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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 ?

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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.

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  • $\begingroup$ What do you mean by MCT does hold ? That if $\psi_n\searrow f$ then $\lim_{n\to \infty }\int \psi_n=\int f$ ? $\endgroup$
    – user649261
    Mar 3 '19 at 9:28
  • $\begingroup$ No of course I don't mean that. MCT is a theorem, and that thing you suggest is false. $\endgroup$ Mar 3 '19 at 14:56

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