[proofplan] We first fix the notation for Gaussian measure, Gaussian boundary content, and the Gaussian isoperimetric profile, including the endpoint convention. The analytic input is the Bakry-Ledoux functional Gaussian isoperimetric inequality assumed in the statement. Applying that functional inequality to Lipschitz cutoffs of a smooth set gives the sharp Gaussian surface-area bound. [/proofplan]

[step:Define the Gaussian profile and record the functional inequality] Let $\mathcal L^n$ denote $n$-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $\mathbb R^n$. Let $\gamma_n$ denote the standard Gaussian probability measure on $\mathbb R^n$, defined on Borel sets $E \subseteq \mathbb R^n$ by \begin{align*} \gamma_n(E) = (2\pi)^{-n/2} \int_E e^{-|x|^2/2}\,d\mathcal L^n(x). \end{align*} For a $C^1$ hypersurface $\Sigma \subseteq \mathbb R^n$, its Gaussian surface measure is \begin{align*} \int_\Sigma (2\pi)^{-n/2} e^{-|x|^2/2}\,d\mathcal H^{n-1}(x), \end{align*} where $\mathcal H^{n-1}$ denotes $(n-1)$-dimensional [Hausdorff measure](/page/Hausdorff%20Measure). For a measurable set $E \subseteq \mathbb R^n$, define its lower Gaussian Minkowski content by \begin{align*} \gamma_n^+(E) = \liminf_{r \downarrow 0} \frac{\gamma_n(E_r)-\gamma_n(E)}{r}, \end{align*} where $E_r = \{x \in \mathbb R^n : \operatorname{dist}(x,E) < r\}$. Let $\Phi: \mathbb R \to (0,1)$ be the one-dimensional Gaussian distribution function \begin{align*} \Phi(t) = (2\pi)^{-1/2}\int_{-\infty}^t e^{-s^2/2}\,d\mathcal L^1(s), \end{align*} and define the Gaussian isoperimetric profile $I: [0,1] \to [0,\infty)$ by $I(0)=I(1)=0$ and \begin{align*} I(a) = \Phi'(\Phi^{-1}(a)) \end{align*} for $0<a<1$. The function $I$ is continuous on $[0,1]$. We will use the Bakry-Ledoux functional Gaussian isoperimetric inequality assumed in the statement: if $f: \mathbb R^n \to [0,1]$ is locally Lipschitz and $|\nabla f| \in L^1(\mathbb R^n,\gamma_n)$, then \begin{align*} I\left(\int_{\mathbb R^n} f\,d\gamma_n\right) \le \int_{\mathbb R^n} \sqrt{I(f)^2+|\nabla f|^2}\,d\gamma_n. \end{align*} The hypotheses above are satisfied for the Lipschitz cutoff functions used below: they take values in $[0,1]$, are locally Lipschitz, and have bounded gradient with support contained in a finite-width tubular neighbourhood whenever the original set is smooth with Gaussian-finite boundary measure. [/step]

[step:Derive the boundary estimate for smooth sets from Lipschitz cutoffs]Assume first that $A \subseteq \mathbb R^n$ has $C^1$ boundary and finite Gaussian surface measure. Let $d_A: \mathbb R^n \to [0,\infty)$ be the distance function $d_A(x)=\operatorname{dist}(x,A)$. For $\varepsilon>0$, define the Lipschitz cutoff $f_\varepsilon: \mathbb R^n \to [0,1]$ by \begin{align*} f_\varepsilon(x) = \max\left\{1-\frac{d_A(x)}{\varepsilon},0\right\}. \end{align*} Let $\mathbb 1_B$ denote the indicator function of a set $B$, and let $\overline A$ denote the topological closure of $A$. Then $f_\varepsilon=1$ on $A$, $f_\varepsilon=0$ outside $A_\varepsilon$, and $|\nabla