A function $f$ is measurable if and only if $\arctan f$ is measurable

Question

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?

Answer 1

(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.

Answer 2

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.