It seems that simply one can't measure it. Here below is briefly described
an example of a nondegenerate indefinite inner product space having
no cardinal-valued Morse index.

Consider $(\ell^{2},<.,.>)$ as
naturally embedded (via Riesz) into its (huge) algebraic dual, say
$\mathcal{A}$, let $\mathcal{F}$ be the real vector space of all
finitely supported functions from $\mathbb{R}$ to $\mathbb{\ell^{\textrm{2}}}$,
and put $V$:= $\mathcal{A\times F}$. Next, write $\mathcal{A}$
as a direct sum
$\mathcal{A}=\ell^2\oplus\{E}$ (hence
$\dim E=2^{c}$),
let $\pi:\mathcal{A}$ $\to$ $\{E}$
be the attached algebraic projection, and let $[.,.]$
be a scalar product on $E\times E$. If $u=$($\varphi,f)$, and $v=$($\psi,g)$
are in $V$, then define the bilinear symmetric pairing

$a(u,v)$:= - $[\pi \varphi,\pi \psi]$ + $\varphi(\sum_{t\in\mathbb{R}}g(t))+\psi(\sum_{t\in\mathbb{R}}f(t))$ + < $ \sum_{t\in\mathbb{R}} f(t),\sum_{t\in\mathbb{R}}g(t))> $
- $\sum_{t\in\mathbb{R}}$ <
$f(t),g(t)$> .

Define also the subspace $W$ of $V$ by $W$ := {$ {\{ (\varphi,f)|\:\varphi=-\sum_{t\in\mathbb{R}}f(t)\} }$}.
Then is not hard to see that:

1) $W$ is negative definite (w.r.t. $a$), and $W^{\bot}$ = { 0 } , hence $a$ is non-degenerate.

2) [Using the C-B-S inequality and Riesz] Any maximal negative
definite subspace $\mathcal{N}$ of $V$ containing $W$ is a linear
subspace of $\ell^{2}\times\mathcal{F}$, hence $\dim$ $\mathcal{N}$
= $c$.

3) Any maximal negative definite subspace $\mathcal{M}$ of $V$ containing
$E\times ${ 0 }$ $ has $\dim\mathcal{M}$ $>$ $c$.

Consequently, $\mathcal{M}$ and $\mathcal{N}$ are not isomorphic
as real vector spaces.