Is there a well-defined completely dephasing channel in infinite dimensions for the "position eigenbasis"?

In a $$d$$-dimensional Hilbert space, a channel is a completely positive trace preserving map, and a completely dephasing channel $$\Delta$$ (implicitly defined with respect to a fixed orthonormal basis $$\{|i\rangle\}_{i=1}^d$$) acts on a state $$\rho$$ according to $$$$\Delta(\rho) = \sum_{i=1}^d |i\rangle \langle i| \langle i | \rho |i\rangle \tag{1}.$$$$ More generally, in a separable Hilbert space $$\mathcal{H}$$ we can define a channel $$\Phi$$ according to its action $$\Phi(|i\rangle \langle j|)$$, where $$|i\rangle, |j\rangle$$ are elements of an orthonormal basis for $$\mathcal{H}$$. So for any fixed basis, we can define a completely dephasing channel analogously to Eq. (1), where the sum possibly extends to infinity.

My question is, does this reasoning break down when we try to define an analogous channel using the "position eigenbasis" $$\{|x\rangle\}_{x \in \mathbb{R}}$$ for states defined on the real line? That is, for a wavefunction $$\langle x|\psi\rangle = \psi(x)$$, suppose I define a channel $$\Delta_{\mathbb{R}}$$ according to $$$$\Delta_{\mathbb{R}}(|\psi\rangle \langle \psi|) = \int_\mathbb{R} |\psi(x)|^2 |x \rangle \langle x|. \tag{2}$$$$ As far as I can tell, the RHS of this equation is a valid state - it corresponds to a unit trace operator if $$\psi(x)$$ is normalized, and seems to be bounded if $$\psi(x)$$ is.

On the other hand, I'm worried a lot could be wrong here. In finite dimensions, $$\Delta$$ can be thought of as a "measure and discard" channel where a projective measurement $$\{|i\rangle \langle i|\}_{i=1}^d$$ is applied to a state $$\rho$$ and then the result is discarded, leaving a stochastic mixture over possible measurement outcomes. From this perspective the RHS of Eq. (2) would correspond to making an infinitely precise position measurement and then discarding the result, which seems suspicious. However this doesn't rule out getting $$\Delta_{\mathbb{R}}$$ to work some other way, since it seems well defined by the collection of operators $$\Delta_{\mathbb{R}}(|i\rangle \langle j|)$$.

Is there a formal statement about whether this map is properly a channel in infinite-dimensional Hilbert spaces?

So the RHS of Eq. (2) is not a valid state, as it is not necessarily trace class. Fixing an orthonormal basis $$\{|\phi_i\rangle\}$$ for $$\mathcal{H}$$, you can evaluate the trace of this operator as, e.g. \begin{align} \text{Tr}\left (\int_{\mathbb{R}} dx |\psi(x)|^2 |x\rangle \langle x| \right) &= \sum_i \int_{\mathbb{R}} dx |\psi(x)|^2 |\langle \phi_i|x\rangle |^2 \\&= \int_{\mathbb{R}} dx |\psi(x)|^2 \sum_i| \phi_i(x)|^2 \end{align} where I exchanged the sum and integral in the second line (see here). The sum in this expression is not necessarily bounded. For instance, if $$\psi$$ has support only on an interval in $$\mathbb{R}$$ we could write $$\sum_i |\phi_i(x)|^2 = \delta(0)$$ by the closure relation for our orthonormal basis (e.g. this link), and one assumes this problem doesn't resolve by expanding the domain.
Since applying the map $$\Delta_{\mathbb{R}}$$ to a state results in an operator that may not be trace-class, the map can't be a valid channel the way its written here.