[proofplan] We prove the varifold statement, since a smooth properly embedded minimal submanifold without boundary in the ball defines a [stationary integral varifold](/page/Stationary%20Integral%20Varifold) with [weight measure](/page/Weight%20Measure) $\mathcal{H}^m\big|_\Sigma$. The proof applies the [first variation formula for varifolds](/page/First%20Variation%20Of%20A%20Varifold) to radial vector fields $X(x)=\gamma(|x-x_0|)(x-x_0)$ and computes the tangential divergence exactly. Instead of differentiating a nonsmooth cutoff at a sphere, we first obtain the identity for intervals whose endpoints are continuity radii of the mass function, and then remove the continuity restriction by one-sided approximation. The nonnegative annular integrand gives monotonicity. [/proofplan] [step:Translate the centre to the origin and define the radial data] By translating $\mathbb{R}^n$, assume $x_0=0$. Let $\mu := \|V\|$ denote the [weight measure](/page/Weight%20Measure) of the [stationary integral varifold](/page/Stationary%20Integral%20Varifold) $V$. Define $r: B(0,R)\setminus\{0\} \to (0,R)$ by $r(x):=|x|$. Define $A:(0,R)\to[0,\infty)$ by $A(s):=\mu(B(0,s))$. Define the density-ratio function $F:(0,R)\to[0,\infty)$ by $F(s):=s^{-m}A(s)$. For $\mu$-almost every $x$, let $T_xV$ denote the [approximate tangent plane](/page/Approximate%20Tangent%20Plane) of $V$ at $x$. Write $x^\top$ and $x^\perp$ for the orthogonal projections of $x$ onto $T_xV$ and $(T_xV)^\perp$, respectively. Then \begin{align*} |x|^2 = |x^\top|^2 + |x^\perp|^2. \end{align*} Let $\theta_m := \mathcal{L}^m(B(0,1))$ denote the volume of the unit ball in $\mathbb{R}^m$. It is enough to prove, for $0<\sigma<\rho<R$, \begin{align*} \frac{A(\rho)}{\rho^m}-\frac{A(\sigma)}{\sigma^m}=\int_{B(0,\rho)\setminus B(0,\sigma)}\frac{|x^\perp|^2}{|x|^{m+2}}\,d\mu(x), \end{align*} because division by $\theta_m$ gives the stated formula. [/step] [step:Compute the tangential divergence of a radial variation] Let $\gamma:[0,R)\to\mathbb{R}$ be a $C^1$ function that is constant on a neighbourhood of $0$ and has compact support in $[0,R)$. Define $X_\gamma:B(0,R)\to\mathbb{R}^n$ by $X_\gamma(x):=\gamma(|x|)x$. Because $\gamma$ is constant near $0$, the formula extends to a $C^1$ vector field at $x=0$, with $D(X_\gamma)_0=\gamma(0)I_n$, where $I_n:\mathbb{R}^n\to\mathbb{R}^n$ denotes the identity linear map. For $\mu$-almost every $x\ne0$, choose an orthonormal basis $e_1,\dots,e_m$ of $T_xV$. The chain rule gives \begin{align*} D(X_\gamma)_x(e_i)=\gamma(|x|)e_i+\gamma'(|x|)\frac{x\cdot e_i}{|x|}x. \end{align*} Taking the tangential trace over $T_xV$ yields \begin{align*} \operatorname{div}_{T_xV}X_\gamma=\sum_{i=1}^m D(X_\gamma)_x(e_i)\cdot e_i=m\gamma(|x|)+\gamma'(|x|)\frac{|x^\top|^2}{|x|}. \end{align*} Using $|x^\top|^2=|x|^2-|x^\perp|^2$, this becomes \begin{align*} \operatorname{div}_{T_xV}X_\gamma=m\gamma(|x|)+|x|\gamma'(|x|)-\gamma'(|x|)\frac{|x^\perp|^2}{|x|}. \end{align*} By the [first variation formula for varifolds](/page/First%20Variation%20Of%20A%20Varifold), stationarity means that the integral of $\operatorname{div}_{T_xV}X_\gamma$ against $\mu$ vanishes for every compactly supported $C^1$ ambient vector field. Hence \begin{align*} \int_{B(0,R)}\left(m\gamma(|x|)+|x|\gamma'(|x|)\right)\,d\mu(x)=\int_{B(0,R)}\gamma'(|x|)\frac{|x^\perp|^2}{|x|}\,d\mu(x). \end{align*} [guided] The