Continuity of Radon transform w.r.t the angle Let $f \in L^1(\mathbb R^n)$ (or in case it helps, actually a probability density on $\mathbb R^n$). Define the Radon transform $R[f]:S_{n-1} \times \mathbb R \to \mathbb R$ of $f$  by
$$
R[f](w,b) := \int_{\mathbb R^n}\delta(x^\top w - b)f(x)\,dx = \int_{H_{w,b}}f(x)\,d\sigma(x),
$$
where $S_{n-1}$ is the unit-sphere in $\mathbb R^n$ and $d\sigma$ is the areal measure on the hyperplane $H_{w,b}:=\{x \in \mathbb R^n \mid x^\top w = b\}$.
Question. Under what minimal additional conditions on $f$ is mapping $w \mapsto R[f](w,b)$ continuous continuous on $S_{n-1}$, for each $b \in \mathbb R$ ?
 A: This is only a comment (but I amn't entitled) and applies also to your related question.  Your formula  involves three operations--composition, multiplication and integration.  You finally ask when Euler type results hold--when do smoothness properties of the integrand carry over to the function defined by the integral.
In order to address your question, it is necessary to clarify the precise definitions in use.  The presence of the $\delta$ function clearly implies that one  has to work in the context of distribution theory--fortunately the requisite (elementary) theory has been around since the 60's.  First composition: you are composing the Dirac function with a very regular function so no problem there.  The product does, however, require more than integrability from $f$, say continuity.  You then have to interpret the integral as a parametrised one in the sense of distributions but this theory is on record (J. Sebastião e Silva). Now comes the good news: any sensible version of Euler-type theorems (differentiating under the integral sign) holds without further restrictions.
Added after comments. Avoiding distributions just begs the question. Or do you really think you can integrate an $L^1$ function along a hyperspace?  And in what sense are you carrying out the final integral? Given a simple and flexible notion of the parametrised integral of distributions which supplies existence without further restrictions and the Göttergeschenk of Euler theorems for free, why would you spurn it?
I will leave (and given the tone of the comments, leave means leave) with my wee conjecture: if $f$  is a continuous integrable function, then your expression is infinitely differentiable with respect to $w$ and the derivatives can be calculated by differentiating under the integral sign.
I am formulating this as a conjecture since I haven't had an opportunity to sit down and write out the details.
A: In the following I shall assume that $f$ is continuous with compact support.
It is known that if $\varphi : \mathbb R^n \to \mathbb R$ has $\nabla \varphi \ne 0$ at all the points of $H = \varphi^{-1} (0)$, then
$$\int _{\mathbb R^n} f(v) \, \delta (\varphi(v)) \, \mathrm d v = \int _H \frac{f(s)} {\|\nabla \varphi(s)\|} \, \mathrm d s \ ,$$
the latter integral being with respect to the natural measure induced on $H$.
In your case $\varphi(x) = \langle w, x \rangle - b$, so $\| \nabla \varphi \| = \| w \| = 1$.
Since $w \ne 0$, let us assume that $w_n \ne 0$. Consider the (global) parametrization $h : \mathbb R^{n-1} \to H_{w,b}$ given by
$$h(u_1, \dots, u_{n-1}) = \left( u_1, \dots, u_{n-1}, \frac 1 {w_n} (b - u_1 \, w_1 - \dots - u_{n-1} \, w_{n-1}) \right) .$$
In this paramerization, the coefficients of the Riemannian metric on $H_{w,b}$ are $g_{ii} = 1 + \frac {w_i ^2} {w_n ^2}$ and $g_{ij} = \frac {w_i w_j} {w_n ^2}$ for $i < j$, so your integral becomes
$$\sqrt {\det (g_{ij})} \int _{\mathbb R ^{n-1}} f \left( u_1, \dots, u_{n-1}, \frac 1 {w_n} (b - u_1 \, w_1 - \dots - u_{n-1} \, w_{n-1}) \right) \ \mathrm d u_1 \dots \mathrm d u_{n-1} \ .$$
Since $f$ is continuous with compact support, the integral written above exists and is finite. Using Lebesgue's dominated convergence theorem, if the sequence $w^{(k)}$ converges to $w$ then you also get the convergence of the corresponding integrals.
The determinant, too, is continuous on some neighbourhood of some given $w$: if the coordinate $w_n$ is non-zero, then it is so on some neighbourhood $U$ of $w$ in $S^{n-1}$ (by the continuity of the coordinate functions), so you may safely divide by it; since all the operations involved in that determinant al algebraic, the determinant will depend continuously on $w$ in the neighbourhood $U$. Since continuity is a local property, this is enough.
To conclude: $f$ being continuous with compact support is enough for the continuity of its Radon transform (with respect to $w$ and $b$ jointly); it is up to you to decide whether this is too restrictive or not. This condition (continuity and compact support) is sufficient to ensure that the restriction of $f$ to any hyperplane $H_{w,b}$ is integrable. Absent it, I do not know what to replace it with.
