Inclusions relating standard norms, in measure theory I know that for finite measure space $(X, \mathcal A ,\mu )$ and $1\leq p< q<\infty $ , the inclusion $\mathcal L^q\subseteq \mathcal L^p\subseteq\mathcal L^1 $ holds true (applying Holder's inequality). I was looking for an example that could prove that this inclusion does not hold true for non-finite measure space. Also I was wondering about the inclusion of  the space $\mathcal L^\infty$.
 A: $\frac{-\log x}{1+x^2}$ is a $L^1(\mathbb{R}^+)$ function that is essentially unbounded, while $\frac{1}{x+1}$ is a bounded function over $\mathbb{R}^+$, but is it not integrable. Hence over $\mathbb{R}^+$, we do not have $L^1\subseteq L^{\infty}$  neither $L^{\infty}\subseteq L^{1}$. Intermediate cases are similar. 
Also notice that if $1\leq p<q\leq \infty$ and the measure of the space is finite, we have:
$$ L^1\color{red}{\supseteq} L^p \color{red}{\supseteq} L^q \color{red}{\supseteq} L^{\infty}$$
by Holder's inequality.
A: As a complement to Jack D'Aurizio's answer, here are examples of functions showing that neither inclusion, $L^p \subseteq L^q$ nor $L^p \supseteq L^q$, are true in general on an infinite measure space. 
Let $q > p \geq 1$, then $f(x) = x^{-1/p}$ is in $L^q( [1,\infty))$ but not in $L^p( [1,\infty))$, since 
$$\int_1^{\infty} x^{-q/p} dx = \frac{p}{q-p} \textrm{ and } \int_1^{\infty} x^{-p/p} dx = \infty.
$$
Furthermore, $g(x) = \frac{1}{(x-1)^{1/q}} \chi_{[1,2)}(x)$ is in $L^p([1,\infty))$ but not in $L^q([1,\infty))$, since 
$$\int_1^{\infty} g(x)^q dx = \int_1^2 \frac{dx}{x-1} = \infty \textrm{ and } \int_1^{\infty} g(x)^p dx = \int_1^2 \frac{dx}{(x-1)^{p/q}} = \frac{q}{q-p}.
$$
In this last example, I am taking advantage of the fact that $L^q$ is not contained in $L^p$ for the finite measure space $[1,2]$.
