[proofplan] The BV isoperimetric inequality follows by combining the [BV Coarea Formula](/theorems/598) with the geometric [Isoperimetric Inequality for Sets of Finite Perimeter](/theorems/600) and a layer-cake representation of the $L^{n/(n-1)}$ norm. The strategy: write $|Du|(\mathbb{R}^n)$ as the integral of perimeters $P(\{u > t\})$ over $t$ via coarea, write $\|u\|_{L^{n/(n-1)}}$ in terms of the layer-cake distribution function $\mathcal{L}^n(\{|u| > t\})$, apply the geometric isoperimetric inequality to each level set to convert measures into perimeters, and finally apply Minkowski's integral inequality (or equivalently, the embedding from the layer-cake distribution into the $L^p$ norm) to recover the desired bound. The argument is performed first on $u \ge 0$ (handling positive and negative parts separately) and the result for general $u$ follows from $|u| = u^+ + u^-$. [/proofplan]

[step:Reduce to non-negative $u$ by splitting positive and negative parts]Write $u = u^+ - u^-$ where $u^+ = \max(u, 0)$ and $u^- = \max(-u, 0)$. Both $u^+, u^- \in BV(\mathbb{R}^n) \cap L^{n/(n-1)}(\mathbb{R}^n)$, with $|Du^+|(\mathbb{R}^n) + |Du^-|(\mathbb{R}^n) = |Du|(\mathbb{R}^n)$ (a standard property of $BV$ under composition with the chain rule for the Lipschitz functions $\max(\cdot, 0)$ and $\max(-\cdot, 0)$, which together with the identity decomposition gives no extra mass at $\{u = 0\}$). Since $|u| = u^+ + u^-$ is a sum of disjoint-support non-negative functions, $\|u\|_{L^{n/(n-1)}}^{n/(n-1)} = \|u^+\|_{L^{n/(n-1)}}^{n/(n-1)} + \|u^-\|_{L^{n/(n-1)}}^{n/(n-1)}$. If we prove the inequality for non-negative $u$, then \begin{align*} \|u\|_{L^{n/(n-1)}} &\le \|u^+\|_{L^{n/(n-1)}} + \|u^-\|_{L^{n/(n-1)}} \\ &\le C_n |Du^+|(\mathbb{R}^n) + C_n|Du^-|(\mathbb{R}^n) = C_n |Du|(\mathbb{R}^n). \end{align*} Hence we may assume $u \ge 0$.[/step]

[guided]We reduce to the case $u \ge 0$ by analysing positive and negative parts. Define \begin{align*} u^+: \mathbb{R}^n \to [0, \infty), &\quad x \mapsto \max(u(x), 0), \\ u^-: \mathbb{R}^n \to [0, \infty), &\quad x \mapsto \max(-u(x), 0). \end{align*} Then $u = u^+ - u^-$ and $|u| = u^+ + u^-$, with $u^+ u^- = 0$ pointwise. *BV regularity.* The truncation map $t \mapsto \max(t, 0)$ is Lipschitz with Lipschitz constant $1$. By the BV chain rule for Lipschitz post-composition (a standard result in BV theory, often called the BV chain rule for monotone Lipschitz functions), $u^+ \in BV(\mathbb{R}^n)$ with $|Du^+|(\mathbb{R}^n) \le |Du|(\mathbb{R}^n)$. Similarly $u^- \in BV(\mathbb{R}^n)$. The decomposition of total variations: since $\{u > 0\}$ and $\{u < 0\}$ are disjoint Borel sets whose union supports $Du$ (the level set $\{u = 0\}$ may contribute mass to $Du$, but a careful chain-rule analysis shows it does not contribute to $Du^+$ or $Du^-$ — this is the content of the chain rule for $\max(\cdot, 0)$ at $t = 0$). The result is \begin{align*} |Du^+|(\mathbb{R}^n) + |Du^-|(\mathbb{R}^n) = |Du|(\mathbb{R}^n). \end{align*} *$L^{n/(n-1)}$ regularity.* Both $u^+, u^- \le |u| \in L^{n/(n-1)}(\mathbb{R}^n)$, hence $u^\pm \in L^{n/(n-1)}(\mathbb{R}^n)$. The disjoint-support identity: \begin{align*} \|u\|_{L^{n/(n-1)}}^{n/(n-1)} = \int_{\mathbb{R}^n} |u|^{n/(n-1)} \, d\mathcal{L}^n = \int_{\mathbb{R}^n} (u^+)^{n/(n-1)} \, d\mathcal{L}^n + \int_{\mathbb{R}^n} (u^-)^{n/(n-1)} \, d\mathcal{L}^n, \end{align*} since for $a, b \ge 0$ with $ab = 0$, $(a + b)^{n/(n-1)} = a^{n/(n-1)} + b^{n/(n-1)}$. Hence \begin{align*} \|u\|_{L^{n/(n-1)}} = \big(\|u^+\|_{L^{n/(n-1)}}^{n/(n-1)} + \|u^-\|_{L^{n/(n-1)}}^{n/(n-1)}\big)^{(n-1)/n} \le \|u^+\|_{L^{n/(n-1)}} + \|u^-\|_{L^{n/(n-1)}}, \end{align*} where the last inequality is the standard $\ell^{p}$-to-$\ell^1$ comparison for two terms, $a^{1/q} + b^{1/q} \ge (a + b)^{1/q}$ when $q \ge 1$, applied with $q = n/(n-1) \ge 1$. If we prove the bound $\|v\|_{L^{n/(n-1)}} \le C_n |Dv|(\mathbb{R}^n)$ for $v \ge 0$ (with $v \in BV \cap L^{n/(n-1)}$), then applying it to $v = u^+$ and $v = u^-$ separately and summing gives \begin{align*} \|u\|_{L^{n/(n-1)}} \le C_n |Du^+|(\mathbb{R}^n) + C_n |Du^-|(\mathbb{R}^n) = C_n |Du|(\mathbb{R}^n). \end{align*} This reduces the problem to the case $u \ge 0$.[/guided]

