Are Pointwise Convergent Functions Limited to Measurable Sets?

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Discussion Overview

The discussion revolves around the properties of pointwise convergent functions and their implications for measurable sets. Participants explore whether the set where a pointwise limit function is less than or equal to a constant can be expressed in terms of the sets defined by the approximating functions. Additionally, there is a question regarding the measurability of certain sets derived from measurable sets.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions if the set where the limit function \( f \) is less than or equal to \( a \) can be expressed as the limit inferior of the sets where the approximating functions \( f_n \) are less than or equal to \( a \).
  • Another participant provides a counterexample using the functions \( f_n(x) = \frac{1}{n}x^2 \) and \( f(x) = 0 \) to illustrate that the proposed equality does not hold in general.
  • There is a suggestion that the inclusion \( \{f \leq a\} \supseteq \liminf \{f_n \leq a\} \) is valid, but the exact equality is disputed.
  • Regarding the second question, one participant asserts that the set derived from a measurable set by subtracting a constant is measurable, referencing the Borel-sigma-algebra.
  • Another participant expresses confusion about the initial claim and seeks clarification on their reasoning, particularly concerning the dependence of \( n \) on \( \varepsilon \) in their argument.
  • A later reply emphasizes that the dependence of \( n \) on \( \varepsilon \) undermines the validity of the argument that \( \varepsilon \) being arbitrary leads to a conclusion.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the first question regarding the equality of sets, as a counterexample is provided that challenges the initial claim. There is agreement on the second question about the measurability of the derived set.

Contextual Notes

The discussion highlights the complexity of pointwise convergence and its implications for set theory, particularly in the context of measure theory. The counterexample provided raises questions about the conditions under which the proposed relationships hold.

tohauz
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Hi,
1)is this true? If f_n(x) -> f(x) pointwise, then
{x:f(x)<=a} = union{k=1^infty}intersection{n=k^infty}{x:f_n(x)<=a}.
2)if A is measurable set, subset of reals, then is A-const set measurable?
Thanks
 
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I'm not getting your question. Could you make it more clear.
 
For the first, you want to know if

[tex]\{f\leq a\}=\liminf\{f_n\leq a\}[/tex]

In other words, you want to know whether [itex]f\leq a[/itex] if and only if [itex]f_n\leq a[/itex] eventually.

This is false, take [itex]f_n(x)=\frac{1}{n}x^2[/itex], then f(x)=0 and take a=0.
Then [itex]\{f\leq 0\}=\mathbb{R}[/itex], but [itex]\liminf\{f_n\leq 0\}=\{0\}[/itex]

As you do notice, you do have the [itex]\supseteq[/itex] inclusion.

(b) is true. Let [itex]\mathbb{B}[/itex] be the Borel-sigma-algebra. Then we put

[itex]\mathcal{A}=\{B-k~\vert~B\in \mathcal{B}\}[/itex]

Try to prove that [itex]\mathcal{A}[/itex] is a sigma-algebra which contains the open intervals...
 
micromass said:
For the first, you want to know if

[tex]\{f\leq a\}=\liminf\{f_n\leq a\}[/tex]

In other words, you want to know whether [itex]f\leq a[/itex] if and only if [itex]f_n\leq a[/itex] eventually.

This is false, take [itex]f_n(x)=\frac{1}{n}x^2[/itex], then f(x)=0 and take a=0.
Then [itex]\{f\leq 0\}=\mathbb{R}[/itex], but [itex]\liminf\{f_n\leq 0\}=\{0\}[/itex]

As you do notice, you do have the [itex]\supseteq[/itex] inclusion.

(b) is true. Let [itex]\mathbb{B}[/itex] be the Borel-sigma-algebra. Then we put

[itex]\mathcal{A}=\{B-k~\vert~B\in \mathcal{B}\}[/itex]

Try to prove that [itex]\mathcal{A}[/itex] is a sigma-algebra which contains the open intervals...

Appreciate your help, you i "proved" my first claim. Could you help to find my error?
So, if for given x [itex]f(x) \leq a[/itex], then for [itex]\forall \varepsilon>0, \forall x[/itex] [itex]\exists k[/itex] such that [itex]\forall n\geq k[/itex]


[tex]f(x) - \varepsilon < f_n(x) < f(x) + \varepsilon \leq a + \varepsilon[/tex]
Since [itex]\varepsilon[/itex] was arbitrary, we're done!
 
The thing is that your n depends on [itex]\varepsilon[/itex]. So if you take [itex]\varepsilon[/itex] smaller, then n will be bigger. So the argument, "[itex]\varepsilon[/itex] is arbitrary" doesn't really work here.
 
micromass said:
The thing is that your n depends on [itex]\varepsilon[/itex]. So if you take [itex]\varepsilon[/itex] smaller, then n will be bigger. So the argument, "[itex]\varepsilon[/itex] is arbitrary" doesn't really work here.

I see it now, thanks
 

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