# 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?

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.)

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.