[step:Apply the BV coarea formula and the geometric isoperimetric inequality at each level]Assume $u \ge 0$. By the [BV Coarea Formula](/theorems/598), \begin{align*} |Du|(\mathbb{R}^n) = \int_0^\infty P(\{u > t\}, \mathbb{R}^n) \, d\mathcal{L}^1(t). \end{align*} For each $t \ge 0$, the superlevel set $E_t := \{u > t\}$ has finite perimeter (by the coarea formula for $\mathcal{L}^1$-a.e. $t$) and finite Lebesgue measure (since $\mathcal{L}^n(E_t) \le t^{-n/(n-1)} \int (u)^{n/(n-1)} \, d\mathcal{L}^n < \infty$ by Markov's inequality). Hence the [Isoperimetric Inequality for Sets of Finite Perimeter](/theorems/600) applies: \begin{align*} \mathcal{L}^n(E_t)^{(n-1)/n} \le c_n P(E_t, \mathbb{R}^n) \quad \text{for } \mathcal{L}^1\text{-a.e. } t > 0, \end{align*} where $c_n = (n\omega_n^{1/n})^{-1}$ is the sharp dimensional constant from the geometric isoperimetric inequality.[/step]

[guided]We now relate $|Du|(\mathbb{R}^n)$ to integrals of $\mathcal{L}^n(E_t)^{(n-1)/n}$ over the level parameter $t$. The two analytical inputs are the coarea formula and the geometric isoperimetric inequality. *Coarea.* The [BV Coarea Formula](/theorems/598) states that for $u \in BV(\mathbb{R}^n)$ with $u \ge 0$, \begin{align*} |Du|(\mathbb{R}^n) = \int_{-\infty}^\infty P(\{u > t\}, \mathbb{R}^n) \, d\mathcal{L}^1(t) = \int_0^\infty P(\{u > t\}, \mathbb{R}^n) \, d\mathcal{L}^1(t), \end{align*} where the second equality uses $u \ge 0$, so $\{u > t\} = \mathbb{R}^n$ for $t < 0$ and the perimeter of $\mathbb{R}^n$ is zero. The hypothesis for coarea is met because $u \in BV(\mathbb{R}^n)$. *Finite measure of level sets.* For each $t > 0$, by Markov's inequality, \begin{align*} \mathcal{L}^n(\{u > t\}) = \mathcal{L}^n(\{u^{n/(n-1)} > t^{n/(n-1)}\}) \le t^{-n/(n-1)} \int_{\mathbb{R}^n} u^{n/(n-1)} \, d\mathcal{L}^n = t^{-n/(n-1)} \|u\|_{L^{n/(n-1)}}^{n/(n-1)}, \end{align*} which is finite by the hypothesis $u \in L^{n/(n-1)}(\mathbb{R}^n)$. *Finite perimeter of level sets.* For $\mathcal{L}^1$-a.e. $t \in (0, \infty)$, the level set $E_t = \{u > t\}$ has finite perimeter by the coarea formula (which states that $P(E_t, \mathbb{R}^n) < \infty$ for $\mathcal{L}^1$-a.e. $t$, since $\int_0^\infty P(E_t, \mathbb{R}^n)\, dt = |Du|(\mathbb{R}^n) < \infty$). *Isoperimetric inequality.* The [Isoperimetric Inequality for Sets of Finite Perimeter](/theorems/600) states that for any Borel set $E \subseteq \mathbb{R}^n$ with finite perimeter and finite Lebesgue measure, \begin{align*} \mathcal{L}^n(E)^{(n-1)/n} \le c_n P(E, \mathbb{R}^n), \end{align*} where $c_n = (n\omega_n^{1/n})^{-1}$. The hypotheses are met for $E = E_t$ at $\mathcal{L}^1$-a.e. $t$ by the previous two paragraphs. Hence \begin{align*} \mathcal{L}^n(E_t)^{(n-1)/n} \le c_n P(E_t, \mathbb{R}^n) \quad \text{for $\mathcal{L}^1$-a.e. } t > 0. \end{align*} Integrating in $t$ and using coarea on the right, \begin{align*} \int_0^\infty \mathcal{L}^n(E_t)^{(n-1)/n} \, d\mathcal{L}^1(t) \le c_n \int_0^\infty P(E_t, \mathbb{R}^n) \, d\mathcal{L}^1(t) = c_n |Du|(\mathbb{R}^n). \end{align*} This is the key estimate. The next step uses a layer-cake-type representation of $\|u\|_{L^{n/(n-1)}}$ to relate the left-hand integral to the $L^{n/(n-1)}$ norm.[/guided]

