real analysis - Prove graph of measurable function is measurable in $sigma$-finite case and that the product measure is $0$.

Let $(X, \mathcal{A}, \mu)$ be a $\sigma$-finite measure space, and let $f: X \to \mathbb{R}$ be measurable. Then, $\Gamma(f)$, the graph of $f$ defined as

$$\Gamma = \{(x,y) \in X \times \mathbb{R}: f(x) = y\}$$

is measurable in the $\sigma$-algebra $\mathcal{A \times L}$ where $(\mathbb{R},\mathcal{L},m)$ is the measure space composed of the Lebesgue $\sigma$-algebra ($\mathcal{L}$) on $\mathbb{R}$ and the Lebesgue measure $m$.

Furthermore, prove that the product measure is $0$.

For the first part, I am trying to find the measurable rectangle to prove this is measurable in the product sigma algebra.

I know that $\Gamma = X \times \{f(x)\}$

$X \in \mathcal{A}$ trivially. Also, is the reason why $\{f(x)\} \in \mathcal{L}$ the fact that $f$ is measurable? I know that $f$ being measurable means that

$$\{x:f(x) > a\} \in \mathcal{A} \ \ \forall a \in \mathbb{R}$$

How does this translate to $\{f(x)\}$ being measurable on $\mathcal{L}$?

Furthermore, assuming this is proven. Let $\chi_A$ be the indicator function of some set $A$.

We have that the measure of $\Gamma$, by definition is

$$(\mu \times m) (\Gamma)=\int_\Gamma \mathrm{d}(\mu \times m) = \int_{X\times\mathbb{R}} \chi_\Gamma ((x,y)) \mathrm{d}(\mu \times m)$$

and since the indicator function is, by definition, non-negative, we can use Fubini's theorem to get

$$(\mu \times m)(\Gamma)=\int_X\int_\mathbb{R} \chi_{\{(x,y):f(x)=y\}} ((x,y)) \mathrm{d}m \mathrm{d}\mu$$
But here I have no idea on how to do the first integral or how to proceed in general from here.

Thank you so much!