1. ## 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,

2. 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?

3. 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, $f^{-1}(]-\infty,b])$ is a Borel set.

In a more general way, since $]-\infty,b]$ 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)

4. 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 $b\in\mathbb R$

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

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

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

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

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

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

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

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

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

Hence for every $b$, $\{f\leq b\}$ is in $\mathcal B(\mathbb R) \Longrightarrow f$ 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 $\sigma$ -algebra $\mathcal B (\mathbb R)$ is the $\sigma-$ 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 $h\mathbb R,\mathcal B(\mathbb R) ) \longrightarrow (\mathbb R,\mathcal B(\mathbb R) )" alt=" h\mathbb R,\mathcal B(\mathbb R) ) \longrightarrow (\mathbb R,\mathcal B(\mathbb R) )" /> and you show that you have $h^{-1}(K)$ is like any of the above type of sets, FOR EVERY K of any of the above, then h is measurable.