# A generalization of conditional expectation to non-integrable random variables

I'm reading about conditional expectations from Achim Klenke's Probability Theory: A Comprehensive Course, which defines conditional expectations for integrable random variables $$X$$ with respect to a $$\sigma$$-algebra $$\mathcal F$$ as the unique (a.s.) $$\mathcal F$$-measurable random variable $$Y$$ for which $$\mathbb E[X\mathbb 1_A] = \mathbb E[Y\mathbb 1_A]$$ for all $$A \in \mathcal F$$. The following remark is made to generalize this definition to some non-integrable random variables:

Let $$X : \Omega \to \mathbb R$$ be a random variable such that $$X^- \in \mathcal L^1(\mathbb P)$$. We can define the conditional expectation as the monotone limit $$\mathbb E[X\,|\,\mathcal F] = \lim_{n \to \infty} \mathbb E[X_n\,|\,\mathcal F]$$ where $$-X^- \leq X_1$$ and $$X_n \uparrow X$$. Due to the monotonicity of the conditional expectation, it is easy to show that the limit does not depend on the choice of the sequence $$(X_n)$$ and that it fulfills the conditions of [the definition of conditional convergence].

(Presumably we want $$(X_n) \subset \mathcal L^1(\mathbb P)$$ to avoid circular definitions.)

My question: Why do we need $$-X^- \leq X_1$$?

It's not hard to verify these claims. By monotonicity of conditional expectation, we get that $$\mathbb E[X_n\,|\,\mathcal F] \uparrow \mathbb E[X\,|\,\mathcal F]$$ (this limit may be infinite). So for any $$A \in \mathcal F$$, by the Beppo-Levi monotone convergence theorem, since the $$\mathbb E[X_n\,|\,\mathcal F]$$ are integrable, $$\mathbb E[\mathbb E[X\,|\,\mathcal F]\mathbb 1_A] = \mathbb E\left[\lim_{n \to \infty} \mathbb E[X_n\,|\,\mathcal F]\mathbb 1_A\right] = \lim_{n \to \infty} \mathbb E\left[\mathbb E[X_n\,|\,\mathcal F]\mathbb 1_A\right] = \lim_{n \to \infty} \mathbb E[X_n\mathbb 1_A] = \mathbb E[X\mathbb 1_A].$$ If $$(\tilde X_n)$$ is another sequence with $$\tilde X_n \uparrow X$$ and $$-X^- \leq \tilde X_1$$, letting $$Y = \lim \mathbb E[X_n \,|\,\mathcal F]$$ and $$\tilde Y = \lim\mathbb E[\tilde X_n\,|\,\mathcal F]$$, then both $$Y$$ and $$\tilde Y$$ are $$\mathcal F$$-measurable (as a limit of $$\mathcal F$$-measurable functions), and this calculation shows $$\mathbb E[Y\mathbb 1_A] = \mathbb E[\tilde Y \mathbb 1_A]$$ for every $$A \in \mathcal F$$. It follows that $$Y = \tilde Y$$.

I can see that we'd want $$X^- \in \mathcal L^1(\mathbb P)$$ because otherwise we can't have that both $$(X_n) \in \mathcal L^1(\mathbb P)$$ and $$X_n \uparrow X$$. But I don't see why we need $$-X^- \leq X_1$$. In fact this seems contradictory if we require $$X_n \uparrow X$$. Am I missing something?

Since your random variables may take negative values you need a generalized version of the monotone convergence theorem. Such a result exists which, in addition to $$X_n\uparrow X$$, assumes that $$X_1^-$$ is integrable. We ensure that this is satisfied by assuming $$-X^-\leq X_1$$.
Note that you can construct a sequence $$(X_n)$$ with the desired properties by taking $$X_n=X1_{\{X\leq n\}}$$.
• The MCT you're talking about generalizing assumes the random variables are nonnegative, but not necessarily integrable. You shouldn't need to generalize the MCT in that way if the random variables $X_n$ are integrable (see Klenke, Theorem 4.20): if $N$ is the null event on which $X_n$ does not converge to $X$, then $X_n' = (X_n - X_1)\mathbb 1_{N^c}$ is nonnegative and $X-X_1$ a.s. By linearity and the MCT for nonnegative measurable random variables, $\mathbb E[X_n] \to \mathbb E[X]$. Jan 19 at 13:42
• I am just trying to guess why one would assume $-X^-\leq X_1$ and this is one possibility. But as you suggest it is sufficient that $X_1$ is integrable. However, it is my experience that people do not always care about assuming too much as long as it solves the problem. The assumptions on $(X_n)$ may be stricter than necessary but as long as such a sequence exists (and it does) it does not really matter. Jan 19 at 14:01
Here is my understanding. If $$-X^{-}\le X_1$$, then we would have $$X_n^-\uparrow X^-$$ and $$X_n^+\uparrow X^+$$. Thus, $$X_n\in\mathcal{L}^1$$ iff $$X_n^+\in \mathcal{L}^1$$. We can take such sequence, say, $$X_n=X\mathbb{I}_{\{|X|\le n \}}$$. If instead $$-X^- > X_1$$, then we would have $$X^- and $$X_n^-\downarrow X^-$$. It is not clear if one can take $$(X_n)$$ such that both $$X_n^\pm\in \mathcal{L}^1$$ since we only know $$X^-\in \mathcal{L}^1$$