Let $f$ be Lebesgue measurable on $[0,1]$. Show that $\|f(\sqrt{x})\|_1\leq\frac{7}{6}\|f\|_2 $
The first part of the problem asks to show the bound $\|f(\sqrt{x})\|_1\leq2\|f\|_1 $ which is easy by doing $u$-sub. To be honest, I am not even sure if we can claim $f(\sqrt{x})$ is measurable. The problems says $f$ is Lebesgue measurable (not Borel measurble).
Any hints would be appreciated, I am not sure how to tackle it. I tried messing around with sets where $|f|\geq1$ and $|f|<1$ but could not get anything.