Say that for each $i=1,\dotsc,m$ we have a function $f_i :(0,1) \to (0,\infty)$ that is continuous, non-increasing, and such that $\int_0^1 f_i(t) dt = \infty$. If we define the pointwise minimum function $f :(0,1) \to (0,\infty)$ via $f(t) = \min_{1\le i \le m}f_i(t)$, then it's easy to show that $f$ is continuous and non-increasing. But do we know that $f$ is also not integrable, i.e. $\int_0^1 f(t) dt =\infty$?
If we drop the monotonicity assumption, this is clearly false, so if it's going to hold then it'll rely crucially on monotonicity. If any one function is minimal in a neighborhood of the origin, then the result is trivially true, so the key is handling the possibility of the minimal functions switching infinitely often. I've tried to build counterexamples with two functions switching places infinitely often, but to no avail.
If anyone has any ideas on how to prove OR disprove this, I would appreciate it. Thanks!
For convenience I'll work on $[1,\infty)$ first. Think about a sequence $a_n$ with $a_1 = 1$ and $a_1 < a_2 < \cdots $ that blasts off to $\infty$ fast. Then define
$$f = \frac{1}{a_2 - a_1}\chi_{[a_1,a_2)} + \frac{1}{a_4 - a_2}\chi_{[a_2,a_4)} + \frac{1}{a_6 - a_4}\chi_{[a_4,a_6)} + \cdots,\,\, \text { and }$$
$$g = \frac{1}{a_3 - a_1}\chi_{[a_1,a_3)} + \frac{1}{a_5 - a_3}\chi_{[a_3,a_5)} + \frac{1}{a_7 - a_5}\chi_{[a_5,a_7)} + \cdots$$
So $f,g$ are step functions with infinitely many steps. Verify that both $\int_0^\infty f, \int_0^\infty g = \infty.$
Are these functions decreasing? They are if we make sure
$$\tag 1\frac{1}{a_2 - a_1} > \frac{1}{a_3 - a_1}> \frac{1}{a_4 - a_2} > \frac{1}{a_5 - a_3} > \cdots.$$
Next note that
$$\min(f,g)= \frac{1}{a_3 - a_1}\chi_{[a_1,a_2)} + \frac{1}{a_4 - a_2}\chi_{[a_2,a_3)} + \frac{1}{a_5 - a_3}\chi_{[a_3,a_4)}+ \cdots $$
Therefore
$$\tag 2 \int_1^\infty \min(f,g) = \frac{a_2-a_1}{a_3 - a_1} + \frac{a_3-a_2}{a_4 - a_2} + \frac{a_4-a_3}{a_5 - a_3} + \cdots$$
Thus we will have a counterexample if $(1)$ holds and $(2) < \infty.$ So if we choose $a_n$ so that $(1)$ holds and also so that $(a_{n+1}-a_n)/(a_{n+2} - a_n) \le 1/n^2,$ we'll be done. Can we do this? Sure, induction works well here. An explicit sequence that works is $a_n = n^{4n}.$
Now the $f,g$ constructed are not continuous, but this is a minor issue. We can find piecewise linear decreasing $F,G$ such that $0\le F \le f, 0\le G \le g$ and $\int_1^\infty F = \infty = \int_1^\infty G.$ So $F,G$ is a counterexample on $[1,\infty).$ Finally, for an example on $(0,1]$ we can make a change of variables and use $F(1/x)/x^2, G(1/x)/x^2.$