Measures with finite mass relative to a fixed measure Fix a function $f\in L^1_\text{loc}(\mathbb{R}^n)$.  Let
$$
L^1_\text{rel}[f]=\{ g\in L^1_\text{loc}(\mathbb{R}^n) : \|g-f\|_1<\infty\}.$$
be space of functions which differ from $f$ by an $L^1$ function. Observe, that for any $g\in L^1_\text{rel}[f]$ there is a well defined notion of relative integral given by
$$
I_\text{rel}(g; f)= \lim_{R\to \infty} \left(\int_{B_R} g\; dL^n- \int_{B_R} f\; dL^n\right).
$$
To see this observe, that for any $\varepsilon>0$, there an $R_\varepsilon$ so that
$$
\int_{\mathbb{R}^n\setminus B_{R_\varepsilon}} |f-g| \; dL^n <\varepsilon
$$
and so for $R_\varepsilon<R_1<R_2$, 
$$
 \left|\int_{B_{R_1}} g\; dL^n- \int_{B_{R_1}} f\; dL^n- \left(\int_{B_{R_2}} g\; dL^n- \int_{B_{R_2}} f\; dL^n \right)\right|=\left|\int_{B_{R_2}\setminus B_{R_1}}f-g\; dL^n \right| < \varepsilon.
$$
My question is what extent is there something analogous when one considers  Radon measures on $\mathbb{R}^n$.  For instance, the above means this should be possible for measures which are absolutely continuous with respect to Lebesgue measure (and more generally absolutely continuous outside of a compact set).  To what extent can one weaken this?
Any references would be appreciated.
Edit 
To clarify what I am asking:
Going back to functions, one could define
$$
\hat{L}_\text{rel}^1[f]=\{g\in L_\text{loc}^1(\mathbb{R}^n): I_\text{rel}(g; f) \mbox{ exists}\}.
$$
Clearly, ${L}_\text{rel}^1[f]\subset \hat{L}_\text{rel}^1[f]$, but the latter
is a bigger space (and also less natural as it depends on the exhaustion of $\mathbb{R}^n$).
With this in mind for a fixed Radon measure $\mu\in M(\mathbb{R}^n)$ one could define
$$
\hat{M}_{\text{rel}}[\mu]=\{ \nu\in M(\mathbb{R}^n): \lim_{R\to \infty} (\nu(B_R)-\mu(B_R)) \mbox{ exists}\}.
$$
so my question is whether there is a natural choice of $M_{\text{rel}}[\mu]\subset \hat{M}_{\text{rel}}[\mu]$.
 A: The first natural thing that come to mind would be to consider the variation of the difference of the measure, i.e. assuming that the limit
$$
\lim_{R\to\infty}|\nu-\mu|(B_R)
$$
exists.
Considering the desired situation as stated in a comment (i.e. Hausdorff measures with disjoint but close supports), let's rephrase this a bit: The total variation of a measure is indeed a metric on the space of measures, so the above can also be written as
$$
\lim_{R\to\infty}d_{TV}(\nu\llcorner B_R,\mu\llcorner B_R)
$$
i.e. as the total variation distance of the measures restricted to the balls. As this metric may be too strong for the applications in mind one could replace the metric $d_{TV}$ by weaker metrics and use the condition that
$$
\lim_{R\to\infty}d(\nu\llcorner B_R,\mu\llcorner B_R)
$$
exists. There are several weaker metrics for measures available, e.g. the ones mentioned in the answers to this question. As an example, one might consider the Kantorovich-Rubinstein metric which is
$$
d_{KR}(\mu,\nu) = \sup\{\int f d(\mu-\nu)\mid \mathrm{Lip}(f)\leq 1,\ \|f\|_\infty\leq 1\}
$$
i.e. in this particular case
$$
d_{KR}(\mu\llcorner B_r,\nu\llcorner B_R) = \sup\{\int_{B_R} f d(\mu-\nu)\mid \mathrm{Lip}(f)\leq 1,\ \|f\|_\infty\leq 1\}.
$$
You may play around with the bounds on the Lipschitz constant and the function $f$ as the former is relevant for the measurement of "geometric nearness" while the latter is relevant for the measurement of the mass mismatch.
A: First of all, your question has nothing to do with the function $f$ - as all the statements are formulated in terms of the difference $f-g$. So, you are asking, given a measure $\mu$ on a space $X$ and a $\mu$-integrable function $\phi$, whether the integrals of $\phi$ over an increasing exhausting sequence of sets $K_n$ converge to the integral of $\phi$ over the whole space. 
