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 $M_{\text{rel}}[\mu]\subset \hat{M}_{\text{rel}}[\mu]$.

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