$
\newcommand{\set}[1]{\left\{ #1 \right\}}
\newcommand{\abs}[1]{\left \lvert #1 \right \rvert}
\newcommand{\N}{\mathbb{N}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\ve}{\varepsilon}
$The Problem: Suppose we have a sequence $\set{f_k}_{k \in \N}$ of functions such that, $\forall k \in \N$,
- $f_k : E \subseteq \R^n \to \R$ with $E$ (Lebesgue) measurable, and of finite measure (i.e. $\mu(E) < +\infty)$
- $f_k$ is (Lebesgue) measurable
- There exist constants $c_x$ such that $\abs{f_k(x)} \le c_x < +\infty$ $\forall k \in \N$ and $\forall x \in E$
On this premise, we want to show that, $\forall \varepsilon > 0$, $\exists F \subseteq E$ a closed subset such that $\mu(E \setminus F) < \varepsilon$, and $\exists c \in \R$ such that $\abs{f_k(x)} \le c$ $\forall k \in \N$ and $\forall x \in F$.
Thoughts & Attempts:
This feels like a sort of "uniform boundedness" thing, in the sense that, even those there's a finite bound for all of the $f_k(x)$, there is one finite one that works for them all. Though obviously that's an intuitive notion, and searching brings up seemingly unrelated results.
That said, I have some ideas floating around, but nothing feels quite concrete in terms of piecing them together, or finding the desired conclusion. For instance, some of the thoughts going around:
-
Of course, $\forall \ve > 0$, we can easily get an $F \subseteq E$ which is closed and $\mu(E \setminus F) < \ve$. (This is one characterization of measurable sets after all.) But this path on its own seems a little fruitless and basic, and doesn't lead anywhere for me.
-
Since $\set{f_k(x)}_{k \in \N}$ is bounded by $c_x$, then $\exists \set{f_{k_\ell}}_{\ell \in \N} \subseteq \set{f_k}_{k\in\N}$ such that $f_{k_\ell}(x)$ converges as $\ell \to \infty$. (This follows from Bolzano-Weierstrass.) We could use this to define some function $f$ as a subsequential limit of $\set{f_k}_{k \in \N}$. (Perhaps $f_{k_\ell}$ could converge a.e. to $f$?)
-
This seems like there would be a chance to apply Egorov's theorem. On the assumption the previous holds, we could find, given $\ve > 0$, $F \subseteq E$ closed where $\mu(E \setminus F) < \ve$, and $f_{k_\ell} \to f$ uniformly on this closed set $F$, all by Egorov.
For reference's sake, here's Egorov's theorem, (mostly) as given in my text (Theorem $4.17$ in Measure & Integral: An Introduction to Real Analysis by Wheeden & Zygmund):
Egorov's Theorem: Suppose that $\set{f_k}_{k \in \N}$ is a sequence of measurable functions converging (Lebesgue-)a.e. in a finite-measure set $E$, to a finite limit $f$. Then $\forall \ve > 0$, $\exists F \subseteq E$ closed such that $\mu(E \setminus F) < \ve$ and $\set{f_k}_{k \in \N}$ converges uniformly to $f$ on $F$.
- The question becomes, however, how to choose the necessary constant $c$? The naive approach would just be to have $c := \sup_{x \in F} c_x$, but would that always be finite? I doubt it. But then the question becomes: what to choose, then? I imagine this has something to do with using the closedness of $F$ (and maybe that would ensure the previously-mentioned $c$ is finite), but how?
Can anyone give me some ideas as to where to go? Be that in terms of fleshing out my ideas and clarifying my problem spots, or putting me on the right path?
Best Answer
$H(x)=\sup_k f_k(x)$ is a measurable function; it is finite a.e. in $E$ by hypothesis. The sets $E_M:=\{x \in E \,:\, |H(x)| \le M \}$ are increasing for $M=1,2,\ldots$, and their union over all positive integer $M$ has full measure in $E$. Thus by continuity from below, given $\epsilon$ there exists $M$ such that $\mu(E_M)>\mu(E)-\epsilon/2$. By inner regularity of Lebesgue measure, there exists a closed subset $F \subset E_M$ with $\mu(F)>\mu(E_M)-\epsilon/2$.