[proofplan] We prove the result from the Borel-set definition of log-concavity, using compact inner approximation only to keep every set to which log-concavity is applied Borel. Given Borel sets downstairs, choose compact subsets of their inverse images upstairs; compactness makes their Minkowski convex combination compact and hence Borel. The linearity of $P$ then places this compact convex combination inside the inverse image of the downstairs convex combination, and inner regularity lets the compact approximations recover the full pushforward measures. [/proofplan]

[step:Verify that the pushforward is a Borel probability measure] Since $P: \mathbb{R}^{n+m} \to \mathbb{R}^n$ is continuous, $P^{-1}(A)$ is a Borel subset of $\mathbb{R}^{n+m}$ for every Borel set $A \subset \mathbb{R}^n$. Therefore the pushforward measure $P_\#\mu$ is well-defined on Borel subsets of $\mathbb{R}^n$ by \begin{align*} (P_\#\mu)(A) := \mu(P^{-1}(A)). \end{align*} Moreover, \begin{align*} (P_\#\mu)(\mathbb{R}^n) = \mu(P^{-1}(\mathbb{R}^n)) = \mu(\mathbb{R}^{n+m}) = 1, \end{align*} so $P_\#\mu$ is a Borel probability measure. [/step]

[step:Place compact approximations upstairs inside the pulled-back convex combination]Let $A,B \subset \mathbb{R}^n$ be Borel sets, let $\lambda \in [0,1]$, and assume that the Minkowski convex combination \begin{align*} (1-\lambda)A+\lambda B := \{(1-\lambda)a+\lambda b : a \in A,\ b \in B\} \end{align*} is Borel. Let $K,L \subset \mathbb{R}^{n+m}$ be compact sets such that $K \subset P^{-1}(A)$ and $L \subset P^{-1}(B)$. Define the continuous map \begin{align*} \Psi: K \times L &\to \mathbb{R}^{n+m} \\ (u,v) &\mapsto (1-\lambda)u+\lambda v. \end{align*} Since $K \times L$ is compact, the set $(1-\lambda)K+\lambda L=\Psi(K\times L)$ is compact and hence Borel in $\mathbb{R}^{n+m}$. We claim that \begin{align*} (1-\lambda)K+\lambda L \subset P^{-1}((1-\lambda)A+\lambda B). \end{align*} Indeed, let $z \in (1-\lambda)K+\lambda L$. Then there exist $u \in K$ and $v \in L$ such that \begin{align*} z = (1-\lambda)u+\lambda v. \end{align*} Since $K \subset P^{-1}(A)$ and $L \subset P^{-1}(B)$, we have $P(u) \in A$ and $P(v) \in B$. Since $P$ is linear, \begin{align*} P(z) = P((1-\lambda)u+\lambda v) = (1-\lambda)P(u)+\lambda P(v) \in (1-\lambda)A+\lambda B. \end{align*} Thus $z \in P^{-1}((1-\lambda)A+\lambda B)$, proving the inclusion.[/step]

