Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $f \colon \mathbb{R} \to \mathbb{R}$ be a measurable function. I want to show that $f^2:x\mapsto (f(x))^2$ is measurable.

Apparently it can be shown using the facts that the sum of two measurable functions is measurable, the composition of a continuous function with a measurable function is measurable and "a couple of simple formulae".

I just do not know how to show this, without the fact that the product of two measurable functions is measurable (This is quoted later on, so I presume there is a way to prove the above without using this.)

Thanks for any little tips.

share|improve this question

3 Answers 3

up vote 2 down vote accepted

The function $x\mapsto x^2$ is measurable. Compose that with $f$...

share|improve this answer
    
Thank you for your help! –  Mt123 Feb 11 '13 at 23:59

Hint: Let $g(x)=x^2$ then $f^2=g\circ f$. Clearly $g$ is continuous, and therefore measurable (it is a polynomial).

share|improve this answer
    
Thanks! I've got it now! –  Mt123 Feb 11 '13 at 23:58

Of course!

So, let $h(x) = x^2$, which is continuous. Then $h(f(x)) = (f(x))^2$ is the composition of a continuous map and a measurable map, and so is measurable.

Thanks!

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.