Why do we need absolute continuity of $\langle M \rangle_t(\omega)$ with repect to the Lebesgue measure? I am trying to understand the proof of proposition 3.2.6 in Stochastic Calculus and Brownian Motion by Karatzas and Shreve. For $X$ bounded they use Lemma 3.2.4 in the same book and eventually claim(the arguments are detailed below) that $[X^{(m_k)}-X] \to 0$ using the bounded convergence theorem and the absolute continuity of $ t \mapsto\langle M \rangle_t$
I do not understand why do we need the  almost sure absolute continuity of the quadratic variation of $M$ with respect to the Lebesgue measure in the proof
Proposition 3.2.6
If the function $t \mapsto \langle M \rangle_t(\omega)$ is absolutely continuous with respect to Lebesgue measure for $P-$a.e $\omega \in \Omega$, then $\mathcal{L}_0$ isn dense in $\mathcal{L}$ with respect to the metric
$$
d(x,y)=\sum_{n=1}^{\infty}2^{-n}\left(1 \wedge [x-y]_n\right)
$$ where
$$
[X]_n^2=E\int_0^n X_u^2 d\langle M \rangle_u
$$
Proof
If $X \in \mathcal{L}$ is bounded ,then Lemma 3.2.4 guarantees the existence of a bounded sequence $\{X^{(m)}\}$ of simple processes satisfying
 $$
 \sup_{T>0} \lim_{m \to \infty} E \int_0^T \vert X_t^{(m)}-X_t \vert^2 dt =0
 $$
That is $X^{(m)}$ converges to $X$ in $L^2$ and therefore in probability and hence from it we can extract a subsequence $\{X^{(m_k)}\}$ such that the set
 $$
 \{(t, \omega) \in [0,\infty) \times  \Omega: \lim_{k \to \infty} X^{(m_k)}_t(\omega)=X_t(\omega)\}^c
 $$
 has product measure zero.
Now looking at $d(X^{(m_k)}, X)=\sum_{n=1}^{\infty}2^{-n}\left(1 \wedge [X^{(m_k)}- X]_n\right)$ and in particular on the sequence below for a fixed $n$
$$
[X^{(m_k)}- X]_n=E\int_0^n (X^{(m_k)}- X)^2 d\langle M \rangle_u
$$
and then taking the limits as $k \to \infty$ and recalling that $X$ is assumed to be bounded and by construction(in Lemma 3.2.4 of the same book)  the sequence $X^{(m_k)}$ is uniformly bounded by the same constant, we can apply the dominated convergence theorem to conclude that
$ \lim_{k \to \infty}[X^{(m_k)}- X]_n =\lim_{k \to \infty} E\int_0^n (X^{(m_k)}- X)^2 d\langle M \rangle_u =E\int_0^n \lim_{k \to \infty} (X^{(m_k)}- X)^2 d\langle M \rangle_u=0$
Where is the absolute continuity being used here? Since the quadratic variation is an increasing right continuous process, it induces a measure, where does the absolute continuity come into play?
 A: Answer: The assumption of absolute continuity is needed in the application of the dominated convergence theorem (DCT) at the end of the quoted proof.
And the elaboration: For simplicity, I'll let $n=1$. The argument doesn't depend on the value of $n$, but this conveniently normalizes $(\text{Leb}\times P)([0,1]\times\Omega)=1$. Also, $(\text{Leb}\times P)(H)=1$ where $H\triangleq\{(t,\omega):\lim_{k\rightarrow\infty}X^{(m_k)}(t,\omega)=X(t,\omega)\}$. I note finally that in Karatzas and Shreve, $M$ is being assumed to be a square-integrable martingale (with continuous paths).
We're looking at 
$$\lim_{k\rightarrow\infty}E\int_0^1(X^{(m_k)}_t-X_t)^2d\langle M\rangle_t=:L.$$ 
Let $Y^k=\int_0^1(X^{(m_k)}_t-X_t)^2d\langle M\rangle_t$. To assert $L=E\lim_{k\rightarrow\infty}Y^k$ by DCT, we need (i) $Y^k$ are dominated by an integrable random variable and (ii) $Y^k$ is convergent a.s.
