Let $(X,\mathcal{A},\mu)$ be a measureable space. Let $u\in \mathcal{M}_{\mathbb{R}}^{+}(\mathcal{A})$ and $\lbrace u_{j}\rbrace_{j\geq 1}$ be a sequence of functions in $\mathcal{M}_{\mathbb{R}}^{+}(\mathcal{A})$ that satisfies $u_{j}\to u$ and $\lim_{j\to\infty}\int_{X}u_{j}\, \mathrm{d}\mu=4$. Note that the convergence is pointwise convergence i.e. $\forall x\in X:u_{j}(x)\to u(x)$.
How do I show that $\int_{X}u\, \mathrm{d}\mu\leq 4$? All I know is
$$\int_{X}u\, \mathrm{d}\mu=\lim_{j\to\infty}\int_{X}u_{j}\, \mathrm{d}\mu=\sup_{j\in \mathbb{N}}\int_{X}u_{j}\, \mathrm{d}\mu$$
I want to give an example that the abovementioned condition is satisfied and where $\int_{X}u\, \mathrm{d}\mu=1$.
This is a conseqeunce of Fatou's Lemma:
$$ \int_X \liminf_{n\to\infty}u_n\,d\mu\le\liminf_{n\to\infty}\int_X u_n\,d\mu, $$ and hence $$ \int_X u\,d\mu=\int_X \liminf_{n\to\infty}u_n\,d\mu\le\liminf_{n\to\infty}\int_X u_n\,d\mu= \lim_{n\to\infty}\int_X u_n\,d\mu=4. $$