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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?

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this is a trick question. you may have $f^2$ integrable but $f$ not measurable. –  mike Oct 8 '12 at 15:23
I'm considering measurable functions. –  Bunny Oct 8 '12 at 15:31
But if $f$ is measurable, you have $f$ integrable since $|f| \le 1 + |f|^p$ for any $p\ge 1$. –  martini Oct 8 '12 at 15:31
@martini But what can we say for $p<1$ ? –  Bunny Oct 8 '12 at 15:41
Then $f$ needn't be integrable, for example is $x^{-1}$ integrable over $[0,1]$, but its square root (i. e. $p = 1/2$) $x^{-1/2}$ is ... –  martini Oct 8 '12 at 15:45

2 Answers 2

up vote 2 down vote accepted

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 $$ \int_{[0,1]} |f| \le \int\left.\begin{cases} |f| & \text{on }\{|f|\le 1\} \\ f^2 & \text{on } \{f>1\} \end{cases}\right\} \le 1 + \int |f|^2. $$

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This uses less background than my answer (+1). However, I would have compared $|f|$ to $1$ on $\{|f|\le1\}$ instead of $|f|$. –  robjohn Jan 21 '13 at 4:34

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 $$ \int_0^1|f|\cdot1\,\mathrm{d}x\le\left(\int_0^1|f|^2\,\mathrm{d}x\right)^{1/2}\left(\int_0^11^2\,\mathrm{d}x\right)^{1/2} $$

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 $\dfrac1p+\dfrac1q=1$.

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