Weak*-continuity on set of probability measures

Question

Let $\mathcal{P}$ be a set of Borel probability measures on $[0, 1]$ equipped with the weak-topology as described here: What is the weak*-topology on a set of probability measures?. Let $J:[0, 1]\to \mathbb{R}_{+}$ be a continuous function and consider the mapping $$\Lambda:\mathcal{P} \to \mathbb{R}_{+}$$ with $$\Lambda:\nu\mapsto \int_{0}^{1}{J(\nu([0, t]))dt}.$$ Why is it true that $\Lambda$ is weak-continuous? I guess one has to show that if $\nu_{n}\to \nu$ in the total variation norm implies that $\Lambda(\nu_{n})\to \Lambda(\nu)$. Is that correct and how can one show this?

My second question: Consider for $\rho\in [0, 1]$, $$\mathcal{P}(\rho)=\left\{\nu\in \mathcal{P}:\int_{0}^{1}{td\nu(t)}=\rho\right\}.$$ Is this set convex and compact for the weak*-topology? The convexity follows as for $\nu_{1}, \nu_{2}\in \mathcal{P}(\rho)$ and $\alpha\in [0, 1]$ we have $$\int_{0}^{1}{td((1-\alpha)\nu_{1}+\alpha \nu_{2})(t)}=(1-\alpha)\int_{0}^{1}{td\nu_{1}(t)}+\alpha \int_{0}^{1}{td\nu_{2}(t)}=\rho.$$ Is this correct and how to show the compactness?

Answer 1

If $\nu_n \to \nu$ in weak* topology then $\nu_n [0,t]\to \nu [0,t] $ for all but countably many values of $t$. Hence, $J(\nu_n [0,t])\to J(\nu [0,t]) $ for all but countably many values of $t$. This implies, by Bounbded Convergece Theorem, that $\Lambda (\nu_n) \to \Lambda (\nu)$.

For the second part you need the fact any sequence of probability measures on $[0,1]$ has a subsequence which converges in weak* topolpogy. So if $\int_0^{1}td\nu_n(t)=\rho$ then we can find a subsequence $\nu_{n_k}$ converging to some $\nu$ and this implies $\int_0^{1}td\nu(t)=\rho$. This proves that $\mathcal P(\rho)$ is relatively compart. I wiil let you verfiy that it is closed is the compact space of probabilty measures on $[0,1]$ so it is compact.