Theorem 11.35.4.6830F8DE9750 For every finite real $L$, the set
\[ \mathcal K_ L=\{ \nu \in \mathcal P(X):H_\varphi (\nu )\leq L\} \]
is weakly compact. Its measures are uniformly absolutely continuous with respect to $\mathrm{d}V$: for every $\epsilon >0$ there is a $\delta >0$ such that
FA8EC628AEAE
11.35.5
\begin{equation} \mathrm{d}V(E)<\delta \quad \Longrightarrow \quad \sup _{\nu \in \mathcal K_ L}\nu (E)<\epsilon \end{equation}
for every Borel set $E$. In particular, writing $\mathrm{d}\nu =h_\nu \mathrm{d}V$, the densities in this set are uniformly integrable in $L^1(\mathrm{d}V)$. On $\mathcal K_ L$, weak convergence of measures also implies weak convergence of their densities in $L^1(\mathrm{d}V)$.
Proof.
Weak compactness follows from lower semicontinuity and compactness of $\mathcal P(X)$. We may assume that the sublevel set is nonempty, so that $L\geq B_\varphi (0)$. Fix $t>0$. By theorem 11.33.5, there is $\delta _ t>0$ such that
\[ 0\leq \chi \leq 1,\quad \chi \in C(X),\quad \int _ X\chi \, \mathrm{d}V<\delta _ t \quad \Longrightarrow \quad B_\varphi (t\chi )>B_\varphi (0)-1. \]
Indeed, otherwise there would be a sequence of such functions $\chi $ tending to zero in $L^1$ with $B_\varphi (t\chi )\leq B_\varphi (0)-1$, contrary to that theorem. For $\nu \in \mathcal K_ L$, the defining supremum gives
\[ t\int _ X\chi \, \mathrm{d}\nu \leq L-B_\varphi (t\chi ) <L-B_\varphi (0)+1. \]
Choose $t$ so large that $(L-B_\varphi (0)+1)/t<\epsilon /2$. If $\mathrm{d}V(E)<\delta _ t/2$, outer regularity gives an open neighborhood $O$ of $E$ with $\mathrm{d}V(O)<\delta _ t$. For any compact $K\subset E$, choose a continuous cutoff $0\leq \chi \leq 1$ equal to one on $K$ and supported in $O$. The last inequality gives $\nu (K)<\epsilon /2$, uniformly in $\nu \in \mathcal K_ L$, and inner regularity yields $\nu (E)\leq \epsilon /2<\epsilon $. This proves (11.35.5).
In particular, every measure in the sublevel set is absolutely continuous, with a density of integral one. Together with the uniform small-set estimate, this is exactly uniform integrability on the finite measure space $(X,\mathrm{d}V)$.
Finally, suppose that $\nu _ j\to \nu $ weakly within the sublevel set. For a bounded measurable $G$, choose continuous $G_ k$ with $|G_ k|\leq \| G\| _\infty $ and $G_ k\to G$ in $\mathrm{d}V$-measure; such approximations exist by Lusin's theorem. Uniform absolute continuity gives
\[ \sup _{\sigma \in \mathcal K_ L}\int _ X|G_ k-G|\, \mathrm{d}\sigma \longrightarrow 0. \]
To see this, split $X$ into $\{ |G_ k-G|\leq a\} $ and its complement; the integral over the first set is at most $a$, and the integral over the second is at most $2\| G\| _\infty $ times a uniformly small measure. For fixed $k$, weak convergence applies to $G_ k$. Letting first $j\to \infty $ and then $k\to \infty $ proves convergence of the integrals against $G$, which is exactly weak $L^1(\mathrm{d}V)$ convergence of the densities. In particular, every sequence in the sublevel set has a subsequence whose densities converge weakly in $L^1$.
$\square$
Pome
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