I have a problem in bridging the gaps in the following lemma related to conditional multivariate random variable. Trouble is in handling proper manipulations
Lemma: Let: $(X,Y)$ - continous multivariate random variables described by uniformly continous PDF $f(x,y)$. Define informally random variable $(X|Y=y)$ as taking value $X(\omega)$ under the condition $Y(\omega)=y$ $(\omega \in \Omega, y \in \mathbb{R})$.
Then: the assumption of uniform continuity of f implies:
$$f_{X|Y}(x|y)=\left\{ \begin{array}{rcl} \frac{f(x,y)}{f_{2}(y)} &:& f_{2}(y) \in \mathbb{R} \setminus \{0\}\\ 0 &:& f_{2}(y)=0\\\ \end{array} \right.$$ where: $f_{2}(y)=\int\limits_{-\infty}^{+\infty}f(x,y)dx$
Proof(my attempt): Let: $B \in \mathcal{B}(\mathbb{R}) $, $\epsilon \in \mathbb{R}_{+}$, $y \in \mathbb{R}$. Then:
$P[X \in B | Y \in (y-\epsilon, y +\epsilon)] = \frac{P[X \in B \wedge Y \in (y-\epsilon, y+\epsilon)]}{P[Y\in(y-\epsilon, y+\epsilon)]} = \frac{P[X \in B \wedge Y \in (y-\epsilon, y+\epsilon)]}{P[X \in \mathbb{R} \wedge Y\in(y-\epsilon, y+\epsilon)]}=\frac{\int\limits_{B}\int\limits_{y-\epsilon}^{y+\epsilon} f(x,y)dydx}{\int\limits_{-\infty}^{+\infty}\int\limits_{y-\epsilon}^{y+\epsilon}f(x,y)dydx} \xrightarrow{\epsilon \to +\infty} \frac{\int\limits_{B}\int\limits_{-\infty}^{+\infty} f(x,y)dydx}{\int\limits_{-\infty}^{+\infty}\int\limits_{-\infty}^{+\infty}f(x,y)dydx}$
I got stuck at last manipulation. I do not feel how to make further manipulations to obtain desired form.
I would be thankful for help!
I managed to see further steps of reasoning:
$F_{X|Y}[B|(y-\epsilon,y+\epsilon)]=P[X \in B | Y \in (y-\epsilon, y +\epsilon)] = \frac{P[X \in B \wedge Y \in (y-\epsilon, y+\epsilon)]}{P[Y\in(y-\epsilon, y+\epsilon)]} = \frac{P[X \in B \wedge Y \in (y-\epsilon, y+\epsilon)]}{P[X \in \mathbb{R} \wedge Y\in(y-\epsilon, y+\epsilon)]}=\frac{\int\limits_{B}\int\limits_{y-\epsilon}^{y+\epsilon} f(x,y)dydx}{\int\limits_{-\infty}^{+\infty}\int\limits_{y-\epsilon}^{y+\epsilon}f(x,y)dydx} \xrightarrow{\epsilon \to +\infty} \frac{\int\limits_{B}\int\limits_{-\infty}^{+\infty} f(x,y)dydx}{\int\limits_{-\infty}^{+\infty}\int\limits_{-\infty}^{+\infty}f(x,y)dydx} = \frac{\int\limits_{-\infty}^{+\infty}\int\limits_{B} f(x,y)dxdy}{\int\limits_{-\infty}^{+\infty}\int\limits_{-\infty}^{+\infty}f(x,y)dxdy}\implies \frac{\partial F}{\partial y}= \frac{\int\limits_{B} f(x,y)dx}{\int\limits_{-\infty}^{+\infty}f(x,y)dx}=\frac{\int\limits_{-\infty}^{+\infty}f(x,y)dx}{f_{2}(y)} \implies \frac{\partial F}{\partial y\partial x} = \frac{\partial}{\partial y}\frac{\partial F}{\partial x} = \frac{f(x,y)}{f_{2}(y)} $