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Suppose $(X,\mathcal{A},\mu)$ is a measure space. Let $\lbrace f_n\rbrace_{n\ge 1}$ and $f$ be extended real-valued $\mathcal{A}$-measurable functions on some set $A\in \mathcal{A}$ such that $\lim_{n\to \infty}f_n=f$ on $A$. Then for $\alpha\in \mathbb{R}$, $$\mu\lbrace x\in A:f(x)>\alpha \rbrace\le \liminf_{n\to \infty}\mu \lbrace x\in A:f_n(x)\ge \alpha \rbrace\tag{$1$}$$

Proof Attempt: (Note: I have yet to take a measure of a sequence of sets that was neither nonascending or nondescending).

Step 1: Since $f_n\to f$, we have $$(*)=\mu \lbrace x\in A:f(x)>\alpha \rbrace =\mu\Bigg( \bigcap_{k\ge 1}\bigcup_{N\ge 1}\bigcap_{n\ge N}\bigg\lbrace x\in A: f_n(x)\ge \alpha + \frac{1}{k} \bigg\rbrace\Bigg)\tag{$2$}$$ This is seen by noting that $f(x)>\alpha$ iff for each $k$ there is an $N$ such that for all $n\ge N$ we have $f_n \ge \alpha + \frac{1}{k}$.

Step 2: The sets we are intersecting over $k$ in the RHS of (2) are increasing with respect to $k$ since if $f_n \ge \alpha + \frac{1}{k}$ then $f_n\ge \alpha + \frac{1}{k+1}$, so we cannot apply continuity of measure, but I suppose we can say $$(*)\le \liminf_{k\to \infty}\mu \Bigg( \bigcup_{N\ge 1}\bigcap_{n\ge N}\bigg\lbrace x\in A: f_n(x)\ge \alpha + \frac{1}{k}\bigg\rbrace \Bigg)=(**)\tag{$3$}$$

Step 3: The sets we are taking the union of in the RHS of (3) are increasing with respect to $N$ since if $f_n \ge \alpha + \frac{1}{k}$ for $n\ge N$, then this also holds for $n\ge N+1$. Hence by continuity of measure we have $$(**)= \liminf_{k\to \infty}\lim_{N\to \infty}\mu \bigg( \bigcap_{n\ge N} \bigg\lbrace x\in A: f_n(x) \ge \alpha + \frac{1}{k}\bigg\rbrace\bigg)\tag{$4$}$$

Step 4: I am not sure whether the intersection in the RHS of (4) is descending, so I assume I can conclude that $$(**)\le \liminf_{k\to \infty}\lim_{N\to \infty}\liminf_{n\to \infty}\mu \bigg\lbrace x\in A: f_n(x)\ge \alpha + \frac{1}{k}\bigg\rbrace\tag{$5$}$$

Since $N$ is not in (5), we can take this a step farther and say $$(**)\le \liminf_{k\to \infty}\liminf_{n\to \infty}\mu \bigg\lbrace x\in A: f_n(x)\ge \alpha + \frac{1}{k}\bigg\rbrace\tag{$6$}$$

Can I just naively pass the $\liminf_{k\to \infty}$ through the measure in the RHS of (6) to conclude that the RHS of (6) is $(\le )$ the RHS of (1)?

Update: All answers are gratefully accepted, but I would like to know in particular if there's a way to get from the RHS of (6) to the RHS of (1).

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  • $\begingroup$ I've made an edit to address the update. I hope it helps. $\endgroup$ – kobe Feb 9 '15 at 16:45
  • $\begingroup$ Nice. Thanks for the alternative solution as well. $\endgroup$ – The Substitute Feb 10 '15 at 0:36
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You can't interchange limits like that without some justification. I suggest going this route: show that if $A_n = \{x : f_n(x) \ge \alpha\}$ and $A = \{x : f(x) \le \alpha\}$, then $A\subseteq \liminf_n A_n$. Then deduce by Fatou's lemma that $\mu(A) \le \liminf_n \mu(A_n)$.

Let $x\in A$. Since $f(x) > \alpha$ and $f_n(x)\to f(x)$, $f_n(x) > \alpha$ for all but finitely many $n$. Thus $x\in \liminf_n A_n$. Therefore, $A\subseteq \liminf A_n$. Consequently, \begin{align}\mu(A) &\le \mu(\liminf A_n) = \int 1_{\liminf_n A_n}\, d\mu\\ &= \int \liminf_n 1_{A_n}\,d\mu\\ &\le \liminf_n \int 1_{A_n}\, d\mu \quad (\text{by Fatou's lemma})\\ &= \liminf_n \mu(A_n).\end{align}


In response to the update, use the fact that for all $k$ and $n$, $$\left\{x : f_n(x) > \alpha + \frac{1}{k}\right\} \subseteq \left\{x : f_n(x) \ge \alpha\right\}.$$ Then you can claim that for all $k$ and $n$, $$\mu\left\{x : f_n(x) > \alpha + \frac{1}{k}\right\} \le \mu \{x : f_n(x) \ge \alpha\}.$$ This implies that for all $k$, $$\liminf_n \mu\left\{x : f_n(x) > \alpha + \frac{1}{k}\right\} \le \liminf_n \mu\{x : f_n(x) \ge \alpha\},$$ and consequently $$\liminf_k\liminf_n \mu\left\{x : f_n(x) > \alpha + \frac{1}{k}\right\} \le \liminf_n \mu\{x : f_n(x) \ge \alpha\}.$$

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