Is there a relation between Bounded convergence theorem and convergence in distribution ?
More specifically, if we have $g \geq$ 0 continuous. and $X_n \to X_{\infty} $ in distribution, can we directly write :
\begin{align} E(g(X_{\infty})) &= \int_{t=0} ^{\infty} P(g(X_{\infty}) \geq t) dt \\ & \leq \int_{t=0} ^{\infty} \liminf P(g(X_n) \geq t) dt \\ &=^{BCT} \liminf \int_{t=0} ^{\infty} P(g(X_n) \geq t) dt \\ &= E(g(X_{n}))? \end{align}
It is indeed correct that $\mathbb E[g(X)]\leqslant \liminf_{n\to \infty} \mathbb E[g(X_n)]$.
In the last step, a $\liminf$ is missing. The bounded convergence theorem is actually Fatou's lemma, and the equal sign has to be replaced by $\leqslant$. And you have to justify the fact that the sequence $(g(X_n))_{n\geqslant 1} $ converges in distribution to $g(X)$.