Problem on Set

1. Aug 27, 2010

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

2. Aug 27, 2010

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
$$t < \alpha \le \beta < u$$

3. Aug 27, 2010

wayneckm

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.

4. Aug 27, 2010

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?

5. Aug 27, 2010

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.

6. Aug 27, 2010

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

7. Aug 27, 2010

wayneckm

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

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

8. Aug 28, 2010

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.

9. Aug 28, 2010

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).

10. Aug 28, 2010

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.