Assume you have a stochastic base $(\Omega, \mathcal{F},P,\mathbb{F})$. A submartingale is usually defined as
an adapted process
for each $t$ $E(|X_t|)<\infty$
and $E(X_t|\mathcal{F}_s)\ge X_s$ a.s.
However, in a book I am reading I have come over another definition, which is almost the same, but it does not require integrability, but reguires that $E(X_t^+)< \infty$. (The author also adds càdlàg paths as a requirement, I am not sure if that is relevant for my question.)
This creates a problem for me. Because I am not sure then how to construct the conditional expectation $E(X_t|\mathcal{F}_s)$ using the Radon-Nikodym theorem. The most natural step is constructing $E(X_t^+|\mathcal{F}_s)$ and $E(X_t^-|\mathcal{F}_s)$ and defining $E(X_t|\mathcal{F}_s)=E(X_t^+|\mathcal{F}_s)-E(X_t^-|\mathcal{F}_s)$.
Because of the integrability, there is no problem using Radon-Nikodym to define $E(X_t^+|\mathcal{F}_s)$, the problem is defining $E(X_t^-|\mathcal{F}_s)$. The first step is defining a measure on $(\Omega,\mathcal{F_s},Q)$ such that $Q(A) = E(X^-_t\mathcal{X}_A), A \in \mathcal{F}_s$. In order to use the Radon-Nikodym theorem we need $\sigma$-finiteness. But I can not see that we have that in this case? We could define $E_n = \{X^-_t<n\}$, since $X_t^-$ are $\mathcal{F}_t$ measurable, these sets are in $\mathcal{F}_t$, so we do have that $(\Omega, \mathcal{F}_t,Q)$ is $\sigma$-finite. But the sets may not be $\mathcal{F}_s$-measurable, so how do we get $\sigma$-finiteness on the space $(\Omega,\mathcal{F}_s,Q)$?
Update: I get the same problem in this book: Continuous Martingales and Brownian motion. However I can't find how they construct the conditional expectation in this book either. There must be a simple answer for this since it is used in many books.
For any non-negative random variable $\xi$, if $E(\xi)<\infty$, the conditional expectation $E(\xi\mid \mathscr{G})$ is well defined, where $\mathscr{G}$ is a sub-$\sigma$-algebra. On the other hand, if $E(\xi)=\infty$, we consider, for $n\ge 1$, the random variable \begin{align*} \xi_n = \xi\, \mathbb{I}_{\xi < n}. \end{align*} Note that, $\{\xi_n\}_{n=1}^{\infty}$ is non-decreasing and \begin{align*} \lim_{n\rightarrow \infty}\xi_n = \xi, \end{align*} $P$-a.s. For each $n$, we define the set function $Q_n$ on the sub-$\sigma$-algebra $\mathscr{G}$ defined by \begin{align*} Q_n(A) = \int_A \xi_n dP, \end{align*} for $A \in \mathscr{G}$. Then $Q_n$ is a finite measure, and the Radon-Nikodym derivative $dQ_n/dP$ exists. That is, the conditional expectation $E(\xi_n \mid \mathscr{G})$ is well defined. Moreover, it can be shown that, for $n \ge 1$, \begin{align*} E(\xi_n \mid \mathscr{G}) \le E(\xi_{n+1} \mid \mathscr{G}), \end{align*} $P$-a.s. See Page 195 of this book. We then define \begin{align*} E(\xi\mid \mathscr{G}) = \lim_{n\rightarrow \infty}E(\xi_n \mid \mathscr{G}). \end{align*}
For any random variable $\eta$, the conditional expectation $E(\eta\mid \mathscr{G})$ is considered to be defined if \begin{align*} \min\left(E(\eta^+\mid \mathscr{G}), \, E(\eta^-\mid \mathscr{G}) \right) < \infty, \end{align*} $P$-a.s.