Prove that if $f^2(x)$ is Lebesgue integrable on E, so is $f(x)$.

Please check my proof, thank you.

Let $$f(x)$$ be a nonnegative and measurable function defined on the set $$E$$ with $$m(E) < \infty$$. Prove that if $$f^2(x)$$ is Lebesgue integrable on E, then so is $$f(x)$$.

Proof.

Let $$E_1 = \{x \in E: 0 \leq f^2(x) < 1\}$$, then for $$x \in E_1$$ we have $$0 \leq f(x) < 1$$. Similarly let $$E_2 = \{x \in E: 1 \leq f^2(x) < M\}$$, then for $$x \in E_2$$ we have $$f(x) < M$$. Also, $$E_1 \cap E_2 = \emptyset \implies$$ $$E_1$$ and $$E_2$$ are measurable. Now we can bound $$\int f$$ as follows: \begin{align} \int_E f(x) \, dx = \int_{E_1 \cup E_2} f(x) \, dx &\leq \int_{E_1}f(x) \, dx + \int_{E_2}f(x) \, dx \\ \\ &< \int_{E_1}dx + \int_{E_2}M \, dx \\ \\ &< m(E_1) + M\cdot m(E_2) \end{align} So, since $$f$$ is nonnegative and measurable, and $$f$$'s Lebesgue integral on the measurable set $$E$$ is bounded, $$f$$ must be Lebesgue integrable.

What I'm worried about is my assertion that the disjointedness of $$E_1$$ and $$E_2$$ is wronge. Counterexample: $$E = [0,1]$$. Let $$E_1 = \mathscr{N}$$, a nonmeasurable set in $$[0,1]$$, then clearly $$E_2 = E \setminus E_1$$ is disjoint from $$E_2$$, their union is $$E_1$$ and they are not both measurable.

So I believe that the way I divided $$E$$ in the proof ensures that $$E_1$$ and $$E_2$$ are measurable but not for the reason I stated.

Unfortunately, your proof is incorrect because there is an implicit assumption that $$f$$ is bounded, which is not a consequence of square-integrability and trivializes the problem (if you assume boundedness of $$f$$, then you don't even need to split into two sets!).
Moreover, the measurability of $$E_1$$ and $$E_2$$ does not follow from their disjointness either; it follows from the fact that $$f$$ is measurable (and so $$\sqrt f$$ is also measurable, and $$E_1$$ and $$E_2$$ are just sub- and super-level sets of $$\sqrt f$$).
To correct the proof, consider the following: On the set $$E_2$$, we have $$0 \le f \le f^2$$, and so $$\int_{E_2} f < \infty$$. For the set $$E_1$$, we merely have $$0 \le f < 1$$ so that $$\int_{E_1} f \le 1\cdot m(E_1) \le m(E) < \infty$$.