Borel measurable function

• Mar 10th 2010, 06:49 PM
koko2009
Borel measurable function
Hi there,
I have a question on real analysis, I will write my try and could you please help me to solve it.
we need to prove that f(x)=X^2 ,Q(x)= tanx, and J(x)= -sqrt(x) are borel measurable by the formal definition {f<= b}is in Borel, for all b in R.
Thanks,
• Mar 10th 2010, 07:02 PM
Drexel28
Quote:

Originally Posted by koko2009
Hi there,
I have a question on real analysis, I will write my try and could you please help me to solve it.
we need to prove that f(x)=X^2 ,Q(x)= tanx, and J(x)= -sqrt(x) are borel measurable by the formal definition {f<= b}is in Borel, for all b in R.
Thanks,

What do you mean?
• Mar 11th 2010, 01:08 PM
Moo
Quote:

Originally Posted by Drexel28
What do you mean?

What do you know about measure theory ?

Any function f is Borel measurable if (with your definition) for any b in R, $\displaystyle f^{-1}(]-\infty,b])$ is a Borel set.

In a more general way, since $\displaystyle ]-\infty,b]$ is a closed set, it is sufficient for f to be continuous. Thus the pre-image of this set will be a Borel set, since the pre-image of a closed set under a continuous function is a closed set. (and a closed set is a Borel set)
• Mar 11th 2010, 03:21 PM
mabruka
Quote:

Hi there,
I have a question on real analysis, I will write my try and could you please help me to solve it.
we need to prove that f(x)=X^2 ,Q(x)= tanx, and J(x)= -sqrt(x) are borel measurable by the formal definition {f<= b}is in Borel, for all b in R.
Thanks,

Fix $\displaystyle b\in\mathbb R$

$\displaystyle \{f\leq b\} = \{x\in\mathbb R : f(x)\in((-\infty,b]\} = f^{-1}((-\infty,b])$

So as you pointed out, you need to see that $\displaystyle f^{-1}((-\infty,b])$ is a borel set.

We have that $\displaystyle f(x)=x^2$ so

$\displaystyle f^{-1}((-\infty,b]) = \{x \in\mathbb R:-\infty<x^2\leq b \}$
There are 3 cases:

If $\displaystyle b<0$ then $\displaystyle f^{-1}((-\infty,b]) =\emptyset$

If $\displaystyle b=0$ then $\displaystyle f^{-1}((-\infty,b]) =\{0\}$

If $\displaystyle b\geq 0$ then $\displaystyle f^{-1}((-\infty,b]) =\{x: -\infty < x^2 \leq b \}=\{x: -\sqrt b \leq x \leq\sqrt b \}=[-b,b]$

The first and last case obviously are borel sets (the empty set is in every $\displaystyle \sigma$-algebra, closed sets are borelian as well (their complement is open) ), what about the second one?

Well, in $\displaystyle \mathbb R$ with the usual metric, singletons are closed, so $\displaystyle \{0\}$ is closed, therefore is borelian.

Hence for every $\displaystyle b$, $\displaystyle \{f\leq b\}$ is in $\displaystyle \mathcal B(\mathbb R) \Longrightarrow f$ is measurable.

You have to do this for every of your functions. Also as Moo already told you, if you already know (i bet you do) that the borel $\displaystyle \sigma$ -algebra $\displaystyle \mathcal B (\mathbb R)$ is the $\displaystyle \sigma-$ algebra generated by all sets of the form :

open (a,b)

OR

closed [a,b]

OR

psemi-open (a,b]

OR
psemi closed [a,b)

(you can even use the above for rational end points still get the same sigma algebra)

So, if $\displaystyle h:(\mathbb R,\mathcal B(\mathbb R) ) \longrightarrow (\mathbb R,\mathcal B(\mathbb R) )$ and you show that you have $\displaystyle h^{-1}(K)$ is like any of the above type of sets, FOR EVERY K of any of the above, then h is measurable.