Proposition 11.39.580C599ADD5CB

Proof. Write $\delta _ j=\Delta (\varphi _ j,\varphi )$. The bounds $|B_{\varphi _ j}-B_\varphi |\leq \delta _ j$ and $|H_{\varphi _ j}-H_\varphi |\leq \delta _ j$, together with the identity $H_{\varphi _ j}(\nu _ j)=B_{\varphi _ j}(G_ j)+\int _ XG_ j\, \mathrm{d}\nu _ j$, give

If $\| G_ j\| _\infty \leq M$, then $H_\varphi (\nu _ j)\leq B_\varphi (0)+2M+2\delta _ j$, so all the measures lie in one finite sublevel set of $H_\varphi $. By theorem 11.35.4, their densities are uniformly integrable, and a weakly convergent subsequence also converges weakly in $L^1$.

Let $\nu $ be the limit of such a subsequence. Uniform integrability, the common bound and convergence in measure give $\int _ X|G_ j-G|\, \mathrm{d}\nu _ j\to 0$; weak $L^1$ convergence applies to the bounded function $G$, and theorem 11.33.5 gives $B_\varphi (G_ j)\to B_\varphi (G)$. Lower semicontinuity of $H_\varphi $ in the last inequality yields

The minimum formula for bounded measurable densities in theorem 11.36.1 gives the reverse inequality, and equality means exactly that $\nu \in \mathcal A_\varphi (G)$. When this set is a singleton, compactness gives convergence of the whole sequence. $\square$