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.



Is the supremum of continuous functions integrable?
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.


