Let $(f_n: \Omega \subseteq \mathbb{R}^n \to \mathbb{R})_{n \in \mathbb{N}}$ be a sequence of continuous functions converging uniformly to $f$, with $\Omega$ measurable. I'd like to prove that in this case $f$ is integrable and $\lim_{n \to \infty} \int_{\Omega} f_n = \int_{\Omega} f$.
If I could show that for any $\epsilon >0$ we must have $\sup f \leq \sup f_n +\epsilon$ and $\inf f \geq \inf f_n – \epsilon$, I'd be done (since for any partition $P$ we'd have $L(f_n,P) -\epsilon K\leq L(f,P) \leq U(f,P) \leq U(f_n,p)+\epsilon K$ for all positive $\epsilon$, where $K$ is constant.
But how can I ensure that?
Best Answer
It is true only when $\Omega$ has finite measure. If you take $\Omega=\mathbb{R}$,$f\left(x\right)=\frac{1}{\left|x\right|+1},f_{n}\left(x\right)=\begin{cases} f\left(x\right) & \left|x\right|<n\\ \frac{\left(n+1-x\right)}{\left|x\right|+1} & n\leq\left|x\right|<n+1\\ 0 & n+1\le x \end{cases} $, then $f_{n} $ converges uniformly to $f $, although $f $ is not integrable.
If $\Omega$ has finite measure. It is easy to sea that $f$ is measurable. We only need to show that for each $\epsilon$ there is some $N_0$ such that for each $n>N_0$ we have that $\int |f-f_n|<\epsilon $. Take $N_0$such that for all $n>N_0$ we have $\sup |f-f_n|<\epsilon/\mu (\Omega)$ then $\int |f-f_n|\leq \int\epsilon/\Omega=\epsilon $