Proposition 12.10.5249445C21482

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

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

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

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. $\square$