It follows straightforwardly from proposition 1.6.1 that ${\mathcal{F}}_{0}$ is a relatively compact subset of $PSH(X,\omega)$. Moreover ${\mathcal{F}}_{0}$ is closed by Hartogs lemma (1.6.2).

Let $(\varphi_{j})\in{\mathcal{F}}_{\mu}^{\mathbb{N}}$. Then $\psi_{j}:=\varphi_{j}-\sup_{X}\varphi_{j}\in{\mathcal{F}}_{0}$ which is relatively compact. Assume first $\mu$ is smooth. Then $(\int_{X}\psi_{j}d\mu)$ is bounded: this is because if $\psi_{j_{k}}\rightarrow\psi$ in $L^{1}(X)$ then $\psi_{j_{k}}\mu\rightarrow\psi\mu$ in the weak sense of (negative) measures hence $\int_{X}\psi_{j_{k}}d\mu\rightarrow\int_{X}\psi d\mu>-\infty$. Now $\int_{X}\psi_{j}d\mu=\int_{X}\varphi_{j}d\mu-\int_{X}\sup_{X}\varphi_{j}d\mu=-\sup_{X}\varphi_{j}$ thus $(\sup_{X}\varphi_{j})$ is bounded and we can apply the previous proposition to conclude that $(\varphi_{j})$ is relatively compact ( it cannot converge uniformly to $-\infty$ since $\int_{X}\varphi_{j}d\mu=0$).

When $\mu$ is not smooth, it only remains to prove that $(\int_{X}\psi_{j}d\mu)$ is bounded. Assume on the contrary that $\int_{X}\psi_{j}d\mu\rightarrow-\infty$. Extracting a subsequence if necessary we can assume $\int_{X}\psi_{j}d\mu\leq-2^{j}$. Set $\psi=\sum_{j\geq 1}2^{-j}\psi_{j}$. This is a decreasing sequence of $\omega$-psh functions, hence $\psi\in PSH(X,\omega)$ or $\psi\equiv-\infty$. Now it follows from the previous discussion that $\int_{X}\psi_{j}dV\geq-C$ if $dV$ denotes some smooth probability measure on $X$. Thus $\int_{X}\psi dV>-\infty$ hence $\psi\in PSH(X,\omega)$. We obtain a contradiction since by the Monotone convergence theorem, $\int_{X}\psi d\mu=\sum_{j\geq 1}2^{-j}\int_{X}\psi_{j}d\mu=-\infty$. ∎