Let $\{f_n\}_n$ a sequence of function on $X\ne\emptyset$ such that $f_n\ge 0$ for all $n\in\mathbb{N}$. We suppose that \begin{equation} \begin{split} 1)&\quad f_n\le f_{n+1}\\ 2)&\quad f_n\;\text{converges pointwise to the function}\;f\;\text{in}\; X. \end{split} \end{equation} We consider \begin{equation} E_n=\{x:f_n(x)\ge\alpha s(x)\}, \end{equation} where $\alpha\in (0,1)$ and $s\colon X\to\mathbb{R_+}$ such that $s\le f$ in $X$.

It is easy to verify that $E_{n}\subseteq E_{n+1}$ and that $X=\bigcup_{n}E_n$.

The question is: $E_n\ne\emptyset$ for all $n\in\mathbb{N}$ or can some $E_n$ be empty?



$E_1= \{x \in X: f_1(x)>\alpha s(x) \}$, for example, could be empty. The point is that there is no condition on $f_1$ except being non-negative. Therefore, we can just have $f_1=0$. In that case, $E_1 = \emptyset$. The same could be true for any finite number of $E_n$. You only know that infinitely many $E_n$ must be non-empty because $f_n(x)$ converges to something larger than $\alpha s(x)$ in the end.

(I read the question so that $\alpha$ and $s$ are fixed before starting, because that made more sense to me.)

  • $\begingroup$ @ Luke Thanks for your answer. Since $s\colon X\to\mathbb{R_+}=[0,+\infty)$, could it not happen that when $f_1=0$, $E_1\ne\emptyset$? $\endgroup$ – Jack J. Jun 15 '18 at 13:46
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    $\begingroup$ $\mathbb{R}_+$ usually means only strictly positive numbers without $0$. But if it may be zero, that does not change the argument so much because it is still very possible that $s(x)>0 \ \forall x$. Depends on the concrete choice of s. $\endgroup$ – Luke Jun 15 '18 at 15:12

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