f_\varepsilon| \le \varepsilon^{-1}\mathbb 1_{A_\varepsilon\setminus A}$ for $\mathcal L^n$-almost every $x$. Applying the functional inequality to $f_\varepsilon$ gives \begin{align*} I\left(\int_{\mathbb R^n} f_\varepsilon\,d\gamma_n\right) \le \int_{\mathbb R^n} \sqrt{I(f_\varepsilon)^2+|\nabla f_\varepsilon|^2}\,d\gamma_n. \end{align*} Because $0 \le f_\varepsilon \le 1$, $f_\varepsilon \to \mathbb 1_A$ pointwise $\gamma_n$-almost everywhere as $\varepsilon \downarrow 0$, and $\gamma_n$ is a probability measure, the [dominated convergence theorem](/theorems/4) gives \begin{align*} \int_{\mathbb R^n} f_\varepsilon\,d\gamma_n \to \gamma_n(A). \end{align*} The continuity of $I$ on $[0,1]$ therefore gives convergence of the left-hand side to $I(\gamma_n(A))$. On the right-hand side, use the pointwise inequality $\sqrt{a^2+b^2}\le a+b$ for $a,b\ge0$ with $a=I(f_\varepsilon)$ and $b=|\nabla f_\varepsilon|$. Since $I(0)=I(1)=0$, the function $I(f_\varepsilon)$ vanishes outside the tubular layer $A_\varepsilon\setminus A$. Hence \begin{align*} \int_{\mathbb R^n} I(f_\varepsilon)\,d\gamma_n \le \|I\|_{L^\infty([0,1])}\gamma_n(A_\varepsilon\setminus A). \end{align*} Because $A_\varepsilon \downarrow \overline A$ as $\varepsilon\downarrow0$ and $A$ has $C^1$ boundary, $\gamma_n(A_\varepsilon\setminus A)\to0$. The gradient bound gives \begin{align*} \int_{\mathbb R^n}|\nabla f_\varepsilon|\,d\gamma_n \le \frac{\gamma_n(A_\varepsilon)-\gamma_n(A)}{\varepsilon}. \end{align*} Combining these estimates and taking the lower limit gives \begin{align*} \liminf_{\varepsilon \downarrow 0} \int_{\mathbb R^n} \sqrt{I(f_\varepsilon)^2+|\nabla f_\varepsilon|^2}\,d\gamma_n \le \gamma_n^+(A). \end{align*} The [Coarea Formula (Classical)](/theorems/24), applied to the Lipschitz distance map $d_A: \mathbb R^n\to[0,\infty)$, identifies this lower Minkowski content with the corresponding Gaussian surface-area limit for $C^1$ boundary. Hence \begin{align*} I(\gamma_n(A)) \le \gamma_n^+(A). \end{align*}[/step]

[guided]We now prove the smooth case rather than merely citing it. The goal is to turn the set $A$ into functions to which the Bakry-Ledoux functional inequality applies. Let $d_A: \mathbb R^n \to [0,\infty)$ be the distance function $d_A(x)=\operatorname{dist}(x,A)$. For each $\varepsilon>0$, define \begin{align*} f_\varepsilon(x) = \max\left\{1-\frac{d_A(x)}{\varepsilon},0\right\}. \end{align*} This map satisfies $f_\varepsilon: \mathbb R^n \to [0,1]$, equals $1$ on $A$, equals $0$ outside the open parallel set $A_\varepsilon$, and is locally Lipschitz. Since the distance function is $1$-Lipschitz, [Rademacher's theorem](/theorems/3069) gives $|\nabla d_A|\le 1$ almost everywhere, and therefore \begin{align*} |\nabla f_\varepsilon| \le \varepsilon^{-1}\mathbb 1_{A_\varepsilon\setminus A} \end{align*} for $\mathcal L^n$-almost every $x$. These are exactly the hypotheses needed for the Bakry-Ledoux functional Gaussian isoperimetric inequality: $f_\varepsilon$ takes values in $[0,1]$, is locally Lipschitz, and has