The first part is covered by the uniform boundedness of $(X^{(m_k)}_t-X_t)^2$, say by $c\ge 0$, and the integrability of $\langle M\rangle_1$: $Y^k\le c\langle M\rangle_1$.
The second part is covered by another application of DCT, i.e., by the assertion that 
$$\lim_{k\rightarrow\infty}Y^k = \int_0^1\lim_{k\rightarrow\infty}(X^{(m_k)}_t-X_t)^2d\langle M\rangle_t=0\quad\text{a.s.}$$ 
To see this, recall first that $(X^{(m_k)}_t-X_t)^2$ is bounded. Next, let $S(\omega)=\{t:(t,\omega)\in H\}$. Then by Fubini's theorem, $\text{Leb}(S)=1$ a.s. ($\int_\Omega \text{Leb}(S)dP=(\text{Leb}\times P)(H)=1$. This is in fact not so much an application of Fubini's theorem as that of a lemma used in its proof.) Since $\langle M\rangle$ is absolutely continuous a.s., $\lim_{k\rightarrow\infty}(X^{(m_k)}_t-X_t)^2=0$ for $\langle M\rangle$-a.e. $t$, a.s. This is where you need the a.s. absolute continuity of $\langle M\rangle$. 
Remark: I don't think we need to invoke the RN derivative. A bounded RN derivative is a sufficient condition for an application of DCT, not a necessary condition.

Addendum: I'll add a few more words regarding the last paragraph.
We know two things here. One, $\text{Leb}(S)=1$ a.s. Two, the measure $\langle M\rangle$ is absolutely continuous with respect to the Lebesgue measure a.s. 
Now let $\Omega'$ be a set of measure one on which both of these conditions hold. Then fix $\omega\in\Omega'$.
Then $\text{Leb}(S(\omega))=1$, which means $(t,\omega)\in H$ for Lebesgue-a.e. $t$. This in turn means, by the definition of $H$, that
$$
\lim_{k\rightarrow\infty}(X^{(m_k)}(t,\omega)-X(t,\omega))^2=0\quad
\text{for Lebesgue-a.e. }t.
$$
At the same time, $\omega\in\Omega'$ means that the measure $\langle M\rangle(\omega)$ is absolutely continuous with respect to the Lebesgue measure, which in turn means that any event that has full measure under the Lebesgue measure must also have full measure under $\langle M\rangle(\omega)$. (The definition of $\langle M\rangle(\omega)$ being absolutely continuous with respect to the Lebesgue measure is that $\text{Leb}(A)=0\Rightarrow \langle M\rangle(\omega)(A)=0$. This implies that $\text{Leb}(A)=1\Leftrightarrow \text{Leb}(A^c)=0\Rightarrow \langle M\rangle(\omega)(A^c)=0\Leftrightarrow \langle M\rangle(\omega)(A)=\langle M\rangle_1(\omega)$. Apparently, I'm denoting the Lebesgue-Stieltjes measure induced by $t\mapsto\langle M\rangle_t(\omega)$ again by $\langle M\rangle(\omega)$.) It follows that
$$
\lim_{k\rightarrow\infty}(X^{(m_k)}(t,\omega)-X(t,\omega))^2=0\quad
\text{for }\langle M\rangle(\omega)\text{-a.e. }t.
$$
Combined with the fact that the integrand $t\mapsto (X^{(m_k)}(t,\omega)-X(t,\omega))^2$ is bounded, then, by DCT we have
$$
\lim_{k\rightarrow\infty}Y^k(\omega)=0.
$$
Since all this holds for all $\omega\in\Omega'$, we finally have the desired result that $\lim_{k\rightarrow\infty}Y^k=0$ a.s.
