Solving Problem on Set: Wayne's Question

[wayneckm]

Hi all,I have the following question:

Suppose there are two functions $$\alpha,\beta$$, which are both mapping $$\Omega \mapsto \mathbb{R}$$ and $$\alpha \leq \beta$$ on every point $$\omega \in \Omega$$.

I am wondering the validity of the following, for $$t < u$$,

$$\{t < \alpha\}\cap\{\beta<u\} = \{ t < \alpha < u\} = \{ t < \beta < u\}$$

Can anyone justify this? Thanks.Wayne

 

[CompuChip]

Can you be a little more specific what something like {t < a} means?

Is t also a function $\Omega \to \mathbb{R}$? Then does t < alpha mean, that $t(\omega) < \alpha(\omega)$ for any $\omega \in \Omega$?
And is
$$\{ t < \alpha \} := \{ \omega \in \Omega \mid t(\omega) < \alpha(\omega) \}$$ ?

Or is, for example, t a real number and is { t < alpha } the set of all lower bounds of alpha, or something like that?In general, I would say that if it is given that
$$t < \alpha, \alpha \le \beta \text{ and } \beta < u,$$
then you can trivially say
$$t < \alpha \le \beta < u$$

 

[wayneckm]

Thanks for the reply.

Sorry for not specifying clearly enough.

Here $$t,u$$ are constants.

The reason for this question is because $$\beta$$ is some measurable function while $$\alpha$$ is an arbitrary function.

So given the information above, I am thinking whether

$$\{t < \alpha\}\cap\{\beta< \alpha < u\} = \{ t < \alpha< u\} = \{ t < \beta < u\}$$

holds. So that I can say this is a measurable set.

Thanks.

 

[HallsofIvy]

You misunderstood Compuchips question and may be misunderstanding the entire problem. It makes no sense to say "$t< \alpha$" or "$t< \beta< u$" for t and u constants (numbers) and $\alpha$ and $\beta$ functions- there is no order relation that order both numbers and functions. Do you mean "$t< \alpha(x)$ for all x" and "$t< \beta(x)< u$" for all x?

 

[wayneckm]

Sorry, I should write clearly again...

$$\{t < \alpha\}\cap\{\beta< u\} = \{ t < \alpha< u\} = \{ t < \beta < u\}$$ means $$\{\omega \in \Omega | t < \alpha(\omega)\}\cap\{\omega \in \Omega |\beta(\omega)< u\} = \{ \omega \in \Omega | t < \alpha(\omega) < u\} = \{ \omega \in \Omega | t < \beta(\omega) < u\}$$ where $$t,u$$ are constants.

Thanks.

 

[CompuChip]

The "elementary" approach here is to show two inclusions by considering arbitrary elements of the set.

One part of the proof would then be to show that
$$\{\omega \in \Omega | t < \alpha(\omega)\}\cap\{\omega \in \Omega |\beta(\omega)< u\} \subseteq \{ \omega \in \Omega | t < \alpha(\omega) < u\} = \{ \omega \in \Omega | t < \beta(\omega) < u\}$$

Indeed, let
$$\omega \in \{\omega \in \Omega | t < \alpha(\omega)\}\cap\{\omega \in \Omega |\beta(\omega)< u\} \subseteq \{ \omega \in \Omega | t < \alpha(\omega) < u\}.$$
Then it is true that both $t < \alpha(\omega)$ and $\beta(\omega) < u$, and because it is given that $\alpha(\omega) \le \beta(\omega)$ for any omega (in particular this one), you get a string of inequalities
$$t < \alpha(\omega) \le \beta(\omega) < u$$
from which the inclusion follows.

Almost the exact same argument in reverse applies to show that
$$\{\omega \in \Omega | t < \alpha(\omega)\}\cap\{\omega \in \Omega |\beta(\omega)< u\} \supseteq \{ \omega \in \Omega | t < \alpha(\omega) < u\} = \{ \omega \in \Omega | t < \beta(\omega) < u\}$$

 

[wayneckm]

Thanks for the reply.

So this means

$$\{\omega \in \Omega | t < \alpha(\omega)\}\cap\{\omega \in \Omega |\beta(\omega)< u\} = \{ \omega \in \Omega | t < \alpha(\omega) < u\} = \{ \omega \in \Omega | t < \beta(\omega) < u\}$$

is wrong. Rather, it should be written as

$$\{\omega \in \Omega | t < \alpha(\omega)\}\cap\{\omega \in \Omega |\beta(\omega)< u\} = \{ \omega \in \Omega | t < \alpha(\omega) < u\} \bigcap \{ \omega \in \Omega | t < \beta(\omega) < u\}$$

 

[CompuChip]

No, it was correct.
If t, omega and u satisfy
$$t < \alpha(\omega) \le \beta(\omega) < u$$
then of course you can leave out any link of the inequality chain and get
$$t < \beta(\omega) < u$$
and
$$t < \alpha(\omega) < u$$
separately.

 

[JCVD]

Wayne, your original equation is wrong but your revised one is correct. To verify that the original is wrong, see the following counterexample.

Let a(w)=w and b(w)=w+1 for real w, and let t=1 and u=2. Then
{w: t<a(w)} = (1,inf) does not intersect {w: b(w)<u} = (-inf,1) while {w: t<a(w)<u} = (1,2) and {w: t<b(w)<u} = (0,1).

 

[wayneckm]

Thanks a lot.

My proof is as follows:

First, note that $$\{ \alpha(\omega) \leq \beta(\omega) \} = \Omega$$

$$\{ \omega \in \Omega | t < \alpha(\omega) < u\} = \{ \omega \in \Omega | t < \alpha(\omega) < u\} \bigcap \Omega = \{ \omega \in \Omega | t < \alpha(\omega) < u\} \bigcap \{ \alpha(\omega) \leq \beta(\omega) \}$$
$$= \{ \omega \in \Omega | t < \alpha(\omega) \leq \beta(\omega) < u\} \bigcup \{ \omega \in \Omega | t < \alpha(\omega) < u \leq \beta(\omega) \}$$

Note that these two are disjoint sets. Then, similarly,

$$\{ \omega \in \Omega | t < \beta(\omega) < u\} = \{ \omega \in \Omega | t < \beta(\omega) < u\} \bigcap \Omega = \{ \omega \in \Omega | t < \beta(\omega) < u\} \bigcap \{ \alpha(\omega) \leq \beta(\omega) \}$$
$$= \{ \omega \in \Omega | t < \alpha(\omega) \leq \beta(\omega) < u\} \bigcup \{ \omega \in \Omega | \alpha \leq t < \beta < u \}$$

Again, these two sets are disjoint.

Finally,
$$\{\omega \in \Omega | t < \alpha(\omega)\}\cap\{\omega \in \Omega |\beta(\omega)< u\} = \{ \omega \in \Omega | t < \alpha(\omega) \leq \beta(\omega) < u \}$$

So it is the intersection of the above two sets.