[|f|>t]}|f|\le3\sqrt C\,\sqrt{P_f(t)}, 
\end{align*}
so that, by (0), $f\in\F$. 
Thus, indeed $\F\supseteq L^{2,w}$. Now (2) yields
\begin{equation*}
 \F=L^{2,w},
\end{equation*}
as claimed.
A: $\newcommand{\F}{\mathcal F}\newcommand{\Om}{\Omega}$Take any $f\in L^2(\Om)$. Then for any measurable subset $U$ of $\Om$, by Hölder's inequality we have 
$$\int_U|f|\le\|f\|_{L^2(\Om)}|U|^{1/2}, 
$$
so that $f\in\F$. Thus, $\F\supseteq L^2(\Om)$, as you noted. 
The purpose of this partial answer is to show that $\F\not\subseteq L^2(\Om)$. Indeed, let $\Om=(0,1)$ and $f(x)=1/\sqrt x$ for $x\in\Om$. Take any measurable subset $U$ of $\Om$ with $u:=|U|\ne0$, so that $u\in(0,1]$. Let $U_t:=\{x\in\Om\colon f(x)>t\}=(0,u)$, where $t:=1/u^2>0$. Then 
$$\int_U|f|-\int_{U_t}|f|=\int_U f-\int_{U_t}f
=\int_{U\setminus U_t} f-\int_{U_t\setminus U}f
\le\int_{U\setminus U_t}t-\int_{U_t\setminus U}t=0. 
$$
So, 
$$\int_U|f|\le\int_{U_t}|f|=\int_0^u\frac{dx}{\sqrt x}=2\sqrt u=2\sqrt{|U|},
$$
so that $f\in\F$. However, $f\notin L^2(\Om)$. 

It also follows from the above reasoning that (generally, for any $\Om$) 
$$\F=\Big\{f\in L^1(\Om)\colon\;\exists C\in(0,\infty)\ \forall t\in(0,\infty)\,\int_{[|f|>t]}|f|\le C\big|[|f|>t]\big|^{1/2}\Big\},
$$
where $[|f|>t]:=\{x\in\Om\colon|f(x)|>t\}$. That is, in the definition of $\F$, instead of arbitrary open or, equivalently, arbitrary measurable subsets $U$ of $\Om$, one may consider the subsets of $\Om$ of the special form $[|f|>t]$. 