point of choosing a radial vector field is that stationarity converts geometry into a one-variable identity in the radius. Let $\gamma:[0,R)\to\mathbb{R}$ be a $C^1$ function that is constant on a neighbourhood of $0$ and has compact support in $[0,R)$. Define $X_\gamma:B(0,R)\to\mathbb{R}^n$ by $X_\gamma(x):=\gamma(|x|)x$. The condition that $\gamma$ is constant near $0$ is what makes $X_\gamma$ a $C^1$ vector field at the origin; near $0$ it is just a constant multiple of the identity map. Fix a point $x\ne0$ where the approximate tangent plane $T_xV$ exists, and choose an orthonormal basis $e_1,\dots,e_m$ of $T_xV$. Differentiating $X_\gamma$ in the direction $e_i$ gives \begin{align*} D(X_\gamma)_x(e_i)=\gamma(|x|)e_i+\gamma'(|x|)\frac{x\cdot e_i}{|x|}x. \end{align*} Taking the tangential trace over $T_xV$ gives \begin{align*} \operatorname{div}_{T_xV}X_\gamma=\sum_{i=1}^m D(X_\gamma)_x(e_i)\cdot e_i=m\gamma(|x|)+\gamma'(|x|)\frac{|x^\top|^2}{|x|}. \end{align*} The normal term appears by decomposing $x=x^\top+x^\perp$. Orthogonality gives $|x^\top|^2=|x|^2-|x^\perp|^2$, so \begin{align*} \operatorname{div}_{T_xV}X_\gamma=m\gamma(|x|)+|x|\gamma'(|x|)-\gamma'(|x|)\frac{|x^\perp|^2}{|x|}. \end{align*} The [first variation formula for varifolds](/page/First%20Variation%20Of%20A%20Varifold) applies because $X_\gamma$ is a compactly supported $C^1$ ambient vector field in $B(0,R)$, and stationarity says that this first variation is zero. It gives \begin{align*} 0=\int_{B(0,R)}\left(m\gamma(|x|)+|x|\gamma'(|x|)-\gamma'(|x|)\frac{|x^\perp|^2}{|x|}\right)\,d\mu(x). \end{align*} Moving the normal term to the right-hand side gives \begin{align*} \int_{B(0,R)}\left(m\gamma(|x|)+|x|\gamma'(|x|)\right)\,d\mu(x)=\int_{B(0,R)}\gamma'(|x|)\frac{|x^\perp|^2}{|x|}\,d\mu(x). \end{align*} This identity is the source of monotonicity: after a suitable radial [test function](/page/Test%20Function) is chosen, the right-hand side records exactly the squared normal part of the radial vector. [/guided] [/step] [step:Derive the distributional derivative of the density ratio] Let $\psi:(0,R)\to\mathbb{R}$ be a nonnegative $C^1$ function with compact support in $(0,R)$. Define the radial coefficient $\gamma_\psi:[0,R)\to\mathbb{R}$ by \begin{align*} \gamma_\psi(s):=-\int_s^R t^{-m-1}\psi(t)\,d\mathcal{L}^1(t). \end{align*} Since $\operatorname{supp}\psi$ is compactly contained in $(0,R)$, this function is constant near $0$, vanishes near $R$, and is $C^1$. Moreover, for $0<s<R$, \begin{align*} \gamma_\psi'(s)=s^{-m-1}\psi(s). \end{align*} Substituting $\gamma_\psi$ into the radial stationarity identity gives \begin{align*} \int_{B(0,R)}\left(-m\int_{|x|}^R t^{-m-1}\psi(t)\,d\mathcal{L}^1(t)+|x|^{-m}\psi(|x|)\right)\,d\mu(x) =\int_{B(0,R)}\psi(|x|)\frac{|x^\perp|^2}{|x|^{m+2}}\,d\mu(x). \end{align*} Define a Radon measure $\lambda$ on $(0,R)$ by \begin{align*} \lambda(E):=\int_{r^{-1}(E)}\frac{|x^\perp|^2}{|x|^{m+2}}\,d\mu(x) \end{align*} for every Borel set $E\subset(0,R)$. The estimate $|x^\perp|^2\le |x|^2$ and separation from $0$ on compact subintervals of $(0,R)$ show that $\lambda$ is locally finite. We now identify the left-hand side with the distributional pairing of $F$ against $-\psi'$. Choose $0<a<b<R$ with $\operatorname{supp}\psi\subset[a,b]$. Since $\mu$ is a Radon measure, $\mu(\overline{B(0,b)})<\infty$, and \begin{align*} \left|\mathbb{1}_{\{|x|<s\}}s^{-m}\psi'(s)\right| \le a^{-m}\|\psi'\|_{C^0}\mathbb{1}_{\overline{B(0,b)}}(x)\mathbb{1}_{[a,b]}(s) \end{align*} is integrable with respect to $d\mu(x)\,d\mathcal L^1(s)$. Thus [Fubini's theorem](/theorems/2961) applies to the signed integrand. Since $A(s)=\mu(B(0,s))$ and $F(s)=s^{-m}A(s)$, it gives \begin{align*} -\int_0^R F(s)\psi'(s)\,d\mathcal{L}^1(s) =\int_{B(0,R)}\left(\int_{|x|}^R -s^{-m}\psi'(s)\,d\mathcal{L}^1(s)\right)\,d\mu(x). \end{align*} For each $a\in[0,R)$, [integration by parts](/theorems/2098) on the compact interval containing $\operatorname{supp}\psi$ gives \begin{align*} \int_a^R -s^{-m}\psi'(s)\,d\mathcal{L}^1(s) =a^{-m}\psi(a)-m\int_a^R s^{-m-1}\psi(s)\,d\mathcal{L}^1(s), \end{align*} with the first term interpreted as $0$ when $a$ lies outside $\operatorname{supp}\psi$. Applying this with $a=|x|$ shows \begin{align*} -\int_0^R F(s)\psi'(s)\,d\mathcal{L}^1(s)=\int_0^R \psi(s)\,d\lambda(s). \end{align*} We extend from nonnegative tests to arbitrary compactly supported $C^1$ tests as follows. Let $\psi:(0,R)\to\mathbb{R}$ be compactly supported and $C^1$. Choose a compactly supported $C^1$ cutoff function $\eta:(0,R)\to[0,1]$ with $\eta=1$ on $\operatorname{supp}\psi$, and choose $C>\|\psi\|_{C^0((0,R))}$. Then $\psi+C\eta$ and $C\eta$ are nonnegative compactly supported $C^1$ test functions, and \begin{align*} \psi=(\psi+C\eta)-C\eta. \end{align*} Applying the identity to the two nonnegative tests and subtracting gives the same identity for $\psi$. Thus the [distributional derivative](/page/Distributional%20Derivative) of $F$ on $(0,R)$ is the nonnegative Radon measure $\lambda$. [guided] The cutoff should be chosen so that the right-hand side of the stationarity identity becomes exactly the normal-error measure. Let $\psi:(0,R)\to\mathbb{R}$ be a nonnegative $C^1$ function with compact support in $(0,R)$, and define \begin{align*} \gamma_\psi(s):=-\int_s^R t^{-m-1}\psi(t)\,d\mathcal{L}^1(t) \end{align*} for $0\le s<R$. Because $\psi$ is supported away from $0$ and $R$, the function $\gamma_\psi$ is constant near $0$, is zero near $R$, and is a valid radial coefficient for the compactly supported vector field $X_{\gamma_\psi}(x)=\gamma_\psi(|x|)x$. Its derivative is \begin{align*} \gamma_\psi'(s)=s^{-m-1}\psi(s). \end{align*} Therefore the radial stationarity identity becomes \begin{align*} \int_{B(0,R)}\left(-m\int_{|x|}^R t^{-m-1}\psi(t)\,d\mathcal{L}^1(t)+|x|^{-m}\psi(|x|)\right)\,d\mu(x) =\int_{B(0,R)}\psi(|x|)\frac{|x^\perp|^2}{|x|^{m+2}}\,d\mu(x). \end{align*} This is the key choice: $\gamma_\psi'(|x|)/|x|=\psi(|x|)|x|^{-m-2}$, so the right-hand side is already the desired measure paired with $\psi$. Define $\lambda$ on Borel sets $E\subset(0,R)$ by \begin{align*} \lambda(E):=\int_{r^{-1}(E)}\frac{|x^\perp|^2}{|x|^{m+2}}\,d\mu(x). \end{align*} This is locally finite because on each compact interval $[a,b]\subset(0,R)$ we have $|x|\ge a$ and $|x^\perp|^2\le |x|^2$, so the integrand is bounded above by $a^{-m}$ on $r^{-1}([a,b])$, and $\mu$ is locally finite. It remains to show that the left-hand side is exactly $-\int F\psi'\,d\mathcal{L}^1$. Choose $0<a<b<R$ with $\operatorname{supp}\psi\subset[a,b]$. The function \begin{align*} (x,s)\mapsto \mathbb{1}_{\{|x|<s\}}s^{-m}\psi'(s) \end{align*} is absolutely integrable on $B(0,R)\times(0,R)$ because it is supported in $\overline{B(0,b)}\times[a,b]$ and bounded there by $a^{-m}\|\psi'\|_{C^0}$, while $\mu(\overline{B(0,b)})<\infty$. Therefore Fubini's theorem applies to this signed integrand, and since $A(s)=\mu(B(0,s))$ it gives \begin{align*} -\int_0^R F(s)\psi'(s)\,d\mathcal{L}^1(s) =\int_{B(0,R)}\left(\int_{|x|}^R -s^{-m}\psi'(s)\,d\mathcal{L}^1(s)\right)\,d\mu(x). \end{align*} The inner one-dimensional integral is computed by [integration by parts](/theorems/210): \begin{align*} \int_{|x|}^R -s^{-m}\psi'(s)\,d\mathcal{L}^1(s) =|x|^{-m}\psi(|x|)-m\int_{|x|}^R s^{-m-1}\psi(s)\,d\mathcal{L}^1(s), \end{align*} again with the first term equal to $0$ when $|x|$ is outside the support of $\psi$. This is precisely the left-hand side produced by stationarity. Hence \begin{align*} -\int_0^R F(s)\psi'(s)\,d\mathcal{L}^1(s)=\int_0^R \psi(s)\,d\lambda(s). \end{align*} To extend the identity from nonnegative tests to all compactly supported $C^1$ tests, we avoid taking positive and negative parts, since those need not be $C^1$. Let $\psi:(0,R)\to\mathbb{R}$ be compactly supported and $C^1$. Choose a compactly supported $C^1$ cutoff function $\eta:(0,R)\to[0,1]$ such that $\eta=1$ on $\operatorname{supp}\psi$, and choose $C>\|\psi\|_{C^0((0,R))}$. Then $\psi+C\eta\ge0$ and $C\eta\ge0$, both are compactly supported and $C^1$, and \begin{align*} \psi=(\psi+C\eta)-C\eta. \end{align*} The identity holds for the two nonnegative tests on the right-hand side, and subtracting the two identities gives it for $\psi$. Thus the distributional derivative of $F$ is the nonnegative Radon measure $\lambda$. [/guided] [/step] [step:Integrate the derivative measure over annuli with the left-continuous representative] The function $A(s)=\mu(B(0,s))$ is left-continuous because $B(0,s_j)\uparrow B(0,s)$ whenever $s_j\uparrow s$. On every compact interval $[a,b]\subset(0,R)$, the Radon property gives $\mu(\overline{B(0,b)})<\infty$, so $F(s)=s^{-m}A(s)$ is bounded by $a^{-m}\mu(\overline{B(0,b)})$ and belongs to $L^1_{\mathrm{loc}}((0,R))$. The distributional identity from the previous step says that this concrete $L^1_{\mathrm{loc}}$ function has distributional derivative $\lambda$, hence its BV equivalence class has a left-continuous representative. Since the pointwise function $F(s)=s^{-m}\mu(B(0,s))$ is itself left-continuous, it is exactly that representative. We use the following representative form of the one-dimensional theorem for [functions of bounded variation](/page/Functions%20Of%20Bounded%20Variation): if $G\in L^1_{\mathrm{loc}}((0,R))$ has distributional derivative equal to a Radon measure $\nu$, and $G_\ell$ denotes the left-continuous representative, then \begin{align*} G_\ell(b)-G_\ell(a)=\nu([a,b)) \end{align*} for every $0<a<b<R$. The half-open interval convention is part of the representative statement and matches left-continuity. Applying this with $G=F$, $G_\ell=F$, $\nu=\lambda$, $a=\sigma$, and $b=\rho$ gives \begin{align*} \rho^{-m}A(\rho)-\sigma^{-m}A(\sigma)=\lambda([\sigma,\rho)). \end{align*} By the definition of $\lambda$ and the identity $r^{-1}([\sigma,\rho))=B(0,\rho)\setminus B(0,\sigma)$, this becomes \begin{align*} \rho^{-m}A(\rho)-\sigma^{-m}A(\sigma) =\int_{B(0,\rho)\setminus B(0,\sigma)}\frac{|x^\perp|^2}{|x|^{m+2}}\,d\mu(x). \end{align*} [/step] [step:Restore the centre and conclude monotonicity] This formula already contains