I apologize if this is a trivial question – GMT is not my area of expertise but I'm working my way through a proof that makes extensive use of GMT and I haven't been able to find an answer to my question in the usual books.
Suppose $(M^n, g)$ is a Riemannian manifold, for $j ≥ 0$ denote by $ℋ^j$ the induced $j$-dimensional Hausdorff measure. Let $T ∈ D_j(M)$ denote an (integer-multiplicity) rectifiable $j$-current in $(M, g)$, that is, for all compactly supported $j$-forms $ω ∈ D^j(M)$:
$$ T(ω) = ∫_A ⟨ ω, \vec{T} ⟩ \, θ \,dℋ^j \;\;, $$
where $A ⊂ M$ is a $j$-rectifiable set (i.e. up to an $ℋ^j$-nullset it is contained in a countable union of $j$-dimensional submanifolds of $M$), $\vec{T}: M → Λ^j(TM)$ is an $ℋ^j$-measurable section (a simple $j$-vector made of orthonormal vectors that orient the approximate tangent space $T_x A$ of $A$ at a.e. $x$) and $θ: A → ℕ$ is locally $ℋ^j$-integrable. (I'm following Leon Simon's definition here.)
Suppose I picked a different Riemannian metric $g'$ on $M$. Is $T$ then still a rectifiable current w.r.t. $(M, g')$ and the induced Hausdorff measure $ℋ'^j$? Heuristically speaking, when the metric changes, both $\vec{T}$ and $dℋ^j$ change and these changes should cancel out. At least for currents given by smooth submanifolds this is definitely the case. Is it true in general, though?
UPDATE: I'm increasingly sure that one can use a parametrization of $A$ by a countable family of Lipschitz functions (living on Borel sets in $\mathbb{R}^j$) together with the area formula to prove this. I'll be looking into this. In the meantime, in case anyone's got a reference or can give a definite answer to my question, I'd still very much appreciate it!
UPDATE 2: The proof indeed goes through by taking local parametrizations of the submanifolds containing the rectifiable set and then considering the pullback of the ambient metric under such parametrizations and using the area formula. I'll post a detailed proof as soon as I find the time.
