# The square of a measurable function is measurable

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.

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

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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).

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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!

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