Continuum limit of $ \sum_{r=1}^\infty \sum_{z=1}^\infty a_r |r z \rangle \langle r z | = \sum_{l=1} b_l |l \rangle \langle l | $?

Question

Background

Using bra-ket notation consider the identity (for integer $r$) :

$$ \sum_{z=1}^\infty |r z \rangle \langle r z | = I \otimes \underbrace{\begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\\ 0 & 0 & \ddots & 0 \\ 0 & 0 &0 & 1 \end{bmatrix}}_{\dim r} $$

with $I$ begin the identity matrix. Multiplying an arbitrary constant $a_r$ both sides:

$$ \sum_{z=1}^\infty a_r |r z \rangle \langle r z | = I \otimes \underbrace{\begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\\ 0 & 0 & \ddots & 0 \\ 0 & 0 &0 & a_r \end{bmatrix}}_{\dim r} $$

Now, summing over $r$:

$$ \sum_{r = 1}^\infty \sum_{z=1}^\infty a_r |r z \rangle \langle r z | dz = I \otimes \sum_{r = 1}^\infty \underbrace{\begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\\ 0 & 0 & \ddots & 0 \\ 0 & 0 &0 & a_r \end{bmatrix}}_{\dim r} $$

Let us explicitly handle the R.H.S:

$$ \begin{bmatrix} a_1 & 0 & 0 & 0 \\ 0 & a_1 & 0 & 0\\ 0 & 0 & \ddots & 0 \\ 0 & 0 &0 & a_1 \end{bmatrix} + \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & a_2 & 0 & 0\\ 0 & 0 & \ddots & 0 \\ 0 & 0 &0 & a_2 \end{bmatrix} + \dots $$

Hence with $b_r = \sum_{i|r} a_i $

$$ \sum_{r=1}^\infty \sum_{z=1}^\infty a_r |r z \rangle \langle r z | = {\begin{bmatrix} b_1 & 0 & 0 & 0 \\ 0 & b_2 & 0 & 0\\ 0 & 0 & b_3 & 0 \\ 0 & 0 &0 & \ddots \end{bmatrix}} = \sum_{l=1} b_l |l \rangle \langle l | $$

Question

Is the above proof correct? With $b_r = \sum_{i|r} a_i$:

$$ \sum_{r=1}^\infty \sum_{z=1}^\infty a_r |r z \rangle \langle r z | = \sum_{l=1} b_l |l \rangle \langle l | $$

Is there a continuum limit of this formula? Does $ \sum_{z=1}^\infty |r z \rangle \langle r z |$ have a continuum analog?