[guided]The Borel sets $P^{-1}(A)$ and $P^{-1}(B)$ need not be compact, and the Minkowski sum of arbitrary Borel sets can create measurability issues. To avoid applying log-concavity to a non-Borel set, we approximate the inverse images from inside by compact sets and only take Minkowski combinations of those compact approximations. Let $A,B \subset \mathbb{R}^n$ be Borel sets, let $\lambda \in [0,1]$, and assume that \begin{align*} (1-\lambda)A+\lambda B := \{(1-\lambda)a+\lambda b : a \in A,\ b \in B\} \end{align*} is Borel. Choose compact sets $K,L \subset \mathbb{R}^{n+m}$ with $K \subset P^{-1}(A)$ and $L \subset P^{-1}(B)$. Define \begin{align*} \Psi: K \times L &\to \mathbb{R}^{n+m} \\ (u,v) &\mapsto (1-\lambda)u+\lambda v. \end{align*} The map $\Psi$ is continuous because addition and scalar multiplication are continuous in Euclidean space. Since $K \times L$ is compact, its image $(1-\lambda)K+\lambda L=\Psi(K\times L)$ is compact. Therefore $(1-\lambda)K+\lambda L$ is a Borel subset of $\mathbb{R}^{n+m}$, so it is a legitimate input for the measure $\mu$. We now compare this compact set upstairs with the desired convex combination downstairs. Take $z \in (1-\lambda)K+\lambda L$. Then for some $u \in K$ and $v \in L$, \begin{align*} z = (1-\lambda)u+\lambda v. \end{align*} Because $K \subset P^{-1}(A)$ and $L \subset P^{-1}(B)$, we know $P(u) \in A$ and $P(v) \in B$. The only structural property of $P$ used here is linearity: \begin{align*} P(z) = P((1-\lambda)u+\lambda v) = (1-\lambda)P(u)+\lambda P(v). \end{align*} The final expression belongs to $(1-\lambda)A+\lambda B$ by the definition of Minkowski convex combination. Hence $P(z) \in (1-\lambda)A+\lambda B$, which means \begin{align*} z \in P^{-1}((1-\lambda)A+\lambda B). \end{align*} This proves \begin{align*} (1-\lambda)K+\lambda L \subset P^{-1}((1-\lambda)A+\lambda B). \end{align*}[/guided]

[step:Pass from compact approximations to arbitrary Borel sets by inner regularity]Let $A,B \subset \mathbb{R}^n$ be Borel sets, let $\lambda \in [0,1]$, and assume that $(1-\lambda)A+\lambda B$ is Borel. With the convention $t^0=1$ for $t \in [0,1]$, the endpoint cases $\lambda=0$ and $\lambda=1$ reduce to identities, so assume $\lambda \in (0,1)$. The sets $P^{-1}(A)$ and $P^{-1}(B)$ are Borel because $P$ is continuous. Since $\mu$ is a Borel probability measure on the Euclidean space $\mathbb{R}^{n+m}$, it is inner regular on Borel sets. Thus, for every $\varepsilon > 0$, there exist compact sets $K,L \subset \mathbb{R}^{n+m}$ such that \begin{align*} K \subset P^{-1}(A), \qquad L \subset P^{-1}(B), \qquad \mu(K) \geq \mu(P^{-1}(A))-\varepsilon, \qquad \mu(L) \geq \mu(P^{-1}(B))-\varepsilon. \end{align*} By the preceding step, $(1-\lambda)K+\lambda L$ is compact and satisfies \begin{align*} (1-\lambda)K+\lambda L \subset P^{-1}((1-\lambda)A+\lambda B). \end{align*} Monotonicity of $\mu$ and log-concavity of $\mu$ applied to the compact Borel sets $K$ and $L$ give \begin{align*} \mu(P^{-1}((1-\lambda)A+\lambda B)) &\geq \mu((1-\lambda)K+\lambda L) \\ &\geq \mu(K)^{1-\lambda}\mu(L)^\lambda \\ &\geq (\mu(P^{-1}(A))-\varepsilon)^{1-\lambda}(\mu(P^{-1}(B))-\varepsilon)^\lambda, \end{align*} where the last inequality is interpreted with the nonnegative parts if either difference is negative. Letting $\varepsilon \downarrow 0$ and using continuity of $(s,t) \mapsto s^{1-\lambda}t^\lambda$ on $[0,1]^2$ yields \begin{align*} \mu(P^{-1}((1-\lambda)A+\lambda B)) \geq \mu(P^{-1}(A))^{1-\lambda}\mu(P^{-1}(B))^\lambda. \end{align*} Using the definition of pushforward measure on the three Borel sets involved, this is exactly \begin{align*} (P_\#\mu)((1-\lambda)A+\lambda B) \geq (P_\#\mu)(A)^{1-\lambda}(P_\#\mu)(B)^\lambda. \end{align*} Hence $P_\#\mu$ is log-concave.[/step]