integrable gradient with respect to $\gamma_n$. Thus \begin{align*} I\left(\int_{\mathbb R^n} f_\varepsilon\,d\gamma_n\right) \le \int_{\mathbb R^n} \sqrt{I(f_\varepsilon)^2+|\nabla f_\varepsilon|^2}\,d\gamma_n. \end{align*} Why choose this particular cutoff? It approximates the indicator $\mathbb 1_A$ while concentrating its gradient in the thin boundary layer $A_\varepsilon\setminus A$, so the right-hand side measures boundary size. Since $0\le f_\varepsilon\le 1$ and $f_\varepsilon(x)\to \mathbb 1_A(x)$ for $\gamma_n$-almost every $x$ as $\varepsilon\downarrow 0$, the dominated convergence theorem gives \begin{align*} \int_{\mathbb R^n} f_\varepsilon\,d\gamma_n \to \int_{\mathbb R^n}\mathbb 1_A\,d\gamma_n = \gamma_n(A). \end{align*} The endpoint convention $I(0)=I(1)=0$ and continuity of $I$ on $[0,1]$ ensure that this remains valid even when $\gamma_n(A)=0$ or $\gamma_n(A)=1$. Hence the left-hand side converges to $I(\gamma_n(A))$. For the boundary term, first separate the two pieces of the square root. The elementary inequality $\sqrt{a^2+b^2}\le a+b$ for $a,b\ge0$ gives \begin{align*} \int_{\mathbb R^n} \sqrt{I(f_\varepsilon)^2+|\nabla f_\varepsilon|^2}\,d\gamma_n \le \int_{\mathbb R^n} I(f_\varepsilon)\,d\gamma_n + \int_{\mathbb R^n}|\nabla f_\varepsilon|\,d\gamma_n. \end{align*} The first term is negligible. Indeed, $I(0)=I(1)=0$, so $I(f_\varepsilon)$ can be nonzero only on $A_\varepsilon\setminus A$. Since $I$ is continuous on the compact interval $[0,1]$, the number $M_I:=\|I\|_{L^\infty([0,1])}$ is finite, and therefore \begin{align*} \int_{\mathbb R^n} I(f_\varepsilon)\,d\gamma_n \le M_I\gamma_n(A_\varepsilon\setminus A). \end{align*} Here $\partial A$ denotes the topological boundary of $A$. Because $A$ has $C^1$ boundary, $\gamma_n(\partial A)=0$, and the sets $A_\varepsilon\setminus A$ decrease to a null boundary layer as $\varepsilon\downarrow0$. Hence this term tends to $0$. The second term is controlled by the Lipschitz slope of the cutoff. From $|\nabla f_\varepsilon|\le \varepsilon^{-1}\mathbb 1_{A_\varepsilon\setminus A}$ almost everywhere, we get \begin{align*} \int_{\mathbb R^n}|\nabla f_\varepsilon|\,d\gamma_n \le \frac{\gamma_n(A_\varepsilon)-\gamma_n(A)}{\varepsilon}. \end{align*} Taking lower limits gives \begin{align*} \liminf_{\varepsilon\downarrow0}\int_{\mathbb R^n} \sqrt{I(f_\varepsilon)^2+|\nabla f_\varepsilon|^2}\,d\gamma_n \le \gamma_n^+(A). \end{align*} The [Coarea Formula (Classical)](/theorems/24), applied to the Lipschitz distance map $d_A: \mathbb R^n\to[0,\infty)$, is the geometric theorem behind this tube estimate: it rewrites integration over $A_\varepsilon\setminus A$ as integration over the level surfaces $\{d_A=t\}$, and the $C^1$ boundary hypothesis ensures that these level surfaces have the expected limiting Gaussian surface measure as $t\downarrow0$. Combining the functional inequality with these two limits yields \begin{align*} I(\gamma_n(A)) \le \gamma_n^+(A). \end{align*} This is the desired Gaussian isoperimetric inequality for smooth sets.[/guided]