Let $\phi:[0,1]^d\to[0,1]^m$ with $m>d$ be continuous everywhere and differentiable almost everywhere. The Hausdorff measure of $\phi([0,1]^d)$ is given by
$$\mathcal{H}^d(\phi([0,1]^d))=\int_{[0,1]^d}\sqrt{\det((\nabla \phi)^T\cdot\nabla\phi)}.$$
Suppose that $\phi$ is injective and $\phi$ is monotone in the sense that all entries of the Jacobian matrix $\nabla\phi$ are have constant sign on $[0,1]^d$ (with each entry having its own sign). I am hoping to determine a bound on the measure that depends on only $m$ and $d$.
Towards this end, I was able to use the fact that $\text{diam}(\phi([0,1]^d))\le \text{diam}([0,1]^m)=1$, $\left|\frac{\partial\phi_j}{\partial x_i}\right|=\pm\frac{\partial\phi_j}{\partial x_i}$, the fundamental Calculus, and Fubini's Theorem to show that $$\int_{[0,1]^d}\left|\frac{\partial\phi_j}{\partial x_i}\right|\le 1.$$
Using the subadditivity of the square root, this is enough to show that, in the case $d=1$, $$\mathcal{H}^1(\phi([0,1]))\le \sqrt{m}.$$
The problem is that for $d>1$ the determinant results in terms of order $\left(\frac{\partial \phi_{j_1}}{\partial x_{i_1}}\right)\cdots\left(\frac{\partial \phi_{j_d}}{\partial x_{i_d}}\right)$ which need not be integrable. I cannot use Haddamard's inequality to bound the determinant by only using the $L^1$ norms of the partials because $\int_{[0,1]^d}\left(|f|^n\right)$ is not bounded by $\left(\int_{[0,1]^d}|f|\right)^n$ for general $f\in L^1([0,1]^d)$.
I believe that the result is still true due to cancellations of "bad" mixed terms when computing the determinant in the specific examples I tried. I know that these cancellations do not occur when computing $\det(A^TA)$ for a general matrix $A$ with entries of bounded $L^1$ norm, so it must somehow use that $\nabla\phi$ is a Jacobian of a map.
I tried to establish a bound for the measure of $\phi([0,1]^d)$ in terms of the $L^1$ norm of $\nabla\phi$ in this question, but I now believe that might not be the best approach.
I am wondering if there is a bound for $\mathcal{H}^d(\phi([0,1]^d))$ in terms of only $m$ and $d$ and how to derive that bound.