Are Pointwise Convergent Functions Limited to Measurable Sets?

[tohauz]

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

 

[Bhargava2011]

I'm not getting your question. Could you make it more clear.

 

[micromass]

For the first, you want to know if

\{f\leq a\}=\liminf\{f_n\leq a\}

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

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

As you do notice, you do have the \supseteq inclusion.

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

\mathcal{A}=\{B-k~\vert~B\in \mathcal{B}\}

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

 

[tohauz]

For the first, you want to know if

\{f\leq a\}=\liminf\{f_n\leq a\}

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

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

As you do notice, you do have the \supseteq inclusion.

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

\mathcal{A}=\{B-k~\vert~B\in \mathcal{B}\}

Try to prove that \mathcal{A} 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 f(x) \leq a, then for \forall \varepsilon&gt;0, \forall x \exists k such that \forall n\geq k


f(x) - \varepsilon &lt; f_n(x) &lt; f(x) + \varepsilon \leq a + \varepsilon
Since \varepsilon was arbitrary, we're done!

 

[micromass]

The thing is that your n depends on \varepsilon. So if you take \varepsilon smaller, then n will be bigger. So the argument, "\varepsilon is arbitrary" doesn't really work here.

 

[tohauz]

The thing is that your n depends on \varepsilon. So if you take \varepsilon smaller, then n will be bigger. So the argument, "\varepsilon is arbitrary" doesn't really work here.

I see it now, thanks