Can rectifiable varifolds be thought of as generalized characteristic functions? Consider an $n$-rectifiable varifold $V$, that is, a pair (or rather an equivalence class of pairs) $(M, θ)$ where $M$ is a countably $n$-rectifiable, $ℋ^n$-measurable subset of $ℝ^N$ and $θ$ is a positive and locally $ℋ^n$-integrable function on $M$. (All definitions as in Leon Simon's 2014 notes on Geometric Measure Theory.)
The associated Radon measure $μ_V$ is given by
$$
  μ_V(A) ≔ ∫_{M ∩ A} θ dℋ^n
$$
for all $ℋ^n$-measurable sets $A$. Consider the support of this measure:
$$
  \operatorname{supp}(μ_V) ≔ \{ x ∈ ℝ^N | ∀ \text{ open neighborhoods U of x: } μ_V(U) > 0 \}
$$
which is a closed set.
Question: (When) Do we have $\operatorname{supp}(μ_V) = M$ up to a set of $ℋ^n$-measure zero?
Attempt at a proof: By remark 1.8(2) and theorem 1.9 in Leon Simon's notes we have that the $n$-dimensional density $Θ^{*n}$ of $μ_V$ is:
$$
  Θ^{*n}(μ_V, x) = θ(x) > 0
$$
for $ℋ^n$-almost every $x ∈ M$. But by definition of the density this means that $ℋ^n$-almost every $x ∈ M$ has a neighborhood with positive $μ_V$-measure and therefore lies in $\operatorname{supp}(μ_V)$. 
Conversely, let $x ∈ \operatorname{supp}(μ_V)$. Then by definition all balls centered at $x$ have non-zero $μ_V$ measure. But by definition of $μ_V$, this means that every ball centered at $x$ has to have non-empty intersection with $M$. Therefore, $\operatorname{supp}(μ_V)$ consists entirely of adherent points of $M$ and is therefore contained in the closure of $M$. It remains to show that $\operatorname{supp}(μ_V) ∖ M$ has vanishing measure. But here I'm not getting any further since, a priori, $\bar{M}∖M$ might easily have non-zero measure. (Consider that $ℝ$ is the closure of $ℚ$ and $ℝ∖ℚ$ has full measure.) I suppose I need to employ rectifiability of $M$ somehow.
Anyway, suppose the proof can be done and indeed $\operatorname{supp}(μ_V) = M$. But in this case we can forget about $M$ entirely: Set $\bar{θ} ≔ θ · χ_{\operatorname{supp}(μ_V)}$ where $χ_{\operatorname{supp}(μ_V)}$ is the characteristic function. Then we obviously have:
$$
  μ_V(A) = ∫_{M ∩ A} \bar{θ} dℋ^n = ∫_A \bar{θ} dℋ^n
$$
Since $θ$ was positive everywhere on $M = \operatorname{supp}(μ_V)$, we have $\operatorname{supp}(\bar{θ}) = \operatorname{supp}(μ_V) = M$. But the latter was $n$-rectifiable. So it seems that any varifold is given entirely in terms of a non-negative function $\bar{θ}$ whose support is $n$-rectifiable.
Question 2: Can $n$-rectifiable varifolds therefore be thought of generalizations of characteristic functions in the sense that they are given by non-negative functions (whose support is $n$-rectifiable)? 
 A: The support of a rectifiable measure could be the whole space. Take an enumeration of $\mathbb{Q}^n$ and at each point $q_i\in\mathbb{Q}^n$ centre a segment of lenght $1/2^i$. This is a $1$ rectifiable set and which support is dense. I’d say that a good example of when the support is a good description of the geometry of the measure is when the its $n$-dimensional lower density is bounded away from $0$. Furthermore, keep in mind that the upper density is just a very weak information: the lower one is the real deal. 
It’s not clear what Q2 means precisely, but my answer tends to be “yes. They are a way of making computations with rectifiable sets, but I prefer currents”.
EDIT: about the easy way of distinguishing when the support of $\mu_V$ has non-empty interior I do not have a really "always working idea". For general rectifiable sets you cannot expect any good behaviour in general. 
If you prove that your set is contained in the complement of a cone, you can show it is a graph, for instance (chapter 15 of Mattila's book).
You can be sure that if the measure $\mu_V$ is such that $\mu_V(B(x,r))=h(r)$ for some increasing $h$ independent on $x$, then $supp(\mu_V)$ is an analytic manifold (Preiss-Kirchheim paper) 
You could direcltly assume that $M$ is a closed set. Indeed, proofs don't care much about the support anyway since you can always restrict your computations to a compact set inside a borel set of the right dimension $supporting$ the measure and make the computations there.
