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 $\forall \varepsilon>0, \forall x$ $\exists k$ such that $\forall n\geq k$


$$f(x) - \varepsilon < f_n(x) < f(x) + \varepsilon \leq a + \varepsilon$$
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