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Let $E$ be a measurable subset of $\mathbb{R}$ and let $F_n :E \to \mathbb{ R} $ be a sequence of measurable functions. Let $$E^*= \{ x\in E : F_n \text{ is convergent } \}$$ Is the set $E^*$ measurable?

To solve this, I need to understand the relation between the set $E^*$ and the functions $F_n$. They are strongly related, but how can I express this relation in terms of sets and measure?

Safeyya Malkawy
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1 Answers1

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I preassume that $E^{*}:=\{x\in E: (F_n(x))_n\text{ is a convergent sequence}\}$ and mention that $\mathbb N$ denotes the set of positive integers in this answer.

Then for $x\in E$:

$$x\in E^{*}\iff\forall k\in\mathbb{N}\exists n\in\mathbb{N}\forall r,s\in\mathbb{N}\left[r,s\geq n\implies\left|F_{r}\left(x\right)-F_{s}\left(x\right)\right|<\frac{1}{k}\right]$$ So if $\mathbb N_n$ denotes $\{n,n+1,\dots\}$ then:

$$E^{*}=\bigcap_{k\in\mathbb{N}}\bigcup_{n\in\mathbb{N}}\bigcap_{\left(r,s\right)\in\mathbb{N_n}^{2}}\left\{ x\in E:\left|F_{r}\left(x\right)-F_{s}\left(x\right)\right|<\frac{1}{k}\right\}$$

The sets $\left\{ x\in E:\left|F_{r}\left(x\right)-F_{s}\left(x\right)\right|<\frac{1}{k}\right\} $ are measurable and we are dealing with countable unions and intersections.

So $E^{*}$ is measurable.

drhab
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  • but what is that third intersection .. I did not get it .. – Safeyya Malkawy Nov 06 '15 at 14:34
  • What intersection? The one with index $(r,s)$? – drhab Nov 06 '15 at 14:43
  • I guess it's a measurable set but is it the same sequence of functions my sequence is Fn(x) what is (Fn(x))n – Safeyya Malkawy Nov 06 '15 at 14:49
  • You are speaking of a sequence, but $F_n(x)$ on its own is no sequence. It is nothing more than one value of function $F_n$. That's why I preassumed that you were taking about the sequence $F_1(x),F_2(x)\dots$ (this for every $x$) wich can be denoted as $(F_n(x))_n$. Also in your title you are calling (wrongly) $F_n$ a sequence of measurable functions. The right notation for a sequence of functions is $(F_n)_n$. – drhab Nov 06 '15 at 15:03
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    @drhab May i ask why are the sets ${x\in E: |F_{r}(x) - F_{s}(x)| < \frac{1}{k} }$ measurable? – teo theo Nov 13 '15 at 14:43
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    @teotheo If $f$ and $g$ are measurable functions then also functions like $f+g$ and $f-g$ and $|f|$ are measurable (so also $|f-g|$, so that the sets that you mention are measurable). If you wonder why then you might ask a question about that (eventually with link to this question). It goes too far to treat this in a comment. – drhab Nov 13 '15 at 17:50
  • @drhab Your appealing to Cauchy sequences is reasonable, but I don't understand why you let $n$ run over all of $\mathbb{N}$ in the union. – Boar Jan 09 '22 at 11:11
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    @Steve Working out the RHS of the equivalence that starts with $\forall k\exists n\dots$ we get automatically $\bigcap_{k\in\mathbb N}\bigcup_{n\in\mathbb N}\dots$. Do you have an alternative? – drhab Jan 09 '22 at 11:52