A famous theorem says that if $f\geq 0$ is measurable, then there is an increasing sequence of simple function $(\varphi _n)$ s.t. $\varphi _n\nearrow f$. Then we can define $$\int_{\mathbb R}f:=\lim_{n\to \infty }\int_{\mathbb R}\varphi _n.$$
Now my question is : in these conditions, (i.e. $\geq 0$ and measurable), is there a decreasing sequence of simple function $(\psi_n)$ s.t. $\psi_n\searrow f$ ?
Attempts
I would say no because otherwise we could define $$\int f=\lim_{n\to \infty }\int_{\mathbb R}\psi_n,$$ and monotone convergence theorem (MCT) would work as well. But I know that MCT doesn't hold, so I guess this result is not correct. Does someone has a counter example ? And if not, does this result hold ?
The answer is no. But your explanation why makes no sense at all: The fact that there would be problems using $(\psi_n)$ to define $\int f$ does not show that $(\psi_n)$ does not exist. You say "no because otherwise[...], and monotone convergence theorem (MCT) would work as well. But I know that MCT doesn't hold," This is unclear. MCT does hold - I can prove it.
Hint: A simple function is bounded.