[proofplan] We regard $\Sigma$ as an integral varifold in the punctured ball and prove that its first variation has no defect at the isolated point. The finite area and density-one hypotheses allow the logarithmic cutoff argument to pass stationarity across $0$. Once the closed varifold is stationary and has density $1$ at $0$, the stationary-varifold monotonicity formula gives a small ball whose normalized mass ratio is close enough to one for Allard regularity. The resulting graph is weakly minimal across $0$, so interior regularity for the minimal graph equation upgrades it to a smooth embedded minimal hypersurface near $0$. [/proofplan]

[step:Extend the punctured hypersurface to a stationary integral varifold across $0$]Let $G_m(B(0,R))$ denote the Grassmann bundle of unoriented $m$-planes over $B(0,R)$. Define the [integral varifold](/page/Integral%20Varifold) $V$ in $B(0,R)$ by \begin{align*} V(\varphi) = \int_{\Sigma} \varphi(x,T_x\Sigma)\, d\mathcal H^m(x) \end{align*} for every $\varphi \in C_c(G_m(B(0,R)))$. The point $0$ has zero $\mathcal H^m$ measure, so adding $0$ to the support does not change the weight measure. The hypothesis \begin{align*} \mathcal H^m(\Sigma\cap B(0,R))<\infty \end{align*} therefore gives $\|V\|(B(0,R))<\infty$. For a compactly supported $C^1$ vector field $Y:B(0,R)\to\mathbb R^{m+1}$, define the [first variation](/page/First%20Variation) functional $\delta V$ by \begin{align*} \delta V(Y)=\int_\Sigma \operatorname{div}_\Sigma Y\,d\mathcal H^m(x), \end{align*} where $\operatorname{div}_\Sigma Y(x)$ denotes the trace of the derivative of $Y$ restricted to the tangent plane $T_x\Sigma$. We prove that $V$ is [stationary](/page/Stationary%20Varifold) in $B(0,R)$, meaning $\delta V(Y)=0$ for every compactly supported $C^1$ vector field $Y:B(0,R)\to\mathbb R^{m+1}$. Let $X:B(0,R)\to\mathbb R^{m+1}$ be a compactly supported $C^1$ vector field. For $0<\varepsilon<1$, choose a smooth cutoff function \begin{align*} \zeta_\varepsilon:(0,\infty)\to[0,1] \end{align*} with $\zeta_\varepsilon(t)=0$ for $0<t\leq \varepsilon$, $\zeta_\varepsilon(t)=1$ for $t\geq \sqrt\varepsilon$, and \begin{align*} |\zeta_\varepsilon'(t)|\leq \frac{2}{t|\log\varepsilon|}. \end{align*} Define $X_\varepsilon:B(0,R)\setminus\{0\}\to\mathbb R^{m+1}$ by \begin{align*} X_\varepsilon(x)=\zeta_\varepsilon(|x|)X(x). \end{align*} For each $x\in\Sigma$, let $\operatorname{proj}_{T_x\Sigma}:\mathbb R^{m+1}\to T_x\Sigma$ denote the Euclidean [orthogonal projection](/theorems/437) onto the tangent plane $T_x\Sigma$. Since $X_\varepsilon$ is compactly supported in $B(0,R)\setminus\{0\}$ and $\Sigma$ is minimal there, the [first variation formula](/theorems/2728) gives \begin{align*} 0=\delta V(X_\varepsilon)=\int_\Sigma \operatorname{div}_\Sigma X_\varepsilon\,d\mathcal H^m(x). \end{align*} For $x\in\Sigma$, expanding the tangential divergence gives \begin{align*} \operatorname{div}_\Sigma X_\varepsilon(x) =\zeta_\varepsilon(|x|)\operatorname{div}_\Sigma X(x) +\zeta_\varepsilon'(|x|)\left\langle \operatorname{proj}_{T_x\Sigma}\frac{x}{|x|},X(x)\right\rangle. \end{align*} The