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) &lt; \alpha(\omega) for any \omega \in \Omega?
And is
\{ t &lt; \alpha \} := \{ \omega \in \Omega \mid t(\omega) &lt; \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 &lt; \alpha, \alpha \le \beta \text{ and } \beta &lt; u,
then you can trivially say
t &lt; \alpha \le \beta &lt; 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 &lt; \alpha\}\cap\{\beta&lt; \alpha &lt; u\} = \{ t &lt; \alpha&lt; u\} = \{ t &lt; \beta &lt; 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&lt; \alpha" or "t&lt; \beta&lt; 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&lt; \alpha(x) for all x" and "t&lt; \beta(x)&lt; u" for all x?

 

[wayneckm]

Sorry, I should write clearly again...

\{t &lt; \alpha\}\cap\{\beta&lt; u\} = \{ t &lt; \alpha&lt; u\} = \{ t &lt; \beta &lt; u\} means \{\omega \in \Omega | t &lt; \alpha(\omega)\}\cap\{\omega \in \Omega |\beta(\omega)&lt; u\} = \{ \omega \in \Omega | t &lt; \alpha(\omega) &lt; u\} = \{ \omega \in \Omega | t &lt; \beta(\omega) &lt; 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
<br /> \{\omega \in \Omega | t &lt; \alpha(\omega)\}\cap\{\omega \in \Omega |\beta(\omega)&lt; u\} \subseteq \{ \omega \in \Omega | t &lt; \alpha(\omega) &lt; u\} = \{ \omega \in \Omega | t &lt; \beta(\omega) &lt; u\}<br />

Indeed, let
\omega \in \{\omega \in \Omega | t &lt; \alpha(\omega)\}\cap\{\omega \in \Omega |\beta(\omega)&lt; u\} \subseteq \{ \omega \in \Omega | t &lt; \alpha(\omega) &lt; u\}.
Then it is true that both t &lt; \alpha(\omega) and \beta(\omega) &lt; 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 &lt; \alpha(\omega) \le \beta(\omega) &lt; u
from which the inclusion follows.

Almost the exact same argument in reverse applies to show that
<br /> \{\omega \in \Omega | t &lt; \alpha(\omega)\}\cap\{\omega \in \Omega |\beta(\omega)&lt; u\} \supseteq \{ \omega \in \Omega | t &lt; \alpha(\omega) &lt; u\} = \{ \omega \in \Omega | t &lt; \beta(\omega) &lt; u\}<br />

 

[wayneckm]

Thanks for the reply.

So this means

<br /> \{\omega \in \Omega | t &lt; \alpha(\omega)\}\cap\{\omega \in \Omega |\beta(\omega)&lt; u\} = \{ \omega \in \Omega | t &lt; \alpha(\omega) &lt; u\} = \{ \omega \in \Omega | t &lt; \beta(\omega) &lt; u\}<br />

is wrong. Rather, it should be written as

<br /> \{\omega \in \Omega | t &lt; \alpha(\omega)\}\cap\{\omega \in \Omega |\beta(\omega)&lt; u\} = \{ \omega \in \Omega | t &lt; \alpha(\omega) &lt; u\} \bigcap \{ \omega \in \Omega | t &lt; \beta(\omega) &lt; u\}<br />

 

[CompuChip]

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

<br /> \{ \omega \in \Omega | t &lt; \alpha(\omega) &lt; u\} = \{ \omega \in \Omega | t &lt; \alpha(\omega) &lt; u\} \bigcap \Omega = \{ \omega \in \Omega | t &lt; \alpha(\omega) &lt; u\} \bigcap \{ \alpha(\omega) \leq \beta(\omega) \} <br />
<br /> = \{ \omega \in \Omega | t &lt; \alpha(\omega) \leq \beta(\omega) &lt; u\} \bigcup \{ \omega \in \Omega | t &lt; \alpha(\omega) &lt; u \leq \beta(\omega) \}<br />

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

<br /> \{ \omega \in \Omega | t &lt; \beta(\omega) &lt; u\} = \{ \omega \in \Omega | t &lt; \beta(\omega) &lt; u\} \bigcap \Omega = \{ \omega \in \Omega | t &lt; \beta(\omega) &lt; u\} \bigcap \{ \alpha(\omega) \leq \beta(\omega) \} <br />
<br /> = \{ \omega \in \Omega | t &lt; \alpha(\omega) \leq \beta(\omega) &lt; u\} \bigcup \{ \omega \in \Omega | \alpha \leq t &lt; \beta &lt; u \}<br />

Again, these two sets are disjoint.

Finally,
<br /> <br /> \{\omega \in \Omega | t &lt; \alpha(\omega)\}\cap\{\omega \in \Omega |\beta(\omega)&lt; u\} = \{ \omega \in \Omega | t &lt; \alpha(\omega) \leq \beta(\omega) &lt; u \}<br /> <br />

So it is the intersection of the above two sets.