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Let $\lambda_r$ be the Lebesgue measure restricted to the interval $[-r,r]\subset \mathbb{R}$. Each $\lambda_r$ can be normalized to $\mu_r=\lambda_r/2r$ which is a probability. The sequence $\lambda_r$ converges for $r \rightarrow \infty$ to the Lebesgue measure on the real line, but the sequence $\mu_r$ does not converge to a countably additive measure but to a finitely additive probability measure. The limit measure $\mu$ in fact satisfies $\mu(A)=0$ for any bounded subset of $\mathbb{R}$ but $\mu(\mathbb{R})=1$.

I can obtain another finitely additive probability measure by considering the sequence $\nu_r=\dfrac{1}{e^r}\int_{-\infty}^r e^x dx$, which is a probability and at the limit satisfies again $\nu(A)=0$ for any bounded subset of $\mathbb{R}$ and $\nu(\mathbb{R})=1$.

There are infinite ways to do that. My question do the two finitely additive probability measures obtained as limits of the two different sequence $\mu_r$ and $\nu_r$ are different? In which subsets of $\mathbb{R}$ do they differ? Do you know any reference that discusses the generation of finitely additive probability measures as a limit of sequences of countably additive measures?

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  • $\begingroup$ What topology do you consider on the set of measures over $\mathbb{R}$? In other words, what do you mean by measures $\lambda_{r}$ converging to the Lebesgue measure? And note that it is an uncountable family of measures, not a sequence, unless you consider $r\in\mathbb{N}$ ofcourse. $\endgroup$
    – T. Eskin
    Apr 23, 2012 at 15:40
  • $\begingroup$ Also $\nu_{r}$, as defined above, is a constant for each $r$. What type of measure should it represent? $\endgroup$
    – T. Eskin
    Apr 23, 2012 at 15:45
  • $\begingroup$ $\mu\ne\nu$ since $\mu(\mathbb R^+)=\frac12\ne1=\nu(\mathbb R^+)$. $\endgroup$
    – Did
    Apr 23, 2012 at 23:36
  • $\begingroup$ I mean $\lambda_r(x)=1$ if $x \in [-r,r]$ and zero otherwise. You are right is not a sequence, I should specify that $ r \in \mathbb{N}$. $\endgroup$
    – vatna
    Apr 24, 2012 at 6:55

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I object to

the sequence $\mu_r$ [converges] to a finitely additive probability measure.

because it suggests we have a limit as $r\to\infty$. In reality, the following happens: every $\mu_r$ is a point of the unit sphere in $L_1(\mathbb R)$, which is a subset of the unit sphere of $L_\infty(\mathbb R)^*$. By weak* compactness, $\mu_r$ have at least one limit point in the unit sphere of $L_\infty(\mathbb R)^*$. (We can think of any element of $L_\infty(\mathbb R)^*$ as a finitely additive measure because it can be applied to characteristic functions of measurable set.) Any such limit point could be called $\mu$ and there are many of them. For example, on the set $A=\bigcup_k [4^k, 2\cdot 4^k]$ the measures $\mu_r$ take values anywhere between $1/6$ and $1/3$. By restricting to a subsequence with $\mu_{r_j}(A)=\gamma$ for a particular $\gamma\in (1/6,1/3)$ we get a finitely additive measure $\mu$ with $\mu(A)=\gamma$. This already gives uncountably many choices for $\mu$.

And whatever choice of $\mu$ and $\nu$ are made, they will not be the same, as Did pointed out: $\mu((-\infty,b])=1/2$ and $\nu((-\infty,b])=0$ for any real $b$.

For more information, search for invariant mean or read about amenable groups.

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