Consider the measurable space $(\mathbb{R}^n, \mathcal{B}(\mathbb{R}^n))$ with Lebesgue measure $\lambda$. I have a submanifold of $\mathbb{R}^n$ defined as the level set of a smooth function $f:\mathbb{R}^n \to\mathbb{R}^m$ with $n > m$ $$ \mathcal{M} = \left\{x\in\mathbb{R}^n\,:\, f(x) = 0\right\} $$ This submanifold is $n - m$ dimensional and is embedded in the ambient space $\mathbb{R}^n$. Let $\mathcal{H}$ be the Hausdorff measure on this manifold.
What is the Radon-Nikodym derivative of the Lebesgue measure with respect to the Hausdorff measure? $$ \frac{d\lambda}{d\mathcal{H}} = ? $$
Attempted Solution
Intuitively, I would guess this is the "jacobian term" $$ \frac{d\lambda}{d\mathcal{H}}(x) = \left|J_f(x) J_f(x)^\top\right|^{1/2} $$ where $J_f(x)$ is the Jacobian matrix of $f$. However I was not able to prove this. To be honest, I am not sure how generally one goes about finding a Radon-Nikodym derivative. What's the strategy one uses, given the two measures?
I know that the definition is $$ \lambda(A) = \int_A \frac{d \lambda}{d \mathcal{H}} d\mathcal{H} $$