first term converges in $L^1(\Sigma,\mathcal H^m)$ to $\operatorname{div}_\Sigma X$ by dominated convergence, because $|\operatorname{div}_\Sigma X|\leq m\|DX\|_\infty$ and $\mathcal H^m(\Sigma\cap\operatorname{supp}X)<\infty$. For the cutoff term, the density assumption implies that there are $r_0>0$ and $C_0>0$ such that \begin{align*} \mathcal H^m(\Sigma\cap B(0,r))\leq C_0 r^m \end{align*} for $0<r<r_0$. Let $N_\varepsilon$ be the largest integer with $2^{N_\varepsilon}\varepsilon\leq \sqrt\varepsilon$, and define annuli \begin{align*} A_k=\Sigma\cap\bigl(B(0,2^{k+1}\varepsilon)\setminus B(0,2^k\varepsilon)\bigr) \end{align*} for $0\leq k\leq N_\varepsilon$. Since $|x|\geq 2^k\varepsilon$ on $A_k$ and $2^{k+1}\varepsilon<r_0$ for sufficiently small $\varepsilon$, the area-growth estimate gives Using the lower bound $|x|\geq 2^k\varepsilon$ on $A_k$, we first obtain \begin{align*} \int_{A_k}\frac{1}{|x|}\,d\mathcal H^m(x)\leq \frac{1}{2^k\varepsilon}\mathcal H^m(\Sigma\cap B(0,2^{k+1}\varepsilon)). \end{align*} Using the area-growth estimate on the ball $B(0,2^{k+1}\varepsilon)$ gives \begin{align*} \int_{A_k}\frac{1}{|x|}\,d\mathcal H^m(x)\leq C_0 2^m (2^k\varepsilon)^{m-1}. \end{align*} Summing the annular estimates gives \begin{align*} \int_{\Sigma\cap (B(0,\sqrt\varepsilon)\setminus B(0,\varepsilon))}\frac{1}{|x|}\,d\mathcal H^m(x) \leq C_0 2^m\sum_{k=0}^{N_\varepsilon}(2^k\varepsilon)^{m-1}. \end{align*} Since $m\geq2$ and $2^{N_\varepsilon}\varepsilon\leq\sqrt\varepsilon$, the geometric-series estimate gives \begin{align*} C_0 2^m\sum_{k=0}^{N_\varepsilon}(2^k\varepsilon)^{m-1} \leq \frac{C_0 2^m}{1-2^{1-m}}\varepsilon^{(m-1)/2}. \end{align*} Set $C_1= C_0 2^m/(1-2^{1-m})$. Therefore \begin{align*} \left|\int_\Sigma \zeta_\varepsilon'(|x|)\left\langle \operatorname{proj}_{T_x\Sigma}\frac{x}{|x|},X(x)\right\rangle d\mathcal H^m(x)\right| \leq \frac{2\|X\|_\infty C_1}{|\log\varepsilon|}\varepsilon^{(m-1)/2}\to0. \end{align*} Letting $\varepsilon\downarrow0$ gives \begin{align*} \delta V(X)=\int_\Sigma \operatorname{div}_\Sigma X\,d\mathcal H^m(x)=0. \end{align*} Thus the varifold obtained by adding the isolated point is stationary in $B(0,R)$.[/step]

[guided]Let $G_m(B(0,R))$ denote the Grassmann bundle of unoriented $m$-planes over $B(0,R)$. We associate to $\Sigma$ the [integral varifold](/page/Integral%20Varifold) $V$ in $B(0,R)$ defined by \begin{align*} V(\varphi)=\int_\Sigma \varphi(x,T_x\Sigma)\,d\mathcal H^m(x) \end{align*} for every $\varphi\in C_c(G_m(B(0,R)))$, and we write $\|V\|$ for its weight measure. Since a single point has zero $\mathcal H^m$ measure and $\mathcal H^m(\Sigma\cap B(0,R))<\infty$, the varifold has finite mass on $B(0,R)$ even after $0$ is added to the support. The only possible obstruction to stationarity is a hidden boundary term at the deleted point. For a compactly supported $C^1$ vector field $Y:B(0,R)\to\mathbb R^{m+1}$, the [first variation](/page/First%20Variation) is the functional \begin{align*} \delta V(Y)=\int_\Sigma \operatorname{div}_\Sigma Y\,d\mathcal H^m(x), \end{align*} where $\operatorname{div}_\Sigma Y(x)$ is the trace of the derivative of $Y$ along the tangent plane $T_x\Sigma$. Stationarity means that this quantity is zero for every such $Y$. To rule out a defect at $0$, we test stationarity with vector fields that vanish near $0$ and then let the vanishing region shrink. Let $X:B(0,R)\to\mathbb R^{m+1}$ be a compactly supported $C^1$ vector field. For $0<\varepsilon<1$, define a logarithmic radial cutoff \begin{align*} \zeta_\varepsilon:(0,\infty)\to[0,1] \end{align*} so that $\zeta_\varepsilon(t)=0$ for $0<t\leq\varepsilon$, $\zeta_\varepsilon(t)=1$ for $t\geq\sqrt\varepsilon$, and \begin{align*} |\zeta_\varepsilon'(t)|\leq \frac{2}{t|\log\varepsilon|}. \end{align*} Define $X_\varepsilon:B(0,R)\setminus\{0\}\to\mathbb R^{m+1}$ by \begin{align*} X_\varepsilon(x)=\zeta_\varepsilon(|x|)X(x). \end{align*} For each $x\in\Sigma$, let $\operatorname{proj}_{T_x\Sigma}:\mathbb R^{m+1}\to T_x\Sigma$ denote the Euclidean orthogonal projection onto the tangent plane $T_x\Sigma$. This vector field is compactly supported away from $0$, so the minimality of $\Sigma$ in the punctured ball applies without any singular term: \begin{align*} 0=\delta V(X_\varepsilon)=\int_\Sigma \operatorname{div}_\Sigma X_\varepsilon\,d\mathcal H^m(x). \end{align*} Expanding the tangential divergence separates the desired term from the cutoff error: \begin{align*} \operatorname{div}_\Sigma X_\varepsilon(x) =\zeta_\varepsilon(|x|)\operatorname{div}_\Sigma X(x) +\zeta_\varepsilon'(|x|)\left\langle \operatorname{proj}_{T_x\Sigma}\frac{x}{|x|},X(x)\right\rangle. \end{align*} The first term converges to $\operatorname{div}_\Sigma X$ in $L^1(\Sigma,\mathcal H^m)$: indeed $|\operatorname{div}_\Sigma X|\leq m\|DX\|_\infty$, and the finite-area hypothesis gives finite $\mathcal H^m$ measure on the support of $X$. It remains to show that the cutoff error vanishes. The density assumption $\Theta(\Sigma,0)=1$ gives local Euclidean area growth: for some $r_0>0$ and $C_0>0$, \begin{align*} \mathcal H^m(\Sigma\cap B(0,r))\leq C_0r^m \end{align*} whenever $0<r<r_0$. Since the derivative of the cutoff is supported in $B(0,\sqrt\varepsilon)\setminus B(0,\varepsilon)$, we estimate that annulus explicitly. Let $N_\varepsilon$ be the largest integer with $2^{N_\varepsilon}\varepsilon\leq \sqrt\varepsilon$, and set \begin{align*} A_k=\Sigma\cap\bigl(B(0,2^{k+1}\varepsilon)\setminus B(0,2^k\varepsilon)\bigr) \end{align*} for $0\leq k\leq N_\varepsilon$. On $A_k$ we have $|x|\geq 2^k\varepsilon$, while the area-growth bound gives $\mathcal H^m(\Sigma\cap B(0,2^{k+1}\varepsilon))\leq C_0(2^{k+1}\varepsilon)^m$ for sufficiently small $\varepsilon$. Hence \begin{align*} \int_{A_k}\frac{1}{|x|}\,d\mathcal H^m(x) &\leq C_0 2^m(2^k\varepsilon)^{m-1}. \end{align*} Summing over $k$ gives \begin{align*} \int_{\Sigma\cap (B(0,\sqrt\varepsilon)\setminus B(0,\varepsilon))}\frac{1}{|x|}\,d\mathcal H^m(x)\leq C_0 2^m\sum_{k=0}^{N_\varepsilon}(2^k\varepsilon)^{m-1}. \end{align*} Since $m\geq2$, the finite geometric sum is bounded by the terminal scale with factor $(1-2^{1-m})^{-1}$, and therefore \begin{align*} \int_{\Sigma\cap (B(0,\sqrt\varepsilon)\setminus B(0,\varepsilon))}\frac{1}{|x|}\,d\mathcal H^m(x)\leq \frac{C_0 2^m}{1-2^{1-m}}\varepsilon^{(m-1)/2}. \end{align*} Thus the displayed estimate holds with $C_1=C_0 2^m/(1-2^{1-m})$. Because $m\geq2$, the right-hand side tends to $0$ faster than the logarithmic denominator can obstruct. Thus \begin{align*} \left|\int_\Sigma \zeta_\varepsilon'(|x|)\left\langle \operatorname{proj}_{T_x\Sigma}\frac{x}{|x|},X(x)\right\rangle d\mathcal H^m(x)\right| \leq \frac{2\|X\|_\infty C_1}{|\log\varepsilon|}\varepsilon^{(m-1)/2}\to0. \end{align*} Passing to the limit in $\delta V(X_\varepsilon)=0$ gives \begin{align*} \delta V(X)=\int_\Sigma \operatorname{div}_\Sigma X\,d\mathcal H^m(x)=0. \end{align*} Thus the closed varifold has no first-variation defect at the isolated point.[/guided]

[step:Apply density-one regularity to obtain a graphical hypersurface through $0$] The varifold $V$ is an integral stationary varifold in $B(0,R)$, and $0\in\operatorname{spt}\|V\|$ because the hypothesis $\Theta(\Sigma,0)=1$ implies positive limiting mass in every sufficiently small ball about $0$. Here $\operatorname{spt}\|V\|$ denotes the [support](/page/Support) of the weight measure $\|V\|$. Let $\mathcal L^m$ denote $m$-dimensional [Lebesgue measure](/page/Lebesgue%20Measure), and define $\omega_m=\mathcal L^m(B^m(0,1))$. The [density](/page/Density) of $V$ at $0$ is defined by \begin{align*} \Theta(V,0)=\lim_{r\downarrow0}\frac{\|V\|(B(0,r))}{\omega_m r^m} \end{align*} whenever this limit exists. Since $V$ is obtained from $\Sigma$ by adding a single point of zero $\mathcal H^m$ measure, the density agrees with the stated hypersurface density: \begin{align*} \Theta(V,0)=\Theta(\Sigma,0)=1. \end{align*} Because $V$ is stationary, its generalized mean curvature is zero and its first variation is a Radon measure with zero total variation on compact subsets of $B(0,R)$. The [monotonicity formula for stationary varifolds](/page/Monotonicity%20Formula%20for%20Stationary%20Varifolds) implies that the normalized mass ratio \begin{align*} \omega_m^{-1}r^{-m}\|V\|(B(0,r)) \end{align*} is monotone nondecreasing in $r$ and has limit $1$ as $r\downarrow0$. We use the following density-one stationary integral-varifold regularity corollary of [Allard's regularity theorem](/page/Allard%20Regularity%20Theorem): for each dimension $m$ there are constants $\varepsilon_A>0$ and $\alpha\in(0,1)$ such that if an integral $m$-varifold $W$ is stationary in a ball $B(a,r)$, satisfies $a\in\operatorname{spt}\|W\|$, has density $\Theta(W,a)=1$, and has normalized mass ratio \begin{align*} \omega_m^{-1}r^{-m}\|W\|(B(a,r))<1+\varepsilon_A, \end{align*} then, after possibly shrinking to a smaller concentric ball, $\operatorname{spt}\|W\|$ is a single multiplicity-one $C^{1,\alpha}$ graph over an $m$-plane through $a$. This corollary follows from Allard regularity together with tangent-cone compactness: density one forces every tangent cone at $a$ to be a multiplicity-one plane, and convergence to that cone gives the small height and tilt excess required by Allard's theorem. For this dimensional constant $\varepsilon_A>0$, choose $\rho>0$ so that \begin{align*} \omega_m^{-1}\rho^{-m}\|V\|(B(0,\rho))<1+\varepsilon_A. \end{align*} The corollary applies at the support point $0$: the varifold $V$ is integral, stationary in $B(0,R)$, has bounded first variation with zero generalized mean curvature, satisfies $\Theta(V,0)=1$, and has the required normalized mass ratio in $B(0,\rho)$. Hence, after reducing $\rho$ if necessary, there exist an $m$-plane $P\subset\mathbb R^{m+1}$, an [open set](/page/Open%20Set) $\Omega\subset P$, and a function $u:\Omega\to P^\perp$ of class $C^{1,\alpha}$ such that, after a rigid motion preserving $0$, \begin{align*} \operatorname{spt}\|V\|\cap B(0,\rho)=\{y+u(y):y\in\Omega\}\cap B(0,\rho). \end{align*} The graph has multiplicity one because the regularity corollary gives a multiplicity-one graph and $\Theta(V,0)=1$. Since $\Sigma$ is embedded away from $0$, this single graph represents $(\Sigma\cap B(0,\rho))\cup\{0\}$ as an embedded $C^{1,\alpha}$ hypersurface near $0$. [/step]

[step:Use the weak minimal graph equation to remove the analytic singularity] Choose a fixed linear isometry $I:P\to\mathbb R^m$ and choose a unit vector $\nu\in P^\perp$, giving the linear isometry $J:P^\perp\to\mathbb R$ defined by $J(a\nu)=a$. Under the identifications by $I$ and $J$, regard $u:\Omega\to\mathbb R$ as a $C^{1,\alpha}$ function. The stationarity of $V$ implies that $u$ satisfies the weak minimal graph equation on all of $\Omega$: for every [test function](/page/Test%20Function) $\phi\in C_c^1(\Omega)$, \begin{align*} \int_\Omega \frac{\nabla u(y)\cdot \nabla\phi(y)}{\sqrt{1+|\nabla u(y)|^2}}\,d\mathcal L^m(y)=0. \end{align*} Define the coefficient map $A:\Omega\to\mathbb R^m$ by \begin{align*} A(y)=\frac{\nabla u(y)}{\sqrt{1+|\nabla u(y)|^2}}. \end{align*} The map $A$ is $C^{0,\alpha}$ because $u\in C^{1,\alpha}(\Omega)$. The linearized coefficient matrix associated with the minimal graph operator is uniformly elliptic on compact subsets of $\Omega$, since $\nabla u$ is continuous and therefore bounded there. By the Schauder interior [regularity theorem](/theorems/2750) for the quasilinear uniformly elliptic minimal graph equation, as recorded in [interior elliptic regularity for the minimal graph equation](/page/Interior%20Elliptic%20Regularity), applied on each compact subset of $\Omega$ where the coefficient matrix has a positive ellipticity constant and $C^{0,\alpha}$ coefficients, we get $u\in C^{2,\alpha}_{\mathrm{loc}}(\Omega)$. Bootstrapping the now smooth uniformly elliptic equation gives $u\in C^\infty(\Omega)$. [/step]

[step:Conclude smooth embedded minimality after adding the point] Choose $0<\rho_1<\rho$ so that the graph description above holds in $B(0,\rho_1)$. The set \begin{align*} (\Sigma\cap B(0,\rho_1))\cup\{0\} \end{align*} is the graph of the smooth function $u$ over the $m$-plane $P$ in that ball. Hence it is a smooth embedded hypersurface. Since $u$ satisfies the minimal graph equation, its mean curvature vanishes, so the resulting smooth hypersurface is minimal. This proves that $0$ is a removable singularity. [/step]