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On Karatzas and Shreve - Brownian motion and Stochastic Processes pg 146 one reads

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I don't understand how saying that $\mathcal{P}^* \subset \mathcal{P}$ is an abuse of notation.

$X \sim Y$ if and only if $\int_0^T (X_u - Y_u)\, d\langle M \rangle_u = 0$ for every $T \in \Bbb{R}_+$.

This being the case, consider

\begin{align} &A = \bigg\{X \text{ measurable and adapted process} \mid \Bbb{P}\bigg[\int_0^T X_t^2 \, d\langle M \rangle_t < \infty\bigg] \text{for every} T \in [0, \infty)\bigg\}\\ &B = \big\{X \in A \mid X \text{ progressively measurable } \big\} \end{align}

$\mathcal{P} = A \big/\sim$, $\mathcal{P}^* = B \big/\sim$. Since $B \subset A$ why is it an abuse of notation to say that $\mathcal{P}^* \subset \mathcal{P}$?

The same question applies to $\mathcal{P} \subset \mathcal{L}$ ($\mathcal{P}^* \subset \mathcal{L}^*$).

I don't understand why this is an abuse of notation.

Conrado Costa
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1 Answers1

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It might not be that obvious but it is indeed an abuse of notation. First of course, and this might be the problem with the intuition here, it is clear, that under the given conditions we have $$ X_t \text{ progressively measurable} \Rightarrow X_t \text{ adapted} $$ So let's assume there are two processes, $X_t$ is not progressively measurable and $Y_t$ is progressively measurable such that we have $X_t\sim Y_t $, which means we have $[X_t]=[Y_t]$. Now we only want the progressively measurable processes in the equivalence class of $[Y_t]\in\mathcal{P}^*$, while the equivalence class $[X_t]\in\mathcal{P}$ still contains all the adapted processes and we then have certainly \begin{align} [Y_t]\subsetneq[X_t] \tag 1 \end{align} But this means, that the elements of $\mathcal{P}^*$ (only progressively measurable processes) which are eqiuvalence classes are not the same as the elements of $\mathcal{P}$ (the adapted processes) and therefore it is not correct - or better an abuse of notation in the words of Shreve and Karatzas - to write $$ \mathcal{P}^*\subset\mathcal{P} $$ while it is true that each element of $\mathcal{P}^*$ itself is contained in the corresponding element of $\mathcal{P}$ as in $(1)$.

user190080
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