$t \geq 1$.

\[lem\_1010\]

Lemma \[lem\_1010\] is the well-known Borell’s lemma (see its elegant proof in [@borell_what] or [@MS Theorem III.3]). Let $f: \RR^n \rightarrow [0, \infty)$ be an integrable function. For $\theta \in S^{n-1}$ and $t \in \RR$ we define $H_{\theta, t}
= \{ x \in \RR^n ; \langle x, \theta \rangle \leq t \}$ and $$M_f(\theta, t) = \int_{H_{\theta,t}} f(x) dx. \label{eq_146}$$ The function $M_f$ is continuous in $\theta$ and $t$, non-decreasing in $t$, and its derivative $\frac{\partial
M_f}{\partial t}$ is the Radon transform of $f$. Thus, in principle, one may recover the function $f$ from a complete knowledge of $M_f$. Clearly, for any subspace $E \subset \RR^n$, $$M_{\pi_E(f)}(\theta, t) = M_f(\theta, t) \ \ \ \text{for all} \ \ \ \theta \in S^{n-1} \cap E, t \in \RR.
\label{proj_ok}$$ Moreover, let $\theta \in S^{n-1}$, let $E = \RR \theta$ be the one-dimensional subspace spanned by $\theta$, and denote $g = \pi_E(f)$. Then $$g(t \theta) = \frac{\partial}{\partial t}
M_{\pi_E(f)} (\theta, t) = \frac{\partial}{\partial t} M_{f}
(\theta, t) \label{eq_621_}$$ for all points $t \in \RR$ where, say, $g(t \theta)$ is continuous.

Let $n \geq 1$ be an integer, and let $f: \RR^n \rightarrow [0, \infty)$ be an isotropic, log-concave 