I was assigned this excercise:
Determine $p$ so that $f(x)=\frac{1}{x^3 - 1}$ belongs in the $\mathfrak{L}^P(E)$ space, where $E=\mathbb{R}-\{1\}$
I've proceeded along this direction:
$E$ is measurable because, for any given interval in $\mathbb{R}$
$m(I)=m(I/E)+m(I\cap E)$
Even when $I$ contains the element $\{1\}$ we find that $m(I/E)=0$ because I/E is a set with a finite number of elements. On the other hand $m(I\cap E)=m(I)$ in any case.
Since $E$ is measurable, LP spaces are embedded in such a way that: $\mathfrak{L}^{\infty}(E)\subset...\subset\mathfrak{L}^2(E)\subset\mathfrak{L}^1(E)$
If $f\notin\mathfrak{L}^{1}$(E), then it doesn't belong in any LP space over $E$. The function $|f(x)|$ aprroaches $+\infty$ as $x$ approaches $1$, therefore neither: $\int_{-\infty}^{1}|f(x)|dx$ or $\int_{1}^{+\infty}|f(x)|dx$ converge.
As a consequence, no LP space over $E$ contains $f$
Is my conclusion correct? Can you think of a more rigourous solution? Can it be done in a subtler way?
Thank you in advance
The inclusion of $L^p(E)$ spaces which you stated is correct provided the ambient set (the $E$ in our notation) has finite measure. When the measure of $E$ is unbounded, the inclusion does not hold in general. For example, take $E = \mathbb{R}$ and $f(x) = \frac{1}{1+|x|}$ , $x\in \mathbb{R}$. Then $f\notin L^1(\mathbb{R})$, however, for any $p>1$ we have $f\in L^p(\mathbb{R})$ (to see these claims one can use the monotonicity of $f$ and a comparison of the integral of $f$ with the series $\sum\limits_{n=1}^\infty \frac{1}{n^p}$, $p \geq 1$).
In you example of $f(x) = \frac{1}{x^3 - 1}$ on $E:=\mathbb{R}\setminus \{1\}$ the integrability fails near $1$ (in a neighborhood of $\pm \infty$ the function $f$ is integrable with any exponent $p\geq 1$ as discussed above). Indeed, $$ \int\limits_{0}^1 \frac{dx}{|x^3 - 1|^p } =\int\limits_0^1 \frac{dx}{(1-x)^p (x^2 + x + 1)^p} \geq \frac{1}{3^p} \int\limits_0^1 \frac{dx}{(1-x)^p} , $$ but the last integral converges if and only if $p < 1$, hence $f$ is not integrable on $E$ with any exponent $p\geq 1$.