Is $\mathbb{E}[X|\mathscr{B}]$ well-defined when $\mathbb{E}[X|\mathscr{A}]$ is, where $\mathscr{A}\subset\mathscr{B}$ and $X^\pm$ isn't integrable?

Question

Let $X$ be a random variable on a probability space $(\Omega,\mathscr{F},\mathbb{P})$, and let $\mathscr{A}$ and $\mathscr{B}$ be sub-$\sigma$-fields of $\mathscr{F}$ such that $\mathscr{A}\subset\mathscr{B}$.

Question: Is $\mathbb{E}[X|\mathscr{B}]$ well-defined when $\mathbb{E}[X|\mathscr{A}]$ is, but when both the negative part $X^-$ and the positive part $X^+$ of $X$ are not integrable?

For what I understand by well-defined, let's consider the definition of conditional expectation as given by e.g. Shiryaev (Probability, 2013; p. 213):

The conditional expectation of $X^\pm$ with respect to $\mathscr{A}$ is an $\mathscr{A}$-measurable random variable $\mathbb{E}[X^\pm|\mathscr{A}]$ satisfying for every $A\in\mathscr{A}$ that \begin{equation*} \int_A\mathbb{E}[X^\pm|\mathscr{A}]d\mathbb{P} = \int_A X^\pm d\mathbb{P}. \end{equation*} If $\mathbb{E}[X^-|\mathscr{A}]\wedge\mathbb{E}[X^+|\mathscr{A}] < \infty$ almost surely then the conditional expectation $\mathbb{E}[X|\mathscr{A}]$ of $X$ with respect to $\mathscr{A}$ is said to be well-defined and is given by \begin{equation*} \mathbb{E}[X|\mathscr{A}] = \mathbb{E}[X^+|\mathscr{A}] - \mathbb{E}[X^-|\mathscr{A}]. \end{equation*}

In other words, does $\mathbb{E}[X^-|\mathscr{A}]\wedge\mathbb{E}[X^+|\mathscr{A}] < \infty$ almost surely imply $\mathbb{E}[X^-|\mathscr{B}]\wedge\mathbb{E}[X^+|\mathscr{B}] < \infty$ almost surely?

My attempt so far:

Let $B$ be the event in $\mathscr{B}$ on which $\mathbb{E}[X^-|\mathscr{B}] = \mathbb{E}[X^+|\mathscr{B}] = \infty$. For every $A\in\mathscr{A}$ satisfying $B\subseteq A$ it holds that \begin{equation*} \infty\mathbb{P}(B) =\int_B\mathbb{E}[X^\pm|\mathscr{B}]d\mathbb{P} =\int_B X^\pm d\mathbb{P} \leq \int_A X^\pm d\mathbb{P} =\int_A\mathbb{E}[X^\pm|\mathscr{A}]d\mathbb{P}. \end{equation*} Suppose that $\mathscr{A}$ is generated by a finite or countable partition. Then there is a smallest $A\in\mathscr{A}$ satisfying $B\subseteq A$. Either $\mathbb{P}(A) = 0$, or on $A$ we have that $\mathbb{E}[X^\pm|\mathscr{A}]$ equals $\frac{1}{\mathbb{P}(A)}\int_A X^\pm d\mathbb{P}$. Since by assumption $\mathbb{E}[X|\mathscr{A}]$ is well-defined, either $\mathbb{P}(A) = 0$ or at least one of $\int_A X^-d\mathbb{P}$ and $\int_A X^+d\mathbb{P}$ is finite. Both possibilities imply that $B$ must have probability zero, which means that $\mathbb{E}[X|\mathscr{B}]$ is well-defined.

Is there always a set $A\in\mathscr{A}$ satisfying $B\subseteq A$ and small enough such that at least one of $\int_A X^-d\mathbb{P}$ and $\int_A X^+d\mathbb{P}$ is finite? This would solve the problem. Or is there some other approach?

(In the first version of this question I only demanded $X$ to be non-integrable, but this would still allow for one of $\int_\Omega X^-d\mathbb{P}$ and $\int_\Omega X^+d\mathbb{P}$ to be finite, which by the above argument would then imply the well-definedness of $\mathbb{E}[X|\mathscr{B}]$. So I restricted to both $X^-$ and $X^+$ being non-integrable.)

Answer 1

I believe I found a proof that the event $B$ in $\mathscr{B}$ on which $\mathbb{E}[X^-|\mathscr{B}] = \mathbb{E}[X^+|\mathscr{B}] = \infty$ has probability zero, thereby showing that $\mathbb{E}[X|\mathscr{B}]$ is well-defined.

Let $A^\pm_\infty = \{\mathbb{E}[X^\pm|\mathscr{A}] = \infty\}$ and for each nonnegative integer $n$ let $A^\pm_n = \{n\leq\mathbb{E}[X^\pm|\mathscr{A}]\leq n+1\}$. These events lie in $\mathscr{A}$ and satisfy $\Omega = \big(\bigcup^\infty_{n=0}A^\pm_n\big)\cup A^\pm_\infty$, hence \begin{equation*} \begin{aligned} \mathbb{P}(B) &= \sum^\infty_{n=0} \mathbb{P}(B\cap A^-_n) + \mathbb{P}(B\cap A^-_\infty) \\ &= \sum^\infty_{n=0} \mathbb{P}(B\cap A^-_n) + \sum^\infty_{n=0} \mathbb{P}(B\cap A^-_\infty\cap A^+_n) + \mathbb{P}(B\cap A^-_\infty\cap A^+_\infty) . \end{aligned} \end{equation*} By assumption $\mathbb{E}[X|\mathscr{A}]$ is well-defined, which means that $\mathbb{P}(A^-_\infty\cap A^+_\infty) = 0$, and for each $n\in\mathbb{N}$ we have \begin{multline*} \infty\mathbb{P}(B\cap A^-_n) = \int_{B\cap A^-_n} \mathbb{E}[X^-|\mathscr{B}] d\mathbb{P} = \int_{B\cap A^-_n} X^- d\mathbb{P} \\ \leq \int_{A^-_n} X^- d\mathbb{P} = \int_{A^-_n} \mathbb{E}[X^-|\mathscr{A}] d\mathbb{P} \leq (n+1)\mathbb{P}(A^-_n) < \infty \end{multline*} and \begin{multline*} \infty\mathbb{P}(B\cap A^-_\infty\cap A^+_n) = \int_{B\cap A^-_\infty\cap A^+_n} \mathbb{E}[X^+|\mathscr{B}] d\mathbb{P} = \int_{B\cap A^-_\infty\cap A^+_n} X^+ d\mathbb{P} \\ \leq \int_{A^+_n} X^+ d\mathbb{P} = \int_{A^+_n} \mathbb{E}[X^+|\mathscr{A}] d\mathbb{P} \leq (n+1)\mathbb{P}(A^+_n) < \infty , \end{multline*} which means that $\mathbb{P}(B\cap A^-_n) = \mathbb{P}(B\cap A^-_\infty\cap A^+_n) = 0$ for each $n\in\mathbb{N}$. From this we can conclude that $B$ has probability zero, and thus $\mathbb{E}[X|\mathscr{B}]$ is well-defined.