Why can't the strength of superacids be measured in water?
Any acid-base reaction is always an equilibrium:

$$\ce{HA^1 + (A^2)- &lt;=&gt; (A^1)- + HA2}\tag{1}$$

and for each pair of acids $\ce{HA^1}$ and $\ce{HA^2}$ you could calculate a $K_\mathrm{a}$ value to determine one acid’s strength with respect to the other. This $K_\mathrm{a}$ value is typically calculated according to equation $(2)$ if $\ce{(A^2)-}$ (which does not have to feature a negative charge; I just wanted to avoid different descriptions for the two acids) is the solvent.

$$K_\mathrm{a} = \frac{[\ce{HA^2}][\ce{(A^1)-}]}{[\ce{HA^1}]}\tag{2}$$

If $\ce{(A^2)-}$ is $\ce{H2O}$ (i.e. $\ce{HA^2}$ is $\ce{H3O+}$) and the acid $\ce{HA^1}$ is a strong acid or a superacid, then this equation is of little use; the equilibrium $(1)$ will be strongly shifted towards the product side, the concentration $[\ce{HA^1}]$ will be very small and thus $K_\mathrm{a}$ very large. With ‘very small’, I mean that $[\ce{HA^1}]\approx10^{-10}$ — acid $\ce{HA^1}$ is said to be fully dissociated.

Exact measurements of $K_\mathrm{a}$ become very hard if not impossible because you would need to determine extremely minute concentrations. The concentrations $[\ce{HA^2}]$ and $[\ce{(A^1)-}]$ are of little help since:

$$[\ce{HA^2}] = [\ce{(A^1)-}] \approx c_0(\ce{HA^1})\tag{3}$$

This is the levelling effect of water: no matter how strong the acid, there is no acidic species stronger than $\ce{H3O+}$ that can survive in aqueous solution for extended time. Most importantly, consider two very strong acids with different $K_\mathrm{a}$ — e.g. one with $\mathrm{p}K_\mathrm{a} = -15$, the other with $-10$. They have a very different acidity, yet the concentrations of all relevant species in aqueous solutions will be nigh identical.

Therefore, a new, more expandable way to measure and compare the strengths of very strong acids was required.