[step:Bound $\|u\|_{L^{n/(n-1)}}$ by the layer-cake integral via the embedding inequality]The layer-cake representation for $u \ge 0$: \begin{align*} u(x) = \int_0^\infty \mathbb{1}_{\{u > t\}}(x) \, d\mathcal{L}^1(t). \end{align*} Taking the $L^{n/(n-1)}$ norm and applying Minkowski's integral inequality (which says $\|\int f_t \, dt\|_{L^q} \le \int \|f_t\|_{L^q} \, dt$ for $q \ge 1$): \begin{align*} \|u\|_{L^{n/(n-1)}(\mathbb{R}^n)} &= \left\|\int_0^\infty \mathbb{1}_{\{u > t\}} \, d\mathcal{L}^1(t)\right\|_{L^{n/(n-1)}(\mathbb{R}^n)} \\ &\le \int_0^\infty \|\mathbb{1}_{\{u > t\}}\|_{L^{n/(n-1)}(\mathbb{R}^n)} \, d\mathcal{L}^1(t) \\ &= \int_0^\infty \mathcal{L}^n(\{u > t\})^{(n-1)/n} \, d\mathcal{L}^1(t). \end{align*} Combining with the bound from Step 2, \begin{align*} \|u\|_{L^{n/(n-1)}(\mathbb{R}^n)} \le \int_0^\infty \mathcal{L}^n(E_t)^{(n-1)/n} \, d\mathcal{L}^1(t) \le c_n |Du|(\mathbb{R}^n), \end{align*} which is the desired inequality with constant $C_n = c_n = (n\omega_n^{1/n})^{-1}$.[/step]

[guided]We connect $\|u\|_{L^{n/(n-1)}}$ to the integral $\int \mathcal{L}^n(E_t)^{(n-1)/n} \, dt$ via a layer-cake / Minkowski's inequality argument. *Layer-cake representation.* For $u \ge 0$ and any $x \in \mathbb{R}^n$, \begin{align*} u(x) = \int_0^{u(x)} 1 \, d\mathcal{L}^1(t) = \int_0^\infty \mathbb{1}_{\{u(x) > t\}} \, d\mathcal{L}^1(t) = \int_0^\infty \mathbb{1}_{\{u > t\}}(x) \, d\mathcal{L}^1(t). \end{align*} This is the standard layer-cake formula for non-negative functions, expressing $u$ as a vertical integral of indicator functions of its superlevel sets. *Minkowski's integral inequality.* For a measurable family of functions $f_t: \mathbb{R}^n \to [0, \infty)$ indexed by $t \in (0, \infty)$, and for $q \ge 1$, \begin{align*} \left\|\int_0^\infty f_t \, d\mathcal{L}^1(t)\right\|_{L^q(\mathbb{R}^n)} \le \int_0^\infty \|f_t\|_{L^q(\mathbb{R}^n)} \, d\mathcal{L}^1(t). \end{align*} This is the standard Minkowski integral inequality. The hypothesis $q \ge 1$ is met by $q = n/(n-1)$ since $n \ge 2$ (the case $n = 1$ requires separate but easier treatment). The functions $f_t = \mathbb{1}_{\{u > t\}}$ are measurable in $(t, x)$ jointly (the set $\{(t, x) : u(x) > t\}$ is the hypograph of $u$, which is Borel by measurability of $u$). Applying Minkowski with $q = n/(n-1)$ and $f_t = \mathbb{1}_{\{u > t\}}$: \begin{align*} \|u\|_{L^{n/(n-1)}(\mathbb{R}^n)} = \left\|\int_0^\infty \mathbb{1}_{\{u > t\}} \, d\mathcal{L}^1(t)\right\|_{L^{n/(n-1)}(\mathbb{R}^n)} \le \int_0^\infty \|\mathbb{1}_{\{u > t\}}\|_{L^{n/(n-1)}(\mathbb{R}^n)} \, d\mathcal{L}^1(t). \end{align*} For an indicator function, $\|\mathbb{1}_E\|_{L^p}^p = \mathcal{L}^n(E)$, so $\|\mathbb{1}_{\{u > t\}}\|_{L^{n/(n-1)}} = \mathcal{L}^n(\{u > t\})^{(n-1)/n}$. Substituting: \begin{align*} \|u\|_{L^{n/(n-1)}(\mathbb{R}^n)} \le \int_0^\infty \mathcal{L}^n(\{u > t\})^{(n-1)/n} \, d\mathcal{L}^1(t). \end{align*} *Combination.* From Step 2, $\int_0^\infty \mathcal{L}^n(E_t)^{(n-1)/n} \, dt \le c_n |Du|(\mathbb{R}^n)$. Combining, \begin{align*} \|u\|_{L^{n/(n-1)}(\mathbb{R}^n)} \le c_n |Du|(\mathbb{R}^n). \end{align*} This is the BV isoperimetric inequality for non-negative $u$ with constant $C_n = c_n = (n\omega_n^{1/n})^{-1}$. By Step 1, the same bound (possibly with constant doubled, but still depending only on $n$) holds for general $u \in BV(\mathbb{R}^n) \cap L^{n/(n-1)}(\mathbb{R}^n)$. The dependence of $C_n$ on $n$ comes through the geometric isoperimetric constant $c_n$ — explicitly $C_n = (n\omega_n^{1/n})^{-1}$, where $\omega_n$ is the volume of the unit ball in $\mathbb{R}^n$.[/guided]

