Let $(X,\mathcal{A},\mu)$ be a probability space, let $(u_n)_{n\geq 1}\subseteq \mathcal{M}^+(\mathcal{A})$ and $u\in \mathcal{M}^+(\mathcal{A})$. Suppose that $u_n\to u$ uniformly. The problem is to show that $\lim_{n\to\infty}\int_X u_n \,d\mu=\int_X u\, d\mu$. Since $v_n\leq ||v_n||:=\sup_{x\in A} |v_n(x)|$, we get $|\int_X u_n \,d\mu|\leq \int_X |u_n| \,d\mu\leq ||v_n||$, because $\mu$ is a probability. Should I use Lebesgue Dominated Convergence here?
Consider the measure space $(\mathbb{R},\mathcal{B}(\mathbb{R}),\lambda)$. I need to find a sequence $(v_n)_{n\geq 1}\subseteq \mathcal{M}^+(\mathcal{B}(\mathbb{R}))$ such that $v_n\to 0$ uniformly, but $\lim_{n\to\infty}\int_\mathbb{R} v_n \,d\lambda\neq 0$.
The notations used here are from Schilling's "Measures, Integrals and Martingales".
Edit: $\mathcal{M}^+(\mathcal{A})$ stands for the family of the $\mathcal{A}$-measureable positive real functions. $\mathcal{B}(\mathbb{R})$ is the Borel sets in $\mathbb{R}$.
Let $\varepsilon>0$.
If $u_n\to u$ uniformly, then, there exists an $n_0$, such that $$ n\ge n_0\qquad\Longrightarrow\qquad \sup_{x\in X}|u_n(x)-u(x)|<\varepsilon. $$ Hence, if $n\ge n_0$, then $$ \Big|\int_X u_n\,d\mu-\int_X u\,d\mu\,\Big|\le \int_X \big|u_n-u\big|\,d\mu\le \varepsilon \int_Xd\mu =\varepsilon, $$ and hence $$ \lim_{n\to\infty}\int_X u_n\,d\mu=\int_X u\,d\mu. $$ In case of the Lebesgue measure on $\mathbb R$, consider, $$ u_n=\frac{1}{n}\chi_{[0,n]}. $$ Then $u_n\to 0$,uniformly, but $\int_{\mathbb R}\chi_n\,d\lambda=1$.