Is "$\mathbb{E}(T_n|X)\rightarrow 0 $ a.s." equivalent to a statement that does not involve the Radon–Nikodym derivative as a black box?

Question

Let $\{T_n\}_n$ be a sequence of random variables, and let $X$ be another random variable.

Each $\mathbb{E}(T_n|X)$ is a random variable, therefore the statement "$\mathbb{E}(T_n|X)\rightarrow 0$ almost surely" makes sense in the sense of almost sure convergence of random variables.

What makes this difficult to unpack is that almost sure convergence is phrased in terms of individual $\omega$'s in the probability space, but the definition of $\mathbb{E}(T_n|X)$ is via the existence of the Radon-Nikodym derivative, and so is non-constructive...

The question is whether you can have an equivalent definition of this statement that does not use the Radon-Nikodym derivative as a black box. For example, is it equivalent to the statement: "for every measurable set $U$ in the co-domain of $X$, the mean of $T_n$ over $X^{-1}(U)$ converges to $0$"?

Answer 1

You are asking: Is the statement "$E(T_n|X)\to0$ almost surely [a.s.]" equivalent to the statement "for every measurable set $U$ in the co-domain of $X$, the mean of $T_n$ over $X^{-1}(U)$ converges to $0$"?

The answer is no in general. First here, to talk about the mean of $T_n$ over $X^{-1}(U)$, you need to assume that $P(X^{-1}(U))>0$. Taking this into account, we can restate your question as follows: Are the following two statements equivalent:

(i) $Y_n\to0$ a.s.

(ii) $EY_n\,1_{X\in U}\to0$ for all measurable sets $U$ in the co-domain of $X$,

where $Y_n:=E(T_n|X)$. Neither one of the implications (i)$\implies$(ii) or (ii)$\implies$(i) holds in general.

Indeed, let $X$ be uniformly distributed on $[0,1]$, and then let $T_n:=nX1_{X<1/n}$, so that $Y_n=T_n$ and, for $U=[0,1]$, $EY_n\,1_{X\in U}=ET_n=1\not\to0$, whereas $Y_n=T_n\to0$ a.s. So, the implication (i)$\implies$(ii) fails to hold.

On the other hand, let again $X$ be uniformly distributed on $[0,1]$, and then let $T_n:=X1_{X\in\delta_n}$, where $\delta_1,\delta_2,\dots$ are subintervals of $[0,1]$ such that $T_n\to0$ in probability but not a.s. Then again $Y_n=T_n$ and hence (i) fails to hold, whereas (ii) holds by dominated convergence. So, the implication (ii)$\implies$(i) fails to hold.