Vanishing tails are equivalent to continuity of the threshold under uniformly bounded multiplicative perturbations of the density that converge in measure. This explains the role of the tail condition: if it fails, a perturbation supported on sets of shrinking volume can lower the threshold by a fixed factor.
Proposition 12.10.5.249445C21482 Let $0\leq p\in \mathcal D_\omega $ be nonzero. The following are equivalent:
$p\in \mathcal D_\omega ^0$;
the densities $p k_ T$, for $T\in \mathscr G_\omega $, are uniformly integrable;
whenever $F_ j$ are smooth real functions, uniformly bounded and tending to zero in measure, one has
\[ b_\omega (\mathrm{e}^{F_ j}p^ n)\longrightarrow b_\omega (p^ n). \]
If the first condition holds, the convergence in the third also holds for bounded measurable $F_ j$ with the same uniform bound and convergence in measure. If it fails, there are smooth $F_ j\geq 0$, uniformly bounded and tending to zero in measure, such that
\[ \mathrm{e}^{F_ j}p^ n\longrightarrow p^ n\quad \text{in }L^1, \qquad b_\omega (\mathrm{e}^{F_ j}p^ n)\leq 2^{-n}b_\omega (p^ n). \]
Proof.
The first condition implies the second by theorem 12.10.1. Conversely, $\mathrm{d}V(\{ p>K\} )\to 0$ as $K\to \infty $, so uniform integrability and (12.10.2) give
\[ \rho _\omega (p\mathbf1_{\{ p>K\} }) =\sup _{T\in \mathscr G_\omega }\int _{\{ p>K\} }p k_ T\, \mathrm{d}V \longrightarrow 0. \]
This is the first condition by proposition 15.32.9.
If $p\in \mathcal D_\omega ^0$ and $|F_ j|\leq A$, then $p(\mathrm{e}^{F_ j/n}-1)$ tends to zero in measure and its absolute value is at most $(\mathrm{e}^{A/n}+1)p$. The dominated-convergence criterion of proposition 15.32.9 makes its root norm tend to zero, and the triangle inequality gives $\rho _\omega (p\mathrm{e}^{F_ j/n})\to \rho _\omega (p)>0$. Taking negative $n$th powers proves the third condition, also for bounded measurable $F_ j$.
Suppose now that the tails do not tend to zero, and put $R=\rho _\omega (p)>0$. The formula of theorem 15.29.1, which uses only smooth tests, yields $\delta >0$, numbers $K_ j\to \infty $ and smooth strictly positive $T_ j\in \mathscr G_\omega $ such that
\[ \int _{\{ p>K_ j\} }p k_{T_ j}\, \mathrm{d}V>\delta . \]
Since the finite measure $p k_{T_ j}\, \mathrm{d}V$ is regular, we can choose a compact set $C_ j\subset \{ p>K_ j\} $ of mass greater than $\delta /2$ for this measure. By outer regularity of $\mathrm{d}V$, choose an open set $O_ j\supset \{ p>K_ j\} $ with $\mathrm{d}V(O_ j)\leq \mathrm{d}V(\{ p>K_ j\} )+1/j$, and a smooth cutoff $0\leq \chi _ j\leq 1$ that equals one on $C_ j$ and has support in $O_ j$. Fix $A>0$ with $\mathrm{e}^{A/n}\delta /2>2R$, and set $F_ j=A\chi _ j$. Then $F_ j\to 0$ in measure, while
\[ \rho _\omega (p\mathrm{e}^{F_ j/n}) \geq \int _ Xp\mathrm{e}^{F_ j/n}k_{T_ j}\, \mathrm{d}V >2R. \]
Since $b_\omega (\mathrm{e}^{F_ j}p^ n)=\rho _\omega (p\mathrm{e}^{F_ j/n})^{-n}$, this gives the asserted uniform gap. Finally, $|\mathrm{e}^{F_ j}-1|p^ n\leq (\mathrm{e}^ A-1)p^ n\mathbf1_{O_ j}$, whose integral tends to zero by absolute continuity of the integral. This proves the last assertion and the converse to the third condition.
$\square$
For $f=p^ n$, the perturbed densities satisfy $\mathrm{e}^{-A}f\leq \mathrm{e}^{F_ j}f\leq \mathrm{e}^ Af$ with a fixed $A$, so the continuity statement concerns only such order intervals. The instability under general $L^1$ approximation in section 15.32 involves approximations without these bounds.
For attainment at one density, uniform integrability along a single maximizing sequence is enough, and this weaker condition also identifies the limit of its weighted measures.
Proposition 12.10.6.B62E802ED631 Let $0\leq p\in \mathcal D_\omega $ be nonzero and put $R=\rho _\omega (p)$. Suppose $T_ j\in \mathscr G_\omega $ converge weakly to $T$ and satisfy $D_ p(T_ j)\to R$. If the densities $p k_{T_ j}$ are uniformly integrable, then
\[ D_ p(T)=R,\qquad p k_{T_ j}\, \mathrm{d}V\rightharpoonup p k_ T\, \mathrm{d}V. \]
Consequently, a maximizing sequence for $D_ p$ with uniformly integrable densities $pk_{T_ j}$ has a maximizing current among its weak cluster limits.
Proof.
For positive integers $K$, put $p_ K=\min (p,K)$. Since $p$ is finite almost everywhere, uniform integrability gives
\[ \varepsilon _ K:=\sup _ j\int _{\{ p>K\} }p k_{T_ j}\, \mathrm{d}V \longrightarrow 0. \]
For every nonnegative continuous function $h$, the weight $hp_ K$ is bounded and measurable, so the upper semicontinuity of $D_{hp_ K}$ proved in theorem 12.10.1 gives
\[ \limsup _ j\int _ Xhp k_{T_ j}\, \mathrm{d}V \leq \int _ Xhp_ K k_ T\, \mathrm{d}V +\| h\| _\infty \varepsilon _ K. \]
Now let $K\to \infty $. For $h=1$ this yields $R\leq D_ p(T)$, and the supremum formula (12.10.2) gives the reverse inequality. If $\Lambda $ is a weak cluster limit of the measures $p k_{T_ j}\, \mathrm{d}V$, the same estimate for all nonnegative continuous $h$ gives $\Lambda \leq p k_ T\, \mathrm{d}V$. Both measures have mass $R$, so they agree. Since these measures have bounded masses, they form a weakly relatively compact family, and every cluster limit has now been identified. The final assertion follows from the compactness of $\mathscr G_\omega $.
$\square$
Pome
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