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 $f(x)$.


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 $\int_{E_1} f \le 1\cdot m(E_1) \le m(E) < \infty$.

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  • $\begingroup$ What are "sub-level" and "super-level" sets $\endgroup$ – Zduff Nov 14 '18 at 23:09
  • 1
    $\begingroup$ The sets where a function is below or above a given value. $\endgroup$ – user296602 Nov 14 '18 at 23:18

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