Let $\phi:[0,1]^n\to [0,1]^m$ be continuous everywhere and smooth almost everywhere.
The Hausdorff measure of $\phi([0,1]^n)$ is given by $$\mathcal{H}^n\left(\phi\left([0,1]^n\right)\right)=\int_{[0,1]^n}\sqrt{\det\left(\nabla \phi^T\cdot\nabla \phi\right)}$$
I am wondering if possible to say that
$$\mathcal{H}^n\left(\phi\left([0,1]^n\right)\right)~\le~C\left[\int_{[0,1]^n}\left|\nabla\phi\right|\right]^n$$ for some $C>0$. I have done a lot of test cases and the result seems true but I am having trouble proving this.