Why do we need absolute continuity of the mapping $t \mapsto \langle M \rangle_t(\omega)$ wr.t to the Lebesgue measure for almost all $\omega$ to show that the class of simple functions($\mathcal{L}_0$) is dense in the class of all adapted measurable processes for which $[X]_T <\infty$ for all $T>0$( $\mathcal{L}$)?
Note that $$ [X]:=\sum_{n=1}^{\infty}2^{-n}(1 \wedge [X]_n) $$ where $$ [X]_n=E \int_0^T X_t^2 d\langle M\rangle_t(\omega) $$
I don't understand why do we need absolute continuity of the quadratic variation process paths in the last step of proposition 2.6 ? My first idea was that in order for the integral to be well defined we need the absolutely continuity but since any continuous increasing process induces a measure and since X is bounded by assumption in part a), the integral is well defined for all $\omega \in \Omega$.
Why isnt just bounded convergence theorem just enough to take the limit inside the integral and conclude?
Attempted Answer:
I think we need the absolute continuity of the quadratic variation so that we can use the randon nikodym theorem to define a density of the quadratic variation with respect to the lebesgue measure . This is crucial because once inside the intgral we only know that $\mu\{ (t, \omega) \in [0,\infty) \times \Omega:\lim_{k \to \infty} X_t^{(m_k}(\omega)=X_t(\omega)\}=1$ only for the measure $\mu=dt \times P$ where $dt$ is the lebesgue measure and not for the measure $d\langle M \rangle \times P$.
And we have to go through the density to achieve this which raise another question of the uniform boundedness of the density function for all $\omega$ in order to apply the bounded convergence theorem. $ \vert\langle M \rangle \vert \leq K$
Does this make sense? How could I show the uniform boundedness of the derivative of the quadratic variation? Note that if $M$ were the Brownian motion it wouldbe true(bounded above by a constant 1 for all $\omega$)
Any help or hint would be appreciated.