# Identity for a measurable set

Let $$f: \Omega\to [0,\infty[$$ be a simple function and $$(f_n)_{n\in \mathbb{N}}$$ a monotone increasing sequence of simple functions $$f_n: \Omega\to [0,\infty[$$ such that $$f\leq \lim\limits_{n\to \infty} f_n$$. Assume $$f\neq 0$$ and let are $$0<\alpha_1<\alpha_2<...<\alpha_m$$ the value's of $$f$$ and define the set's $$A:=\{x\in \Omega: f(x)>0\}\\A_n:=\{x\in A:f_n(x)\geq f(x)-\varepsilon\}$$ with $$0<\varepsilon<\alpha_1$$.

Then $$A=\bigcup\limits_{n=0}^{\infty} A_n.$$ The inclusion $$\supseteq$$ is clear for me because this follows by the construction of the set's $$A_n$$. But how can I prove the other inclusion $$\subseteq$$? Let be $$x\in A$$.I want to show there exist's an index $$n\in \mathbb{N}$$ such that $$f_n(x)\geq f(x)-\varepsilon$$.

• I have a hard time understanding what $\alpha_i$ is. What is the definition of $\alpha_i$? Commented Nov 23, 2022 at 15:47
• I defined it in my describtion. They are the values of $f$. Commented Nov 23, 2022 at 15:48
• Okay. Since you write $f\not=0$, why not just let $f$ map into $(0,\infty)$? Commented Nov 23, 2022 at 15:55
• Because $f\neq 0$ doesn't imply $f(x)=0$ for all $x\in \Omega$. Commented Nov 23, 2022 at 15:58

Suppose $$x\in A$$. Then by definition of $$A$$, $$f(x)>0$$. We thus have that $$0 < \epsilon < \alpha_1 \leq f(x)\leq \lim_{n\to\infty}f_n(x).$$ If the limit is $$\infty$$ then $$x\in\cup_nA_n$$ is trivial. Supposing then that it is finite, the inequalities above imply that
$$\lim_{n\to\infty} [f_n(x) - f(x)] \geq 0.\tag{1}\label{eq1}$$
Moreover, for any integer $$n$$, we have the equivalence
$$f_n(x) \geq f(x) - \epsilon \iff f_n(x)-f(x) \geq -\epsilon\tag{2}\label{eq2}.$$ From \eqref{eq1} we can infer that there is some $$n_0$$ satisfying the RHS of \eqref{eq2}. Hence $$x\in A_{n_0}$$, implying the other inclusion.