Now, how can one go to prove that, when $1\leq p<q<\infty$, the inclusion $\cal{L}^q\subset\cal{L}^q$ holds for the corresponding spaces? We need to assume that the underlying measure (assumed to be the same for both) is finite, right?
A proof that mentions also the way to prove that for the norms, $\|f\|_p\leq\|f\|_q$ for every $f$ in the particular case when $\mu(X)=1$ would be doubly appreciated. Thanks!
[Reference: Cohn, Measure Theory ($2^{nd}$ ed.), Exercise 3.3.(9)]
I will provide one possible way of dealing with this problem. Let $X$ be a space with $\mu (X)<\infty$, and $p,q\in\Bbb R$ satisfying $1\le p\le q< \infty$. Let $f\in \mathcal L^q(X)$, then $$\begin{align} \int_X |f|^p d\mu &= \int_{|f|\le 1} |f|^p d\mu + \int_{|f|>1} |f|^p d\mu \\ &\le \int_{|f|\le 1} 1\ d\mu + \int_{|f|>1} |f|^q d\mu \\ &\le \int_{X} 1\ d\mu + \int_{X} |f|^q d\mu \\ \end{align}$$ Which implies that
Since $||f||_q$ is finite, then so is $||f||_p$. Therefore we have $f\in \mathcal L^p(X)$ for any $f\in \mathcal L^q(X)$, which implies that $$ \mathcal L^q(X) \subset \mathcal L^p(X) $$ for $p\le q$, $\mu(X)<\infty$.
Now, what went wrong if $\mu(X)=\infty$? It is not hard to see that the yellowish inequality still holds but it no longer implies that $||f||_p$ is finite. This can be realized on $X=\Bbb R$ and a function $f$ of the form $f=\frac1{|1+x|^r}$ for some suitably chosen $r$.