I need to prove that if $f: \mathbb{R} \to \mathbb{R}$ is a measurable function with $\int_{[0,1]}(f(x))^{2} dm(x) < \infty$ then $\int_{[0,1]}|f(x)| dm(x) < \infty$.
I've started by saying that over the set $E=\{x\in [0,1] : |f(x)| \leq (f(x))^{2}\}$ we have $\int_{E}|f(x)| dm(x) \leq \int_{[0,1]}(f(x))^{2} dm(x) < \infty$ by monotonicity.
Now let $x \in E^{c} \cap [0,1]$ so that $|f(x)| > (f(x))^{2}$.
Then we must have $|f(x)|<1$ so by monotonicity $\int_{E^{c} \cap [0,1]}|f(x)|dm(x)<1$.
$E$ and $E^{c}$ are disjoint measurable sets by the measurability of $f$, so:
$\int_{[0,1]}|f(x)| dm(x) = \int_{E}|f(x)| dm(x) + \int_{E^{c} \cap [0,1]}|f(x)| dm(x)$ $\leq \int_{[0,1]}(f(x))^{2}dm(x) + 1 < \infty$
I'm not sure if the my attempt at the proof is correct, and I'd appreciate it if someone could take a look and let me know if its right, and if I would need to add any further justification.
A similar approach to yours: First off, note that $f(x)^2$ is always $\geq 0$. So $|f(x)|\le f(x)^2$ if and only if $|f(x)|\geq 1$. Also, $f(x)^2<|f(x)|\iff |f(x)|< 1$.
Thus, $$\int_{[0,1]} |f(x)|= \underbrace{\int_{\displaystyle\{x\in[0,1]\mid |f(x)|< 1\}} |f(x)|}_{\displaystyle< 1} + \underbrace{\int_{\displaystyle\{x\in[0,1]\mid |f(x)|\geq 1\}}|f(x)|}_{\displaystyle\le\int_{[0,1]}f(x)^2<\infty}<\infty.$$