[proofplan] The proof is an immediate consequence of the monotonicity formula for stationary $m$-varifolds. First, we apply monotonicity at the base point $x_0$ and compare the density ratios at radii $\rho$ and $R$. Then, for a point $y$ lying in the half-ball $B(x_0,R/2)$, we apply the same monotonicity formula at $y$ between the radii $\rho$ and $R/2$ and use the geometric containment $B(y,R/2) \subset B(x_0,R)$. This gives an explicit constant depending only on $m$, $R$, and the total mass in the fixed ball. [/proofplan]

[step:Apply monotonicity at $x_0$ to compare the density ratios]We use the monotonicity formula for stationary varifolds, which applies to the rectifiable stationary $m$-varifold associated to $\Sigma$. The hypothesis $\mu_\Sigma(B(x_0,R))<\infty$ ensures that the comparison mass on the right-hand side is finite. The formula states that if $z \in B(x_0,R)$ and $0 < r < s$ with $B(z,s) \subset B(x_0,R)$, then the density ratio \begin{align*} r \mapsto \frac{\mu_\Sigma(B(z,r))}{r^m} \end{align*} is nondecreasing in $r$. Equivalently, \begin{align*} \frac{\mu_\Sigma(B(z,r))}{r^m} \leq \frac{\mu_\Sigma(B(z,s))}{s^m}. \end{align*} Taking $z := x_0$ and $r := \rho$, the formula gives, for every $s$ with $\rho < s < R$, \begin{align*} \frac{\mu_\Sigma(B(x_0,\rho))}{\rho^m} \leq \frac{\mu_\Sigma(B(x_0,s))}{s^m}. \end{align*} Since $B(x_0,s) \subset B(x_0,R)$, monotonicity of the measure $\mu_\Sigma$ with respect to set inclusion gives \begin{align*} \frac{\mu_\Sigma(B(x_0,\rho))}{\rho^m} \leq \frac{\mu_\Sigma(B(x_0,R))}{s^m}. \end{align*} Letting $s \uparrow R$ uses only the convergence $s^{-m} \to R^{-m}$ and yields \begin{align*} \frac{\mu_\Sigma(B(x_0,\rho))}{\rho^m} \leq \frac{\mu_\Sigma(B(x_0,R))}{R^m}. \end{align*} Multiplying both sides by $\rho^m$ gives \begin{align*} \mu_\Sigma(B(x_0,\rho)) \leq \left(\frac{\rho}{R}\right)^m \mu_\Sigma(B(x_0,R)). \end{align*}[/step]

[guided]The monotonicity formula for stationary varifolds is the only geometric input. The hypothesis that $\Sigma$ is rectifiable lets us regard $\Sigma$ as determining a rectifiable $m$-varifold with mass measure $\mu_\Sigma$, and stationarity is exactly the first-variation hypothesis required by the formula. The added assumption $\mu_\Sigma(B(x_0,R))<\infty$ is used to make the final comparison constant finite; without it, the displayed inequality could have an infinite right-hand side. The formula says that the mass of $\Sigma$ inside balls grows at least like the model power $r^m$, in the precise sense that the ratio \begin{align*} r \mapsto \frac{\mu_\Sigma(B(z,r))}{r^m} \end{align*} is nondecreasing as long as the ball $B(z,r)$ remains inside the region where the varifold is stationary. We apply this with center $z := x_0$. Since $0 < \rho < R$, monotonicity gives, for every $s$ with $\rho < s < R$, \begin{align*} \frac{\mu_\Sigma(B(x_0,\rho))}{\rho^m} \leq \frac{\mu_\Sigma(B(x_0,s))}{s^m}. \end{align*} Because $B(x_0,s) \subset B(x_0,R)$, monotonicity of the measure $\mu_\Sigma$ with respect to set inclusion gives \begin{align*} \mu_\Sigma(B(x_0,s)) \leq \mu_\Sigma(B(x_0,R)). \end{align*} Therefore, for every $s$ with $\rho < s < R$, \begin{align*} \frac{\mu_\Sigma(B(x_0,\rho))}{\rho^m} \leq \frac{\mu_\Sigma(B(x_0,R))}{s^m}. \end{align*} Now let $s \uparrow R$. No continuity property of the measure is needed at this point, because the numerator on the right-hand side has already been bounded by the fixed finite quantity $\mu_\Sigma(B(x_0,R))$, which is finite by hypothesis; only the scalar denominator changes, and $s^{-m} \to R^{-m}$. Hence \begin{align*} \frac{\mu_\Sigma(B(x_0,\rho))}{\rho^m} \leq \frac{\mu_\Sigma(B(x_0,R))}{R^m}. \end{align*} Finally, multiplying by $\rho^m$ gives \begin{align*} \mu_\Sigma(B(x_0,\rho)) \leq \left(\frac{\rho}{R}\right)^m \mu_\Sigma(B(x_0,R)). \end{align*} This is exactly the first asserted estimate.[/guided]

