The isoperimetric theorem [71, Thm 6.1] guarantees the existence of an integral current $A^{\prime}$ supported in $B_{R}$ such that $\partial A^{\prime}=\partial A$ and for which

$$Mass(A^{\prime})^{(n-1)/n}\leq CMass(\partial A).$$

Notice the metric uniform equivalence means we do not need to be careful to distinguish the Hausdorff measure for the Euclidean metric and the Calabi-Yau metric. Let $T$ denote the cone over the current $A-A^{\prime}$, then $\partial T=A-A^{\prime}$, and thus by the quantitative calibrated condition,

$$\begin{split}&Mass(A)\leq\frac{1}{\sin\epsilon}\int_{A}\text{Re}\Omega=\frac{1}{\sin\epsilon}\int_{A^{\prime}}\text{Re}\Omega+\frac{1}{\sin\epsilon}\int_{\partial T}\text{Re}\Omega\\ \leq&\frac{1}{\sin\epsilon}Mass(A^{\prime})+\frac{1}{\sin\epsilon}\int_{T}d\text{Re}\Omega\leq\frac{1}{\sin\epsilon}CMass(\partial A)^{n/(n-1)},\end{split}$$

which is the isoperimetric inequality. ∎

We first isometrically embed $X$ into an ambient Euclidean space $\mathbb{R}^{N}$, so $A$ can be regarded as an integral current compactly supported in $X$. Fix a small number $\rho_{0}>0$ such that over $U$ the $\rho_{0}$-neighbourhood in $\mathbb{R}^{N}$ is isomorphic to the normal bundle, so there is a smooth retraction map $F$ back to $U$. The Lipschitz norm of $F$ is approximately one.

Applying the deformation theorem for $\mathbb{R}^{N}$ [71, section 5.3] to the current $\partial A$, with a parameter $\rho\leq\rho_{0}$ to be fixed, we can write

$$\partial A=P+\partial R,$$

where $P,R$ are integral currents inside $\mathbb{R}^{N}$, supported in the $O(\rho)$ neighbourhood of $\text{supp}(\partial A)$, with

$$Mass(R)\leq C\rho Mass(\partial A),\quad Mass(P)\leq CMass(\partial A),$$

where the constant $C$ depends only on $N,m$. Morever, $P$ is an integral linear sum of $m$-dimensional faces in the standard grid decomposition of $\mathbb{R}^{N}$ with cube size $\rho$. We now push forward via $F$:

$$F_{*}P+\partial F_{*}R=F_{*}(\partial A)=\partial A,$$

since $\partial A\subset U\subset X$ is fixed by $F$. Note that $F_{*}P,F_{*}R$ both live inside $U$, and their mass bounds are essentially the same as $P,R$ respectively.

Suppose first that $Mass(\partial A)\ll 1$. If $P$ is nonzero, then by the grid description of $P$,

$$\rho^{m}\leq Mass(P)\leq CMass(\partial A)$$

So by choosing $\rho=2(CMass(\partial A))^{1/m}$ in the above, we force $P=0$, so $\partial A=\partial F_{*}R$, with mass bound $Mass(F_{*}R)\leq\text{const}Mass(\partial A)^{(m+1)/m}$, so it suffices to take $A^{\prime}=F_{*}R$.

Now suppose $Mass(\partial A)\gtrsim 1$, then we choose $\rho=\rho_{0}$. Without loss of generality, we can replace $\partial A$ by $F_{*}P$, and pretend $R=0$. We know

• •

  $P$ is an integral linear combination of grid cube faces, where all the cubes lie in a bounded region of a fixed grid,

• •

  $F_{*}P$ is an exact current on $X$.

The set of all such $P$ form a finitely generated abelian group, which by classification is isomorphic to the direct sum of $\oplus_{1}^{r}\mathbb{Z}e_{i}$ and a finite abelian group. For any given element

$$P=\sum a_{i}e_{i}+\text{finite group part},$$

the linear coefficients $a_{i}$ of $e_{i}$ for $i=1,\ldots r$ are bounded by $|a_{i}|\lesssim Mass(P)\lesssim Mass(\partial A)$. Each $e_{i}$ gives rise to an exact simplicial chain $F_{*}e_{i}$ inside $X$, which is the boundary of a finite mass integral current $Q_{i}$. Thus

$$Mass(\sum_{1}^{r}a_{i}Q_{i})\leq\sum_{1}^{r}|a_{i}|Mass(Q_{i})\leq\text{const}\cdot Mass(\partial A).$$

The finite group part gives rise to another simplicical chain inside $X$ which is the boundary of some finite mass integral current. Thus we have produced an integral current $A^{\prime}$ with $\partial A^{\prime}=\partial A$, and mass bound

$$Mass(A^{\prime})\leq C(Mass(\partial A))+C\leq\text{const }Mass(\partial A).$$

since we are in the $Mass(\partial A)\gtrsim 1$ case. ∎