[guided]The previous step gives an inclusion for compact subsets $K \subset P^{-1}(A)$ and $L \subset P^{-1}(B)$. We now explain why this is enough for arbitrary Borel sets $A$ and $B$. The point is that Borel probability measures on Euclidean spaces are inner regular: the measure of a Borel set can be recovered as the supremum of the measures of compact subsets contained in it. Let $A,B \subset \mathbb{R}^n$ be Borel sets, let $\lambda \in [0,1]$, and assume that $(1-\lambda)A+\lambda B$ is Borel. The convention in the statement is $t^0=1$ for every $t \in [0,1]$. If $\lambda=0$, the desired inequality reads $(P_\#\mu)(A) \geq (P_\#\mu)(A)$; if $\lambda=1$, it reads $(P_\#\mu)(B) \geq (P_\#\mu)(B)$. Therefore we may assume $\lambda \in (0,1)$. Because $P$ is continuous, $P^{-1}(A)$ and $P^{-1}(B)$ are Borel subsets of $\mathbb{R}^{n+m}$. Since $\mu$ is a Borel probability measure on the Euclidean space $\mathbb{R}^{n+m}$, regularity of finite Borel measures on Euclidean spaces gives inner regularity on Borel sets. Hence, for every $\varepsilon > 0$, there are compact sets $K,L \subset \mathbb{R}^{n+m}$ such that \begin{align*} K \subset P^{-1}(A), \qquad L \subset P^{-1}(B), \qquad \mu(K) \geq \mu(P^{-1}(A))-\varepsilon, \qquad \mu(L) \geq \mu(P^{-1}(B))-\varepsilon. \end{align*} These compact sets are the objects to which we apply log-concavity, so we never need to apply log-concavity to a possibly non-Borel Minkowski sum of arbitrary Borel sets. By the inclusion from the previous step, \begin{align*} (1-\lambda)K+\lambda L \subset P^{-1}((1-\lambda)A+\lambda B). \end{align*} The set $(1-\lambda)K+\lambda L$ is compact, hence Borel. Therefore monotonicity of the measure $\mu$ gives \begin{align*} \mu(P^{-1}((1-\lambda)A+\lambda B)) \geq \mu((1-\lambda)K+\lambda L). \end{align*} Since $K$ and $L$ are compact Borel sets and $\mu$ is log-concave, we may apply the log-concavity inequality for $\mu$ to obtain \begin{align*} \mu((1-\lambda)K+\lambda L) \geq \mu(K)^{1-\lambda}\mu(L)^\lambda. \end{align*} Combining the last two inequalities and using the compact approximations gives \begin{align*} \mu(P^{-1}((1-\lambda)A+\lambda B)) &\geq \mu((1-\lambda)K+\lambda L) \\ &\geq \mu(K)^{1-\lambda}\mu(L)^\lambda \\ &\geq (\mu(P^{-1}(A))-\varepsilon)^{1-\lambda}(\mu(P^{-1}(B))-\varepsilon)^\lambda, \end{align*} with the harmless convention that negative factors are replaced by their nonnegative parts for large $\varepsilon$; as $\varepsilon \downarrow 0$, this convention disappears. Letting $\varepsilon \downarrow 0$ and using continuity of $(s,t) \mapsto s^{1-\lambda}t^\lambda$ on $[0,1]^2$ gives \begin{align*} \mu(P^{-1}((1-\lambda)A+\lambda B)) \geq \mu(P^{-1}(A))^{1-\lambda}\mu(P^{-1}(B))^\lambda. \end{align*} Finally, by the definition of pushforward measure, \begin{align*} (P_\#\mu)((1-\lambda)A+\lambda B) &= \mu(P^{-1}((1-\lambda)A+\lambda B)), \\ (P_\#\mu)(A) &= \mu(P^{-1}(A)), \\ (P_\#\mu)(B) &= \mu(P^{-1}(B)). \end{align*} Substituting these identities yields \begin{align*} (P_\#\mu)((1-\lambda)A+\lambda B) \geq (P_\#\mu)(A)^{1-\lambda}(P_\#\mu)(B)^\lambda. \end{align*} This is the Borel-set log-concavity inequality for $P_\#\mu$.[/guided]