Riemann sum of completeness relation in continuous basis Suppose I have a wave function $\psi $ we express it in a continous states as 
$$\psi= \int_{-\infty}^{\infty} dxC (x)\rvert x\rangle = \int_{-\infty}^{\infty} dx\rvert x\rangle \langle  x \rvert \psi (x)$$
This can be expanded in Riemann sum as 
$$   \psi= \lim_{\Delta x\to 0} \sum_{i=-\infty}^{\infty} \Delta xC (x_i)\rvert x_i\rangle.$$ 
This is not symmetric with the expression for $\psi $ in other  discrete Hilbert spaces.
Which is 
$$\psi= \sum_{i=-\infty}^{\infty} C (\phi_i)\rvert \phi_i\rangle,$$ where this  $i$ takes discrete values.
Symmetry is lost due to the appearance of the factor $\Delta x$.
Can any one please help me to solve this paradox?
 A: There is no paradox. If you have the continuous relation as 
$$| \psi \rangle = \int_{-\infty}^\infty dx \ \psi(x) |x \rangle = \lim_{\Delta x \rightarrow 0} \sum_i \Delta x \psi(x_i) |x_i \rangle,$$
to draw an analogy with the discrete basis expansion $\sum_i C_i |e_i \rangle$, you need to realize that the analogy goes as $C_i \leftrightarrow \Delta x \psi(x_i)$, not $C_i \leftrightarrow  \psi(x_i)$.
For a better intuition, think of it this way, the continuous basis has infinitely many more basis vectors than a discrete basis, meaning that the component of any state $|\psi \rangle$ along any of these basis vectors $|x\rangle$ has to be infinitesimal for the expansion to make sense. So $\psi(x) =\langle x | \psi \rangle$ is more of a component density (for lack of a better term).
This is similar to any other part of physics where densities are involved. For example, the total mass of a system of discrete point masses is:
$$M = \sum_i m_i,$$
whereas for a continuous mass distribution it's
$$M = \int_\mathcal D d^3\mathbf x \rho(\mathbf x) = \lim_{\Delta V \rightarrow 0} \sum _{i} \Delta V \rho(\mathbf{x}_i).$$
Here you can make the analogy $m_i \leftrightarrow \Delta V \rho(\mathbf x_i)$, i.e. the mass of an infinitesimal volume element in the continuous distribution is $\Delta V \rho(\mathbf x_i)$ (and not $\rho$ itself).
Similarly, the component of $|\psi \rangle$ along the basis vector $|x \rangle$ is $\Delta x \psi(x)$, with $\psi(x)$ playing the role of a "density".
A: There is no paradox. You're comparing two different formulas: one is an approximation that applies to a continuous basis of states $|x\rangle$, the other is an exact formula that applies to a discrete Hilbert space. Just because two formulas look different doesn't mean one of them is wrong.
