Why $\int_{E} g(x|y) dx$ is Borel-measurable for any $E\in \mathcal{B}(\mathbb{R})?$

Question

Let $(X, Y)$ be a random variable with an absolutely continuous distribution function given by probability density function $f$. Let $f_{1}$ and $f_{2}$ be the marginal probability density functions of $X$ and $Y$, respectively. We show that for any $E \in \mathcal{B}(\mathbb{R})$ $$ P(X \in E \mid Y=y )=\int_{E}\frac{f(x, y)}{f_2(y)} dx \quad\text {a.s. }\left(P_{Y}(y)\right),$$

that is, $\int_{E} g(x|y) dx$, where $g(x|y)=f(x, y) / f_{2}(y)$, is a version of $P(X \in E | Y=y).$ First note that $g(x|y)$ is defined only for $f_{2}(y) \neq 0$, but this set has probability $0$.

Clearly ${\color{DarkRed} {\int_{E} g(x|y) dx} }$ ${\color{DarkRed} {\text{is Borel-measurable for any}} }$ ${\color{DarkRed} {E\in \mathcal{B}(\mathbb{R}).} }$ .........

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I don't know why $\int_{E} g(x|y) dx$ is Borel-measurable for any $E\in \mathcal{B}(\mathbb{R})?$

I'm not sure if my following approach is correct.

Let's define $\tilde{g}(x|y):=\begin{cases} \frac{f(x,y)}{f_2(y)}& \text{ if } f_2(y)>0 \\ \quad 0 & \text{ if } f_2(y)=0 \end{cases}.$

Since the density function $f$ is a $\mathcal{B}(\mathbb{R^2})/\mathcal{B}(\mathbb{R})$-measurable function from $\mathbb{R^2}$ to $\mathbb{R}$, $f_1(x)=\int_{\mathbb{R}}f(x,y)dy,f_2(y)=\int_{\mathbb{R}}f(x,y)dx$ and $\tilde{g}(x|y)$ are $\mathcal{B}(\mathbb{R^2})/\mathcal{B}(\mathbb{R})$-measurable functions.

For any fixed $E\in \mathcal{B}(\mathbb{R})$, $\chi_{E}(x)\cdot \tilde{g}(x|y)$ is a $\mathcal{B}(\mathbb{R^2})/\mathcal{B}(\mathbb{R})$-measurable function from $\mathbb{R^2}$ to $\mathbb{R}$. By Tonelli's Theorem, $\int_{\mathbb{R}}\chi_{E}(x)\cdot \tilde{g}(x|y)dx=\int_{E}\tilde{g}(x|y)dx$ is a $\mathcal{B}(\mathbb{R})$-measurable function.

Next,I only get $\int_{E} g(x|y) dx=\int_{E} \tilde{g}(x|y) dx \quad \text {a.s.}\left(P_{Y}(y)\right) $ for any fixed $E\in \mathcal{B}(\mathbb{R})$, i.e. $\int_{E} g(x|y) dx$ is almost surely of $\mathcal{B}(\mathbb{R})$-measurable function on $\left(\mathbb{R},\mathcal{B}(\mathbb{R}),P_Y\right)$ for any fixed $E\in \mathcal{B}(\mathbb{R}).$ But why the author said $\int_{E} g(x|y) dx$ is $\mathcal{B}(\mathbb{R})$-measurable for any $E\in \mathcal{B}(\mathbb{R})?$ Any help you can provide would be welcomed.