I will try to answer my question above; it would be great if someone can confirm (since again I did not find any textbook describing this, except the one Exercise in Billingsley mentioned below)!
To set things up, let $(\Omega, \mathcal{F}, \mathbb P)$ be a probability space. $A \in \mathcal{F}$ is an event with probability $\mathbb P(A) > 0$, $X: \Omega \to \mathbb R$ is a random variable and $\mathcal{G} \subset \mathcal{F}$ is a sub-$\sigma$-algebra.
We are interested in defining: $\mathbb E[X \mid A, \mathcal{G}]$. There are two "natural" ways to do this.
First, we will do this by just using the standard definition of conditional expectation but with respect to the measure $\mathbb P_A$, where this is just the conditional probability measure with mass $P_A(B)$ on $\mathcal{F}$-measurable sets $B$:
$$ \mathbb P_A(B) = \frac{\mathbb P(A \cap B)} {\mathbb P(A)}$$
Thus we define $\mathbb E[X \mid A, \mathcal{G}]$ for $X \in L^1(\mathbb P_A)$ by the following properties:
- $\mathbb E[X \mid A, \mathcal{G}]$ is $\mathcal{G}$ measurable.
- $\int_B \mathbb E[X \mid A, \mathcal{G}] d\mathbb P_A = \int_B X d\mathbb P_A $ for all $\mathcal{G}$-measurable sets $B$.
We can quickly see that to check $X \in L^1(\mathbb P_A)$, it is sufficient to check $X \in L^1(\mathbb P)$ while for property 2. we can just check:
$\int_B \mathbb E[X \mid A, \mathcal{G}] d\mathbb P = \int_B X d\mathbb P $ for all sets $B \in \{ G \cap A \mid G \in \mathcal{G}\}$
The second way of defining $\mathbb E[X \mid A, \mathcal{G}]$ is by defining it for indicator variables of $\mathcal{F}$-measurable sets $B$ as (also see related math.se post):
$$ \mathbb E[ \mathbf{1}_{B} \mid A , \mathcal{G}] = \frac{\mathbb E[ \mathbf{1}_{B}\mathbf{1}_{A} \mid \mathcal{G}] }{\mathbb E[ \mathbf{1}_{A} \mid \mathcal{G}]}$$
By exercise 34.4 a) in the book "Probability and Measure" by Billingsley, we get that in fact these two definitions are equivalent. So we are good to go.
Now we are still interested in the calculus of such conditional expectations. It turns out to be simple, since we can just use the standard calculus where the expectations are taken w.r.t. to the measure $\mathbb P_A$! Also properties such as "on the event $A$, $X$ is independent of $\mathcal{G}$", also just mean that $X$ is independent of $\mathcal{G}$ under the measure $\mathbb P_A$.
First observe that since the event $\{ \|X - x\| \leq \epsilon\}$ is $\sigma(X)$-measurable,
$$
\mathbb E(Y \mid \|X - x\| \leq \epsilon)=\frac 1{\mathbb P\left(\|X - x\| \leq \epsilon\right)}\mathbb E(Y \mathbf{1}\{ \|X - x\| \leq \epsilon\})=\frac 1{\mathbb P\left(\|X - x\| \leq \epsilon\right)}\mathbb E(\mu(X)\mathbf{1}\{ \|X - x\| \leq \epsilon\}).$$
Therefore,
$$
\lvert \mu(x)-\mathbb E(Y \mid \|X - x\| \leq \epsilon)\rvert=\left\lvert \frac 1{\mathbb P\left(\|X - x\| \leq \epsilon\right)}\mathbb E(\left(\mu(x)-\mu(X)\right)\mathbf{1}\{ \|X - x\| \leq \epsilon\})\right\rvert
\leqslant \sup_{y:\|y - x\| \leq \epsilon}\lvert \mu(x)-\mu(y)\rvert$$
and we conclude by continuity of $\mu$ at $x$.
If $\mu$ is uniformly continuous, then the convergence is uniform in $x$.
Best Answer
By definition: $$\mathsf P(X>x\mid X>Y)=\dfrac{\mathsf P(X>\max\{x,Y\})}{\mathsf P(X>Y)}\\\mathsf E(\mathbf 1_{\small X>x}\mid X>Y)=\dfrac{\mathsf E(\mathbf 1_{\small X>\max\{x,Y\}})}{\mathsf E(\mathbf 1_{\small X>Y})}\quad$$
So, when $X$ and $Y$ are continuous random variables with joint density function $f_{\small X,Y}$, then :$$\begin{align}\mathsf P(X>x\mid X>Y)&=\dfrac{\iint_{s>\max\{x,t\}}f_{\small X,Y}(s,t)\,\mathrm d \langle s,t\rangle}{\iint_{s>t}f_{\small X,Y}(s,t)\,\mathrm d \langle s,t\rangle}\\[3ex]&=\dfrac{\displaystyle\int_{-\infty}^x\int_x^\infty f_{\small X,Y}(s,t)\,\mathrm d s\,\mathrm d t+\int_{x}^\infty\int_{t}^\infty f_{\small X,Y}(s,t)\,\mathrm d s\,\mathrm d t}{\displaystyle\int_{-\infty}^\infty\int_t^\infty f_{\small X,Y}(s,t)\,\mathrm d s\,\mathrm d t}\end{align}$$
Note: Do not use the same token, $x$, as both a free variable and the integral's bound variable. Always use fresh tokens as an integrals bound variable.