So here is my question,
I wanted to generalize, under what assumptions for some $f$ we have $f\in L^p(\mathbb R)\;\forall p\in\mathbb N.$
And I found the following,
Let $f\in L^p(\mathbb R)$ for some $p\in\mathbb N$ and assume that for $A:=\{x\;:|f(x)|<1\}$ we have,
$$\mu(A)<\infty$$
for an arbitrary measure. Furthermore assume that $\exists C\in\mathbb R$ s.t $|f(x)|\geq C\;\forall x\mathbb\in R$ i.e $f\in L^{\infty}(\mathbb R)$
Then $f\in L^p(\mathbb R)\;\forall p\in\mathbb N.$
Proof:
Let $f\in L^p(\mathbb R)$ for some $p\in\mathbb N$ and assume that for $A:=\{x\;:|f(x)|<1\}$ we have,
$$\mu(A)<\infty$$
Then for any $1\leq q\leq p$ the following holds,
$$\int_{A^C}|f|^qd\mu\leq\int_{A^C}|f|^pd\mu\leq\int_{\mathbb R}|f|^pd\mu<\infty$$ and,
$$ \int_{A}|f|^qd\mu\leq\int_{A}1d\mu=\mu(A)<\infty$$ Hence,
$$\int_{\mathbb R }|f|^qd\mu=\int_{A^C}|f|^qd\mu+\int_{A}|f|^qd\mu\leq\mu(A)+\int_{\mathbb R}|f|^pd\mu<\infty$$ This yields $f\in L^q$ for all $q\leq p$.
Now let $k>p$. Since $k\in\mathbb N$ there exist an positive integer $r$ s.t $k\mod p=r$. Moreover there exists an $n$ s.t $k=pn+r$.
Hence,
$$\int_{\mathbb R}|f|^kd\mu=\int_{\mathbb R}|f|^{np}|f|^rd\mu\leq C^{np}\int_{\mathbb R}|f|^r<\infty$$ since $r\leq p.$
Hence $f\in L^k$ and since $k$ was arbitrary the claim follows.
Is this proof correct? And If yes, is there a possibility to relax the assumption that $f$ has to be bounded?
Thanks!
EDIT
I just see that the second part is redundant because every bounded and integrable $f$ is contained in $L^p$ for all $p$...