If $f$ and $\operatorname{sign}(f)\in L^2([0,1])$, is $f\in L^1([0,1])$?

Question

Is the following argument correct?

By Cauchy-Schwarz:

$$\int_{[0,1]}|f(x)|\,dy=\left|\int_{[0,1]}f(x)\cdot\operatorname{sign}(f(x))\,dy\right| \le \Vert f\Vert_2\cdot 1\le \infty$$

Does this mean that if $f\in L^2([0,1])$, and $f\not\in L^1([0,1])$, then $\operatorname{sign}(f)$ is not measurable?