I wanted to show $L^2_{\mathbb{R}}(M)$ is a closed subset of $L^2_{\mathbb{C}}(M)$, but since $L^2_{\mathbb{R}}(M)$ is not a ($\mathbb{C}$-)subspace of $L^2_{\mathbb{C}}(M)$ I am not too sure if the following is correct.
I thought it'd be alright as I don't think the scalar field is affecting the topology (we are using the metric space-ness, aren't we?), but wanted to see others agree.
Let $( M,\Omega ,\mu )$ be a measure space let $L^{2}_{\mathbb{C}}( M)$ be the $\mathbb{C}$-vector space consisting of $f:M\rightarrow \mathbb{C}$ such that \begin{equation*} \| f\| _{L^{2}_{\mathbb{C}}} :=\left(\int _{M} f\overline{f} d\mu \right)^{1/2} < \infty . \end{equation*} Let $V$ be the subset of $L^{2}_{\mathbb{C}}( M)$ consisting of real-valued functions. I want to show that $V$ is a closed subset of $L^{2}_{\mathbb{C}}( M)$.
Note that $V$ is clearly not a subspace of $L^{2}_{\mathbb{C}}( M)$, since it is not closed under the complex number multiplication.
We show the closedness. To see this, first note that for $f\in L^{2}_{C}( M)$ we have
\begin{align*}
\| f\| _{L^{2}_{C}} < \infty \ \ \Leftrightarrow & \ \ \| f\| _{L^{2}_{C}}=\left(\int _{M} Re( f)^{2} d\mu +\int _{M} Im( f)^{2} d\mu \right)^{1/2} < \infty \\
\Leftrightarrow & \int _{M} Re( f)^{2} d\mu < \infty \ \ \text{and} \ \ \int _{M} Im( f)^{2} d\mu < \infty .
\end{align*}
Now, we have
\begin{equation*}
V=\left\{f:M\mapsto \mathbb{R} \mid \ \| f\| _{L^{2}_{\mathbb{C}}} =\int _{M} Re( f)^{2} d\mu =\int _{M} f^{2} d\mu =:\| f\| _{L^{2}} < \infty \right\} ,
\end{equation*}
with for $f\in V$ the $L^{2}_{\mathbb{C}}( M)$-norm being $\| f\| _{L^{2}_{\mathbb{C}}} =\| f\| _{L^{2}}$. Then, $V$ is just an $L^{2}( M)$-space over $\mathbb{R}$, with the $\mathbb{R}\subset{\mathbb{C}}$ multiplication being inherited from $L^2_{\mathbb{C}}(M)$.
So for $\{f_n\}\subset V$, $\|\cdot\| _{L^{2}_{\mathbb{C}}}$-norm topology convergent is the same as $\|\cdot\| _{L^{2}}$-norm topology convergent, and thus the limit must be in $V$.
The last paragraph seems not true: For simplicity, consider the ambient space ${\bf{R}}$ with the usual norm $|\cdot|$. And consider the norm $|\cdot|_{r}$ on $(0,1)$ defined by the restriction of $|\cdot|$ on $(0,1)$. Then the sequence $\{1/n\}_{n}$ satisfies $|1/n-0|\rightarrow 0$ and at the same time $|1/n-0|_{r}=|1/n-0|$, still, $0\notin (0,1)$.
To tackle this question, for $\{f_{n}\}\subseteq V$ such that $f_{n}\rightarrow f$ in $L_{{\bf{C}}}^{2}$, then there exists a subsequence $\{f_{n_{k}}\}$ such that $f_{n_{k}}\rightarrow f$ a.e., say, $f_{n_{k}}(x)\rightarrow f(x)$ for all $x\in{M}-N$, where $\mu(N)=0$.
Note that $\text{Im}(f_{n_{k}})=0$ and for all $x\in M-N$, we must have $\text{Im}(f_{n_{k}})(x)\rightarrow\text{Im}(f)(x)$, so $\text{Im}(f)(x)=0$. So we conclude that $\text{Im}(f)=0$ a.e.
With the a.e. identification in $L_{{\bf{C}}}^{2}$, then $f=\text{Re}(f)$, so $f\in V$.
Note that if we want to have super rigorous argument, one may assume only that $\text{Im}(f_{n_{k}})=0$ a.e. However, the proof is almost the same, just by collecting all the sets of measure zero regarding the imaginary parts of $f_{n_{k}}$ together with $N$, and finally claim that it is still the case that the imaginary part of $f$ is a.e. zero.