$\int_{s = - \infty}^{\infty} Rf(\varphi,s)h(s) ds = \int_{x \in \mathbb{R}^2} f(x) h(\langle \theta, x \rangle) dx$

Question

I want to show the following: Let $f \in L^1(\mathbb{R^2})$ and let $\varphi \in [0,2\pi]$. Let $h \in L^{\infty}(\mathbb R)$. Then $\int_{s = - \infty}^{\infty} Rf(\varphi,s)h(s) ds = \int_{x \in \mathbb{R}^2} f(x) h(\langle \theta, x \rangle) dx$, where $R$ is the Radon transform. The Radon transform is defined as $Rf(\varphi,s) = \int_{t = - \infty}^{\infty} f(s\theta + t\theta^{\bot}) dt$ and $\theta$ is defined as $\theta = \theta(\varphi) = (\cos\varphi, \sin\varphi), \theta^{\bot} = (-\sin\varphi, \cos\varphi)$.

Answer 1

Making the change of variables $s = \langle x,\theta \rangle$ and $t = \langle x,\theta^\perp \rangle$ so that we have $x = s\theta + t\theta^\perp$. Also note that the Jacobian of the transformation is $1$, i.e., $\,dx = \,ds\,dt$ since it is an orthogonal change of coordinate. Then we may write \begin{align*} \int_{\mathbb{R}^2} f(x) h(\langle \theta, x \rangle) \,dx &= \int_{\mathbb{R}^2} f(s\theta + t\theta^\perp) h(s)\,ds\,dt \\&= \int_{\mathbb{R}} \left(\int_{\mathbb{R}} f(s\theta + t\theta^\perp) \,dt\right) h(s) \,ds \\&= \int_{\mathbb{R}} Rf(\varphi,s)h(s)\,ds\end{align*} where, we used Fubini's theorem in the second line.