# $E[X|\mathcal{Q}] \leq \liminf_nE [X_n\mid\mathcal{Q}]$

If $$(X_n)_n$$ is a sequence of nonnegative random variables and $$\mathcal{Q}$$ is a sub $$\sigma$$-algebra, then the conditional Fatou lemma holds almost surely, $$E[\liminf_n X_n\mid\mathcal{Q}] \leq \liminf_n E[X_n\mid\mathcal{Q}].$$

Let's say that $$(X_n)_n$$ converges in probability to $$X$$. Is it true that, almost surely, $$E[X\mid\mathcal{Q}] \leq \liminf_nE [X_n\mid\mathcal{Q}] \text{ ?}$$

• May I ask if the term "almost surely" here has a rigorous mathematical definition, or is used in plain English sense?
– Anon
Apr 7, 2021 at 0:01
• @Kaind "Almost surely" commonly means "with probability 1". Apr 7, 2021 at 0:03
• @BrianMoehring What is begging the question and answering the wrong question here? You give no comment and you still downvote. Show me where is the error in my argument?
– NN2
Apr 7, 2021 at 1:12
• @BrianMoehring I was modifying "As $X = \lim X_n$ then $\liminf X_n$ exists, hence $\liminf X_n = \lim X_n = X$. So, the hypothese that you said we didn't have, in fact, you have it. Source: math.stackexchange.com/q/122755/195378
– NN2
Apr 7, 2021 at 1:34
• @NN2 The only thing I can say is that if we only have $X = \operatorname{plim} X_n$ and not $X = \lim X_n$, then we may even assume $\liminf X_n \neq \limsup X_n$ almost surely, so there's no way to make the argument valid that you're trying to make. Apr 7, 2021 at 1:56

For an easy counterexample, let $$X_n\geq 0$$ be any sequence of random variables with $$\liminf_n X_n = 0$$ which converges to $$X = 1$$ in probability. Then let $$\mathcal{Q}$$ be large enough so that the $$X_n$$ are $$\mathcal{Q}$$-measurable. The inequality in question then becomes $$X \leq \liminf_n X_n,$$ which is false almost surely.
For instance, this is the case when $$X_n$$ is any sequence of independent random variables supported in $$\{0,1\}$$ such that $$P(X_n = 0) \to 0$$ but $$\sum P(X_n = 0) = \infty$$.
• Do note that the inequality is true in at least certain cases, such as when $\mathcal{Q} = \{0, \Omega\}$ or more generally when $\mathcal{Q}$ is generated by a partition of $\Omega$. I can't give a fine condition on $\mathcal{Q}$ for it to be true, but it's apparently false when $\mathcal{Q}$ is large enough. Apr 7, 2021 at 17:36
• We can always find an example on $([0,1],\mathcal{B}([0,1]),\lambda)$ such that $X_n$ converges in measure to $1$, and take $\mathcal{Q}=\mathcal{B}([0,1])$ (large enough) Apr 7, 2021 at 18:38