Let $(\Omega,\mathcal A,\mu)$ be a measurable space
Suppose $(g_n)_{n\in\mathbb N}$ is a sequence of bounded $\mathcal A$-measurable functions $g_n:\mathbb R\to\mathbb R$ such that $\left(\sqrt{\gamma_n}\right)_{n\in\mathbb N}$ is $L^2(\mu)$-Cauchy. Are we able to conclude that there is a unique $\gamma\in L^1(\mu)$ with $\left\|\gamma_n-\gamma\right\|_{L^1(\mu)}\xrightarrow{n\to\infty}0$?
If $\sqrt{\gamma_n}$ is $L^2(\mu)$-Cauchy, there exists $\delta\in L^2(\mu)$ such that $$ \lim_{n\to\infty}\|\sqrt{\gamma_n}-\delta\|_{L^2}=0. $$ Observe that $$\begin{align*} \int_\Omega |\gamma_n-\delta^2|d\mu &= \int_\Omega |\sqrt{\gamma_n}-\delta| |\sqrt{\gamma_n}+\delta|d\mu\\&\le\left(\int_\Omega |\sqrt{\gamma_n}-\delta|^2 d\mu\right)^{1/2}\left(\int_\Omega |\sqrt{\gamma_n}+\delta|^2d\mu\right)^{1/2}\\ &\le \|\sqrt{\gamma_n}-\delta\|_{L^2}\|\sqrt{\gamma_n}+\delta\|_{L^2}\\&\le \|\sqrt{\gamma_n}-\delta\|_{L^2}\left(\|\sqrt{\gamma_n}\|_{L^2}+\|\delta\|_{L^2}\right). \end{align*}$$ Since $\|\sqrt{\gamma_n}\|_{L^2}\le M$, $\|\delta\|_{L^2}\le M$ for some $M>0$, we have $$\begin{align*} \|\gamma_n -\delta^2\|_{L^1}=\int_\Omega |\gamma_n-\delta^2|d\mu \ \le\ 2M \|\sqrt{\gamma_n}-\delta\|_{L^2}\xrightarrow{n\to\infty} 0 \end{align*}$$ as wanted. This $L^1$-limit is of course unique.