Extending the domains of densely defined bounded integral transforms on $L^2(\Bbb R)$

This is a question I've contemplated for quite some time since it's pretty closely related to Fourier theory (particularly choosing the "right" space to define the Fourier transform on). However I've never been able to come up with anything resembling an answer for this. Nor have I seen it be addressed anywhere.

Let $X$ be dense in $L^2(\Bbb R)$ and $T:X\subseteq L^2(\Bbb R)\to L^2(\Bbb R)$ be the integral operator given by

$$ Tf(x) = \int_{-\infty}^{\infty} k(y,x) f(x)\,dx$$

where the integral is the Lebesgue integral. Assume that $T$ is bounded. If $g\in L^2(\Bbb R)\setminus X$ but we have that

$$ \int_{-\infty}^{\infty} |k(y,x)| |g(x)|\,dx < \infty$$

for each $y$ (i.e. $Tg$ is well-posed). Is it necessarily the case that $Tg$ is in $L^2(\Bbb R)$? Note here that $T$ is meant as an integral transform, not the extension of $T$ (since that would be true trivially). My first approach was to consider some limit of elements in $X$ which approach $g$ in $L^2$, but I couldn't really piece any more of an argument together since it wasn't obvious to me how to proceed.

A partial attempt:

Since $X$ is dense in $L^2(\Bbb R)$, there is a a sequence $(f_m)\subseteq X$ such that $f_m\to g$ in $L^2(\Bbb R)$. Since $L^p$ convergence implies pointwise almost everywhere convergence of a subsequence $(f_{m_k})$ to $g$. Moreover, $f_{m_k}\to g$ in $L^2(\Bbb R)$.

Since $f_{m_k}(x) \to g(x)$ for almost every $x$, we have that $k(y,x) f_{m_k}(x) \to k(y,x)g(x)$ almost everywhere. Assuming that it can be shown that

$$\int_{-\infty}^{\infty} k(y,x) f_{m_k}(x)\,dx\to \int_{-\infty}^{\infty} k(y,x)g(x)\,dx,\tag{1}$$

then since $(f_{m_k})$ is Cauchy and $T$ is bounded, $(Tf_{m_k})$ is Cauchy and thus converges to an element of $L^2(\Bbb R)$. This then says that

$$\int_{-\infty}^{\infty} k(y,x) g(x)\,dx = \lim_k \int_{-\infty}^{\infty} k(y,x) f_{m_k}(x)\,dx = \lim_k Tf_{m_k}(y).$$

Since $\lim_k Tf_{m_k}\in L^2(\Bbb R)$, we'd have that $Tg\in L^2(\Bbb R)$ since they agree almost everywhere.

This is predicated on $(1)$ being true. $(1)$ can be shown if dominated convergence can be applied, though it isn't clear that the $k(y,\cdot)f_{m_k}$ can be bounded uniformly by an integrable function for a fixed but arbitrary $y$. If $g$ could be approximated within by elements in $X$, then this would work, but that is a very strong condition and will not be the case in general.

Here is a silly counterexample which nevertheless shows what is possible.

Let $k(y,x)=1$ for all $x,y$ and define $$ X=\left\{ f \in L^1 \cap L^2 \, \middle| \,\int f(x)\, dx=0\right\}. $$

Note that $X$ is a subspace of $L^2$ which is dense, since for $f\in C_c$, we can set $\alpha =\int f \, dx$ and $f_n = f -\alpha/n\cdot 1_{[0,n]}$. It is easy to see that the "added" term goes to zero in $L^2$ and hence $f_n \to f$ with $f_n \in X$.

Also, on $X$, the integral operator satisfies $T \equiv 0$, which is certainly bounded :)

Finally, for every function $f\in L^2$ with $$ \int |k(y,x) f(x)|\, dx <\infty, $$ we have $f\in L^1$ and applying the integral operator yields the constant function $T f \equiv \alpha$ with $\alpha =\int f \, dx$. Hence, $Tf\in L^2$ iff $\alpha =0$ iff $f \in X$.