Let $\Omega$ be a bounded nonempty domain, assume a family of real valued continous functions on $\Omega$ s.t. $u_{n_k}\rightarrow u$ uniformly on $\Omega$. If $u_{n_k}$ are Lebesgue integrable,
Can we say that $(\int_\Omega u_{n_k})^2 \rightarrow (\int_\Omega u)^2 $ as $k\rightarrow \infty $？
Since $\Omega$ is bounded the Lebesgue measure of omega is finite $\mu (\Omega )<\infty .$ From the uniform convergence follows that for any $\varepsilon > 0$ there exists $k_0 $ such that $$\sup_{t\in\Omega } |u_{n_k } (t) -u(t)|\leq \frac{\varepsilon}{\mu (\Omega )} $$ for $k\geq k_0 $,thus $$\left|\int_{\Omega } u_{n_k } (t) \mu (dt ) -\int_{\Omega } u (t) \mu (dt )\right|\leq \int_{\Omega } |u_{n_k } (t) -u (t) |\mu (dt )\leq \int_{\Omega } \sup_{t\in\Omega } |u_{n_k } (t) -u (t) |\mu (dt ) =\frac{\varepsilon}{\mu (\Omega )} \cdot \mu (\Omega ) =\varepsilon $$ therefore $$a_k =\int_{\Omega } u_{n_k } (t) \mu (dt ) \to \int_{\Omega } u (t) \mu (dt ) =a $$ The function $f(x) = x^2 $ is continuous therefofe $$\left(\int_{\Omega } u_{n_k } (t) \mu (dt )\right)^2 =f(a_k ) \to f(a) =\left(\int_{\Omega } u (t) \mu (dt )\right)^2$$