the multiplicity of the integral varifold because the integration is with respect to the weight measure $\mu$, not with respect to unweighted Hausdorff measure on level sets. If one wants the level-set form, the [coarea formula for countably rectifiable sets](/page/Coarea%20Formula%20For%20Rectifiable%20Sets) applied to the Lipschitz map $r(x)=|x|$ gives sliced level measures induced by $\mu$; for $\mu$-almost every $x\ne0$, $|\nabla^\top r(x)|=|x^\top|/|x|$. Restoring the original centre $x_0$ and dividing by $\theta_m$ gives \begin{align*} \Theta(V,x_0,\rho)-\Theta(V,x_0,\sigma)=\frac{1}{\theta_m}\int_{B(x_0,\rho)\setminus B(x_0,\sigma)}\frac{|(x-x_0)^\perp|^2}{|x-x_0|^{m+2}}\,d\mu(x). \end{align*} The integrand is nonnegative, so $\rho \mapsto \Theta(V,x_0,\rho)$ is nondecreasing on $(0,R)$. [guided] For this step, recall the objects being integrated. The mass function is $A:(0,R)\to[0,\infty)$, $A(s)=\mu(B(0,s))$, the density ratio without the factor $\theta_m^{-1}$ is $F:(0,R)\to[0,\infty)$, $F(s)=s^{-m}A(s)$, and the radial error measure is the Radon measure $\lambda$ on $(0,R)$ defined by \begin{align*} \lambda(E)=\int_{r^{-1}(E)}\frac{|x^\perp|^2}{|x|^{m+2}}\,d\mu(x) \end{align*} for Borel sets $E\subset(0,R)$. The radial stationarity computation and the test-function argument give the distributional identity \begin{align*} -\int_0^R F(s)\psi'(s)\,d\mathcal{L}^1(s)=\int_0^R \psi(s)\,d\lambda(s) \end{align*} for every compactly supported $C^1$ test function $\psi:(0,R)\to\mathbb{R}$. Thus the distributional derivative of $F$ is $\lambda$. Because $A(s)=\mu(B(0,s))$ uses open balls, $A$ is left-continuous: if $s_j\uparrow s$, then $B(0,s_j)\uparrow B(0,s)$, and continuity from below for the Radon measure $\mu$ gives $A(s_j)\uparrow A(s)$. On compact intervals $[a,b]\subset(0,R)$, the bound $F(s)\le a^{-m}\mu(\overline{B(0,b)})$ shows $F\in L^1_{\mathrm{loc}}((0,R))$. The distributional identity was proved for this concrete pointwise function $F$, so its BV equivalence class has derivative $\lambda$; because this same pointwise function is left-continuous, it is the left-continuous representative. We now use the representative form of the one-dimensional theorem for [functions of bounded variation](/page/Functions%20Of%20Bounded%20Variation): if $G$ has distributional derivative $\nu$, then its left-continuous representative $G_\ell$ satisfies \begin{align*} G_\ell(b)-G_\ell(a)=\nu([a,b)) \end{align*} for every $0<a<b<R$. Applying this with $G=F$, $G_\ell=F$, $\nu=\lambda$, $a=\sigma$, and $b=\rho$ gives \begin{align*} F(\rho)-F(\sigma)=\lambda([\sigma,\rho)). \end{align*} The half-open interval is essential: it matches the open-ball convention exactly. Indeed, \begin{align*} r^{-1}([\sigma,\rho))=\{x\in B(0,R):\sigma\le |x|<\rho\}=B(0,\rho)\setminus B(0,\sigma). \end{align*} Using the definition of $\lambda$ gives \begin{align*} \rho^{-m}A(\rho)-\sigma^{-m}A(\sigma)=\int_{B(0,\rho)\setminus B(0,\sigma)}\frac{|x^\perp|^2}{|x|^{m+2}}\,d\mu(x). \end{align*} This proves the annular identity for every pair of radii, including radii for which $\mu(\partial B(0,r))$ may be nonzero. The formula is already the correct varifold formula with multiplicity because $\mu=\|V\|$ is the weight measure. For comparison with a boundary-sphere formulation, the [coarea formula for countably rectifiable sets](/page/Coarea%20Formula%20For%20Rectifiable%20Sets) applied