I have that $f(z)=\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}z^{2}}$ which is stated to be the probability density function of a standard normal distribution on $\mathbb{R}$ and $f_{\alpha}(x|z)=\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}(x-\alpha_{0}-\alpha_{1}z)^{2}}$ for $\alpha = (\alpha_{0},\alpha_{1}) \in \mathbb{R^{2}}$ and stable(constant) $z \in \mathbb{R}$. The latter function is stated to be the probability density function of a normal distribution with mean $\alpha_{0}+\alpha_{1}z$ and variance 1.
Let $h_{\alpha}(x,z) = f_{\alpha}(x|z)f(z)$ and show that this is a probability density function for a regular normal distribution.
According to my textbook both functions are examples of regular normal distributions so $h_{\alpha}(x,z)$ must also be one but I am not sure how to formally show this. I thought about using the Fubini-Tonelli Theorem to work with the integral and show that they should both evaluate to 1. Would that be a viable approach or is there another suitable way?
$\int \int h_{\alpha}(x,z)f(z)dxdz=\int (\int f_{\alpha}(x,z)dx)f(z)dz $ and $\int f_{\alpha}(x,z)dx=1$ for any $z$. Hence, $\int \int h_{\alpha}(x,z)f(z)dxdz=\int f(z)dz=1$.
In the first step I have used Tonelli's Theorem. When you integrate w.r.t. $x$ $f(z)$ acts as a constant so you can pull it out of the integral. $f_{\alpha}(x,z)$ is a normal density function for fixed $z$ (with mean $\alpha_0+\alpha_1 z$ and variance $1$) so its integral w.r.t. $x$ is $1$. Finally, $\int f(z)dz=1$.