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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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    $\begingroup$ Note that $\|a\|\|b\|$, coded as \|a\|\|b\|, has proper spacing and $||a|| ||b||$, coded as ||a|| ||b||, does not. I fixed that and a bunch of other MathJax usage concerns while not logged in. As of a few seconds ago one more approval is awaited. $\endgroup$ – Michael Hardy Oct 31 '16 at 23:35
  • $\begingroup$ Also, $A\backslash B$, coded as A\backslash B, does not have spacing appropriate to binary operation symbols, whereas $A\setminus B$ and $A\smallsetminus B$ do. $\qquad$ $\endgroup$ – Michael Hardy Oct 31 '16 at 23:44
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    $\begingroup$ Wouldn't $|\nu-\mu|(B_R)$ be natural ($|\lambda|$ denoting the variation of the measure $\lambda$)? $\endgroup$ – Dirk Nov 1 '16 at 13:33
  • $\begingroup$ @Dirk That would be natural (and maybe is the answer), but is not so well suited to situations where the measures are concentrated at different (nearby) sets. The examples I had in mind were something like Hausdorff measure of $\{x_n=0\}$ compared to Hausdorff measure of the graph of a function over this plane which rapidly decayed to zero at $\infty$. $\endgroup$ – Rbega Nov 1 '16 at 13:57
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    $\begingroup$ @PietroMajer That't true, but the actual question is about a similar idea for measures and I do not see how that remark helps there. $\endgroup$ – Dirk Nov 1 '16 at 15:27
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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.

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  • $\begingroup$ This seems promising. How big an issue is it that the restricted metrics are not probability metrics (and likely have different masses)? Also, how independent of the exhaustion this would be? $\endgroup$ – Rbega Nov 1 '16 at 14:44
  • $\begingroup$ Oh right - I added something to handle the mismatch. Regarding the dependence on exhaustion, I have no idea (especially since I do not know what you are really after). $\endgroup$ – Dirk Nov 1 '16 at 15:01
  • $\begingroup$ Part of the issue is that I'm not sure exactly what I'm after either -- I have a number of properties I would like to hold, but am not sure if they are consistent. Your $d_{KR}$ seems very close to the right thing, though I wonder whether the restriction to balls is necessary. Do you know whether something like $d_{KR}(\mu,\nu)=\sup\{ \int f d(\mu-\nu): \text{Lip}(f)\leq 1, ||f||_\infty \leq 1, \text{spt}(f) \mbox{ compact}\}$ has ever been considered? $\endgroup$ – Rbega Nov 1 '16 at 16:33
  • $\begingroup$ I don't know. At first glance it looks like the condition on compact support does not change the supremum… You could start looking in the Bogachev books on measure theory. $\endgroup$ – Dirk Nov 1 '16 at 16:40
  • $\begingroup$ I added the compact support just to ensure sure the integrals exist, but they probably aren't necessary. If I'm reading pg.191 of Volume II correctly this is a valid concept in this setting. This gives me something to play around with. Thanks! $\endgroup$ – Rbega Nov 1 '16 at 17:13
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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.

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    $\begingroup$ I don't think this answers what I was trying to ask (which I've clarified) as in general the measure $\mu$ would be signed. So the function $1$ is not integrable (it only exists as a sort of principal value... $\endgroup$ – Rbega Nov 1 '16 at 13:31
  • $\begingroup$ So what? Once again - all what you say is formulated in terms of the difference $f-g$ which is assumed integrable. But apparently you prefer to complicate things. $\endgroup$ – R W Nov 1 '16 at 18:18
  • $\begingroup$ The $L^1$ part was just motivation. The question is what might be the correct notion of "integrable" for the difference of two measures. $\endgroup$ – Rbega Nov 1 '16 at 18:36

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