Lebesgue- integrability of powers 2 and 3 of a function

Question

Let $f:[0,1]\to \mathbb R$ , and $f^2$ is L-integrable, is $f$ also L-integrable? What if $f^3$ is L-integrable?

Answer 1

If $A$ is a non-measurable subset of $[0,1]$ then $$ f(x)=\begin{cases} 1 & \text{if }x\in A \\ -1 & \text{if }x\not\in A \end{cases} $$ is not measurable, but its square is Lebesgue-integrable.

However, if $f$ is measurable, then integrability of $f^2$ does imply that of $f$. To show this you need to rely on the fact that the measure of the domain is finite. You have $$ \int_{[0,1]} |f| \le \int\left.\begin{cases} |f| & \text{on }\{|f|\le 1\} \\ f^2 & \text{on } \{f>1\} \end{cases}\right\} \le 1 + \int |f|^2. $$

Answer 2

Since $x^2$ is a convex function and $[0,1]$ has measure $1$, Jensen's Inequality says that $$ \left(\int_0^1|f|\,\mathrm{d}x\right)^2\le\int_0^1|f|^2\,\mathrm{d}x $$ Furthermore, Cauchy-Schwarz says $$ \int_0^1|f|\cdot1\,\mathrm{d}x\le\left(\int_0^1|f|^2\,\mathrm{d}x\right)^{1/2}\left(\int_0^11^2\,\mathrm{d}x\right)^{1/2} $$

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For any $p\ge1$

Since $x^p$ is a convex function and $[0,1]$ has measure $1$, Jensen's Inequality says that $$ \left(\int_0^1|f|\,\mathrm{d}x\right)^p\le\int_0^1|f|^p\,\mathrm{d}x $$ However, we need to use Hölder's Inequality instead of Cauchy-Schwarz: $$ \int_0^1|f|\cdot1\,\mathrm{d}x\le\left(\int_0^1|f|^p\,\mathrm{d}x\right)^{1/p}\left(\int_0^11^q\,\mathrm{d}x\right)^{1/q} $$ where $\dfrac1p+\dfrac1q=1$.