[step:Contain the comparison ball centered at $y$ inside the fixed ball] Fix $y \in B(x_0,R/2)$. By definition of the Euclidean ball, \begin{align*} |y - x_0| < \frac{R}{2}. \end{align*} If $x \in B(y,R/2)$, then the triangle inequality gives \begin{align*} |x - x_0| \leq |x-y| + |y-x_0| < \frac{R}{2} + \frac{R}{2} = R. \end{align*} Therefore \begin{align*} B(y,R/2) \subset B(x_0,R). \end{align*} [/step]

[step:Apply monotonicity at $y$ and use the fixed mass bound]Fix $0 < \rho < R/2$. Since $y \in B(x_0,R/2) \subset B(x_0,R)$ and $B(y,R/2) \subset B(x_0,R)$, the monotonicity formula applied at the center $y$ with radii $\rho$ and $R/2$ gives \begin{align*} \frac{\mu_\Sigma(B(y,\rho))}{\rho^m} \leq \frac{\mu_\Sigma(B(y,R/2))}{(R/2)^m}. \end{align*} By the containment proved above and monotonicity of the measure $\mu_\Sigma$ with respect to set inclusion, \begin{align*} \mu_\Sigma(B(y,R/2)) \leq \mu_\Sigma(B(x_0,R)). \end{align*} Consequently, \begin{align*} \frac{\mu_\Sigma(B(y,\rho))}{\rho^m} \leq \frac{\mu_\Sigma(B(x_0,R))}{(R/2)^m} = \frac{2^m}{R^m}\,\mu_\Sigma(B(x_0,R)). \end{align*} Multiplying by $\rho^m$ gives \begin{align*} \mu_\Sigma(B(y,\rho)) \leq \frac{2^m}{R^m}\,\mu_\Sigma(B(x_0,R))\,\rho^m. \end{align*} Thus the asserted local area bound holds with \begin{align*} C := \frac{2^m}{R^m}\,\mu_\Sigma(B(x_0,R)). \end{align*} Because $\mu_\Sigma(B(x_0,R))<\infty$ by hypothesis, this constant is finite. It depends only on $m$, $R$, and $\mu_\Sigma(B(x_0,R))$, as required.[/step]

[guided]We now prove the second estimate from the same density-ratio comparison. Fix a point $y \in B(x_0,R/2)$ and a radius $\rho$ satisfying $0<\rho<R/2$. The previous containment step gives \begin{align*} B(y,R/2) \subset B(x_0,R), \end{align*} so the monotonicity formula for stationary varifolds applies at the center $y$ with the two radii $\rho$ and $R/2$. The formula gives \begin{align*} \frac{\mu_\Sigma(B(y,\rho))}{\rho^m} \leq \frac{\mu_\Sigma(B(y,R/2))}{(R/2)^m}. \end{align*} The point of choosing the comparison radius $R/2$ is that the whole ball $B(y,R/2)$ lies inside the fixed ball $B(x_0,R)$. Since $\mu_\Sigma$ is a measure, it is monotone with respect to set inclusion, and therefore \begin{align*} \mu_\Sigma(B(y,R/2)) \leq \mu_\Sigma(B(x_0,R)). \end{align*} Substituting this bound into the density-ratio inequality yields \begin{align*} \frac{\mu_\Sigma(B(y,\rho))}{\rho^m} \leq \frac{\mu_\Sigma(B(x_0,R))}{(R/2)^m} = \frac{2^m}{R^m}\,\mu_\Sigma(B(x_0,R)). \end{align*} Multiplying by $\rho^m$ gives \begin{align*} \mu_\Sigma(B(y,\rho)) \leq \frac{2^m}{R^m}\,\mu_\Sigma(B(x_0,R))\,\rho^m. \end{align*} Thus the local area bound holds with \begin{align*} C := \frac{2^m}{R^m}\,\mu_\Sigma(B(x_0,R)). \end{align*} This constant is finite because $\mu_\Sigma(B(x_0,R))<\infty$ by hypothesis, and it depends only on $m$, $R$, and $\mu_\Sigma(B(x_0,R))$.[/guided]