to $r(x)=|x|$ disintegrates $\mu$ into sliced level measures, and $|\nabla^\top r(x)|=|x^\top|/|x|$ at $\mu$-almost every $x\ne0$. This comparison is not needed for the annular identity, but it explains why the multiplicity remains present in level-set notation. Finally restore the original centre by replacing $x$ with $x-x_0$, and divide by $\theta_m$. We obtain \begin{align*} \Theta(V,x_0,\rho)-\Theta(V,x_0,\sigma)=\frac{1}{\theta_m}\int_{B(x_0,\rho)\setminus B(x_0,\sigma)}\frac{|(x-x_0)^\perp|^2}{|x-x_0|^{m+2}}\,d\mu(x). \end{align*} The integrand is nonnegative, so the density ratio is nondecreasing as a function of $\rho$. [/guided] [/step] [step:Specialize to smooth minimal submanifolds] If $\Sigma^m \subset B(x_0,R)$ is smooth, properly embedded, minimal, and has no boundary in $B(x_0,R)$, the associated varifold $V_\Sigma$ has weight measure \begin{align*} \mu=\mathcal{H}^m\big|_\Sigma, \end{align*} where $\mathcal{H}^m\big|_\Sigma$ denotes the restriction of $m$-dimensional [Hausdorff measure](/page/Hausdorff%20Measure) to the Borel subsets of $\Sigma$. Proper embeddedness ensures local finiteness of this measure in the ball, and the no-boundary condition ensures that compactly supported ambient variations in $B(x_0,R)$ produce no boundary term. Minimality gives zero mean curvature, hence the [first variation](/page/First%20Variation%20Of%20A%20Varifold) of area vanishes for all such variations. Thus $V_\Sigma$ is stationary, and the varifold formula already proved applies. In this smooth case, $T_xV_\Sigma=T_x\Sigma$ for every $x\in\Sigma$, so $(x-x_0)^\perp$ is exactly the normal projection onto $(T_x\Sigma)^\perp$. This proves the stated formula for smooth properly embedded minimal submanifolds. [guided] It remains only to connect the varifold statement back to the smooth formulation in the theorem. A smooth properly embedded minimal submanifold $\Sigma^m\subset B(x_0,R)$ with no boundary in $B(x_0,R)$ defines an integral varifold $V_\Sigma$ whose weight measure is \begin{align*} \mu=\mathcal{H}^m\big|_\Sigma, \end{align*} where $\mathcal{H}^m\big|_\Sigma$ denotes the restriction of $m$-dimensional [Hausdorff measure](/page/Hausdorff%20Measure) to the Borel subsets of $\Sigma$. Proper embeddedness gives local finiteness of $\mathcal{H}^m\big|_\Sigma$ in the ball. The absence of boundary in $B(x_0,R)$ means that compactly supported ambient variations have no boundary contribution. The [first variation formula](/theorems/2728) for a smooth submanifold expresses the [first variation](/page/First%20Variation%20Of%20A%20Varifold) of area as the integral of the mean curvature vector against the ambient variation. Since $\Sigma$ is minimal, its mean curvature vector is zero, so the first variation vanishes for every compactly supported ambient $C^1$ vector field in $B(x_0,R)$. Therefore $V_\Sigma$ is stationary, and the varifold monotonicity identity applies to $V_\Sigma$. For a smooth submanifold there is no distinction between the approximate tangent plane of the varifold and the classical tangent space: $T_xV_\Sigma=T_x\Sigma$ for every $x\in\Sigma$. Hence $(x-x_0)^\perp$ in the varifold formula is exactly the [orthogonal projection](/theorems/437) of $x-x_0$ onto $(T_x\Sigma)^\perp$. This is precisely the formula stated for smooth properly embedded minimal submanifolds. [/guided] [/step]