Pointwise convergence in limit of conditional probabilities implies almost sure convergence? Consider two sequences of real valued random variables $\{X_n\}_{n\in \mathbb{N}}$ and $\{Y_n\}_{n\in \mathbb{N}}$, with, $\forall n \in \mathbb{N}$, $X_n$ and $Y_n$ defined on the probability space $(\Omega, \mathcal{F}, P), $ respectively with support $\mathcal{X}, \mathcal{Y}$. Consider a sequence of real numbers $\{a_n\}_{n\in \mathbb{N}}$. 

Assume that for some $b\in \mathbb{R}$
$$
\lim_{n\rightarrow \infty} | \overbrace{P(X_n\leq b| Y_n=y)}^{\text{scalar}}-a_n|=0 \text{ }\text{ $\forall y \in \mathcal{Y}$}
$$

Does this imply 
$$
\overbrace{P(X_n\leq b| Y_n)}^{\text{Random variable because of $Y_n$ (unless $X_n\perp Y_n$)}}-a_n \rightarrow_{a.s.}0 \text{ as $n\rightarrow \infty$}
$$
?
Given:

$$\forall y \in \mathcal Y, \lim_n \frac{E[1_{B_n} 1_{Y_n=y}]}{P(Y_n=y)}= \lim_n a_n$$

Conjecture:

$$\lim_n E[1_{B_n}| Y_n]= \lim_n a_n$$

Ugh, the fact that we can say '$Y_n=y$' means $Y_n$ is discrete? In that case:

$$E[1_{B_n}| Y_n] = \sum_{y \in \mathcal Y} E[1_{B_n}| Y_n = y]1_{\{Y_n=y\}}$$

$$\to \lim E[1_{B_n}| Y_n] = \lim \sum_{y \in \mathcal Y} E[1_{B_n}| Y_n = y]1_{\{Y_n=y\}}$$

$$= \sum_{y \in \mathcal Y} \lim E[1_{B_n}| Y_n = y]1_{\{Y_n=y\}}$$

$$ = \sum_{y \in \mathcal Y} [\lim E[1_{B_n}| Y_n = y]][\lim[1_{\{Y_n=y\}}]]$$

$$ = \sum_{y \in \mathcal Y} [\lim a_n] \lim[1_{\{Y_n=y\}}]$$

$$ = [\lim a_n] \sum_{y \in \mathcal Y} \lim[1_{\{Y_n=y\}}]$$

Well if I proved for the discrete case, then it would not have been sufficient to disprove as a whole. However, I disproved for the discrete case which is sufficient to disprove as a whole.



Observation: In the discrete case, this sounds like:

$$\lim_n b_{n,k} = \lim_n a_n \to \lim_n \sum_{k} b_{n,k}c_k = \lim_n a_n$$
Given:

$$\forall y \in \mathcal Y, \lim_n \frac{E[1_{B_n} 1_{Y_n=y}]}{P(Y_n=y)}= \lim_n a_n$$

Conjecture:

$$\lim_n E[1_{B_n}| Y_n]= \lim_n a_n$$

Ugh, the fact that we can say '$Y_n=y$' means $Y_n$ is discrete? In that case:

$$E[1_{B_n}| Y_n] = \sum_{y \in \mathcal Y} E[1_{B_n}| Y_n = y]1_{\{Y_n=y\}}$$

$$\to \lim E[1_{B_n}| Y_n] = \lim \sum_{y \in \mathcal Y} E[1_{B_n}| Y_n = y]1_{\{Y_n=y\}}$$

$$= \sum_{y \in \mathcal Y} \lim E[1_{B_n}| Y_n = y]1_{\{Y_n=y\}}$$

$$ = \sum_{y \in \mathcal Y} [\lim E[1_{B_n}| Y_n = y]][\lim[1_{\{Y_n=y\}}]]$$

$$ = \sum_{y \in \mathcal Y} [\lim a_n] \lim[1_{\{Y_n=y\}}]$$

$$ = [\lim a_n] \sum_{y \in \mathcal Y} \lim[1_{\{Y_n=y\}}]$$

Well if I proved for the discrete case, then it would not have been sufficient to disprove as a whole. However, I disproved for the discrete case which is sufficient to disprove as a whole.



Observation: In the discrete case, this sounds like:

$$\lim_n b_{n,k} = \lim_n a_n \to \lim_n \sum_{k} b_{n,k}c_k = \lim_n a_n$$