TITLE: Is the supremum of continuous functions integrable?
QUESTION [11 upvotes]: Let $f_\alpha$ be a family of continuous positive functions $\mathbb R\to \mathbb R$ 
where the index $\alpha$ runs in a compact metric space 
and the map $\alpha\to f_\alpha$ is continuous 
with respect to compact-open topology on the target. Suppose there is a uniform
upper bound on the integrals of $f_\alpha$'s over $\mathbb R$.
Question. Is $\underset{\alpha}{\sup} f_\alpha$ necessarily an integrable function?
Apology. This sure sounds like a homework level question, but after looking at for a while I am not even sure what the answer is.

REPLY [4 votes]: Just one more example: let $\phi \in C_c(\mathbb{R})$ with, say, $\sup_{\mathbb{R}} \phi = 1$.  For $\alpha \in [-\frac{\pi}{2}, \frac{\pi}{2}]$, set
$$f_\alpha(x) = \begin{cases} \phi(x - \tan \alpha), & \alpha \in (-\frac{\pi}{2}, \frac{\pi}{2}) \\\\ 0, &\alpha = \pm \frac{\pi}{2}. \end{cases}$$
Then $\int f_\alpha = \int \phi < \infty$ for each $\alpha \ne \pm \frac{\pi}{2}$, and $\int f_\alpha = 0$ otherwise, but $\sup_\alpha f_\alpha = 1$.