Is the supremum of continuous functions integrable? 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.
 A: 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 $\sup_\alpha f_\alpha = 1$.
A: No. Let the compact metric space be $[0,1]$ with the standard topology and define
$$
f_\alpha(x) =\alpha\max(1-\alpha \vert x\vert,0)
$$
for $\alpha\in[0,1]$. This satisfies the properties asked for, with the upper bound $\int f_\alpha(x)\\,dx\le1$ (and, equality for $\alpha\not=0$).
But,
$$
\sup\_\alpha f_\alpha(x)=1_{\{\vert x\vert\ge1/2\}}\frac{1}{4\vert x\vert}+1_{\{\vert x\vert < 1/2\}}(1-\vert x\vert)
$$
is not integrable.
A: As you stated, no: take e.g. the unit closed interval $[0,1]$ as a compact metric space and define for all $\alpha\in [0,1]$ the function $f_\alpha(x)= \alpha (1-\alpha|x|) _ +$, that depends continuously on $\alpha$ even in the uniform norm on $\mathbb{R}$. Then for $|x|\ge 1$ we have
$ \sup_ {\alpha\in[0,1]} f_ \alpha (x)=1/4x $, which is not integrable.
