The following exercise is in Folland:
Let $\nu$ be a signed measure on $(X, \mathcal{M})$ show that $L^1(\nu) = L^1(|\nu|)$.
My attempt: Consider $f \in L^1(\nu)$ then we know that
$$\int |f| d\nu = \int |f| d\nu^+ - \int |f| d\nu^- < \infty$$
This can only happen iff $\int |f| d\nu^+ < \infty$ and $\int |f| d\nu^- < \infty$. Thus, we see that
$$\int |f| d|\nu| = \int |f| d\nu^+ + \int |f| d\nu^- < \infty$$
So we must have $f \in L^1(|\nu|)$. Since all these steps were two ways, we have also that $f \in L^1(\nu)$.
This "proof" seems too simple and others seem to be giving me the same idea that this is not correct. Where is the flaw and how can I fix it?