# Measurable Functions

How do we prove that a function $f$ is measurable if and only if $\arctan(f)$ is measurable?

If I use the definition of measurable functions, that is, a function is measurable if and only if its inverse is measurable?

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Cite the definition a bit more precisely, for a start.. –  Berci Nov 11 '12 at 13:20
That definition is wrong. –  Jonas Meyer Feb 15 '13 at 3:53

(I will assume we are talking about real-valued functions)

To prove the first implication, assume that $f$ is measurable.

Take $V\subset\mathbb R$ open. Write $g=\arctan$. Then $(g\circ f)^{-1}(V)=f^{-1}(g^{-1}(V))$. As $g$ is continuous, $g^{-1}(V)$ is open. As $f$ is measurable, $f^{-1}(g^{-1}(V))$ is measurable. So $g\circ f$ is measurable.

For the converse, suppose that $g\circ f$ is measurable. Then $f=\tan\circ(g\circ f)$. As $\tan$ is continuous on the image of $g$, we can apply the argument in the first part of the proof to get that $f$ is measurable.

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Thanks a lot! Indeed it is a real valued function. I suppose I can proceed the same way to prove the same but with sine. –  Bhavish Suarez Nov 11 '12 at 14:58
Yes, but with trig functions you always have to be careful with domains: $\arcsin(\sin(x))$ is not necessarily $x$ for an arbitrary $x$. –  Martin Argerami Nov 11 '12 at 16:38

Since f is measurable and arctan is continuous therefore arctan f is Lebesgue measurable. If arctan f is measurable, then tan arctan f is Lebesgue measurable (because tan is continuous) therefore f is Lebesgue measurable.

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thanks! But how I prove that tan and arctan are measurable? –  Bhavish Suarez Nov 11 '12 at 13:25
This is not a correct argument. The composition of Lebesgue measurable functions is not necessarily Lebesgue measurable. What you do have is that if $(X,\Sigma_1)$, $(Y,\Sigma_2)$ and $(Z,\Sigma_3)$ are measurable spaces and $f: X\to Y$ and $g: Y\to Z$ are measurable, then $g \circ f$ is measurable. But we have to be careful when arguing about Lebesgue measurable functions because we give the domain the Lebesgue algebra and the codomain the Borel algebra. –  kahen Nov 11 '12 at 13:34
Oh yes, thank you –  Amr Nov 11 '12 at 13:48