[step:Conclude the proof for general $u$]Combining Steps 1 and 3: for general $u \in BV(\mathbb{R}^n) \cap L^{n/(n-1)}(\mathbb{R}^n)$, \begin{align*} \|u\|_{L^{n/(n-1)}(\mathbb{R}^n)} \le \|u^+\|_{L^{n/(n-1)}(\mathbb{R}^n)} + \|u^-\|_{L^{n/(n-1)}(\mathbb{R}^n)} \le C_n |Du^+|(\mathbb{R}^n) + C_n |Du^-|(\mathbb{R}^n) = C_n |Du|(\mathbb{R}^n), \end{align*} with $C_n = (n\omega_n^{1/n})^{-1}$. The proof is complete.[/step]

[guided]We assemble the full result. The non-negative case (Step 3) gives \begin{align*} \|v\|_{L^{n/(n-1)}(\mathbb{R}^n)} \le C_n |Dv|(\mathbb{R}^n) \quad \text{for } v \ge 0 \text{ in } BV \cap L^{n/(n-1)}. \end{align*} Apply this to $v = u^+$ and $v = u^-$, both of which are non-negative and in $BV(\mathbb{R}^n) \cap L^{n/(n-1)}(\mathbb{R}^n)$ by Step 1: \begin{align*} \|u^+\|_{L^{n/(n-1)}} &\le C_n |Du^+|(\mathbb{R}^n), \\ \|u^-\|_{L^{n/(n-1)}} &\le C_n |Du^-|(\mathbb{R}^n). \end{align*} By the splitting in Step 1, \begin{align*} \|u\|_{L^{n/(n-1)}} \le \|u^+\|_{L^{n/(n-1)}} + \|u^-\|_{L^{n/(n-1)}} \le C_n(|Du^+| + |Du^-|)(\mathbb{R}^n) = C_n |Du|(\mathbb{R}^n), \end{align*} where the final equality is the additive decomposition of the total variation across the disjoint supports of $Du^+$ and $Du^-$ (as established in Step 1). The constant $C_n = (n\omega_n^{1/n})^{-1}$ depends only on the dimension $n$. The proof is complete. This proves the BV Sobolev embedding $BV(\mathbb{R}^n) \cap L^{n/(n-1)}(\mathbb{R}^n) \hookrightarrow L^{n/(n-1)}(\mathbb{R}^n)$ at the critical exponent $n/(n-1)$, with norm controlled by the total variation. The extra hypothesis $u \in L^{n/(n-1)}(\mathbb{R}^n)$ ensures that the right-hand side of the layer-cake integration converges; in fact, the inequality combined with the coarea formula shows that this hypothesis is automatic for $u \in BV(\mathbb{R}^n)$ in suitable settings (e.g., for compactly supported $u$), but stating it as a hypothesis avoids a separate finiteness argument.[/guided]