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Here is an inequality one finds in page 136 of Karatzas & Shreve - Brownian motion and Stochastic Calculus:

$$\begin{align}E\int_0^\infty Y_s^2\,ds&= E\int_0^\infty1_{\{T_s\leq T\}}X_{T_s}^2\,ds\\ &=E\int_0^{A_T+T}X_{T_s}^s\,ds\leq C^2(EA_T+T)<\infty,\end{align}$$

I don't see how $E[A_T] < \infty$.

Among the hypothesis, one does not find this condition:

2.7 Lemma. Let $\{A_t;0\leq t<\infty\}$ be a continuous, increasing (Definition 1.4.4) process adapted to the filtration of the martingale $M=\{M_t,\mathscr{F}_t;0\leq t<\infty\}$. If $X=\{X_t,\mathscr{F}_t;0\leq t<\infty\}$ is a progressively measurable process satisfying $$E\int_0^T X_t^2\,dA_t<\infty$$ for each $T>0$, then there exists a sequence $\{X^{(n)}\}_{n=1}^\infty$ of simple processes such that $$\sup_{T>0}\lim_{n\to\infty}E\int_0^T|X_t^{(n)}-X_t|^2\,dA_t=0.$$ Proof. We may assume without loss of generality that $X$ is bounded (cf. part $(b)$ in the proof of Proposition 2.6), i.e.,

I believe that usually we will consider $A_t = \langle M \rangle_t $ wich in the case when $M$ is a martingale implies that $E[A_T] < \infty$. Nevertheless this is not what is stated in thr Lemma.

Should we assume that $E[A_T] < \infty$? Isn't this a loss of generality?

Am I missing something here?

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$\{A_{t}\}_{t\geq 0}$ is an increasing process so by definition $A_{t}$ is integrable for each $t\geq 0.$

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  • $\begingroup$ Even for local martingales that are not martingales? $\endgroup$ – Conrado Costa Apr 5 '16 at 10:47
  • $\begingroup$ I do not think this is correct. $\endgroup$ – Potato Apr 16 '16 at 20:55
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    $\begingroup$ An adapted process $A$ is called increasing if $A_{t}(\omega)$ is right continuous and nondecreasing for a.s. $\mathbb{P}$ $\omega\in\Omega$ and $\mathbb{E}(A_{t})<\infty$ (see definition 1.4.4). $\endgroup$ – Fabio Andrés Gómez Apr 9 '17 at 22:03
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It seems like a minor mistake to me, but it can be easily fixed: It suffices to prove the lemma for bounded $A$. To see this, define $A^m_t=A_{t\wedge\tau_m}$, where $\tau_m=\inf\{s:A_s\geq m\}$, and for any $k\in\mathbb{N}$ take $m(k)$ such that $$ E\int_{\tau_{m(k)}}^kX_t^2dA_t\leq 1/k. $$ By hypothesis, the lemma is true for $A^m$, so one can find a simple process $X^{(k)}$ such that $$ E\int_0^k|X_t^{(k)}-X_t|^2dA^{m(k)}_t\leq1/k. $$ One can choose $X^{(k)}_s$ to be $0$ for $s\geq \tau_{m(k)}$ (since $dA^{m(k)}=0$ on $[\tau_{m(k)},\infty)$) and by putting the two above together, $$ E\int_0^k|X_t^{(k)}-X_t|^2dA_t\leq2/k, $$ from which the claim follows.

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