$\sigma$-finite measure $\mu$ so that $L^p(\mu) \subsetneq L^q(\mu)$ (proper subset)

Question

I'm looking for a $\sigma$-finite measure $\mu$ and a measure space so that for

$1 \le p <q \le \infty$

$$L^p(\mu) \subsetneq L^q(\mu)$$

I tried the following:

Let $1 \le p <q \le \infty$ and $\lambda$ the Lebesgue measure on $(1,\infty)$ which is $\sigma$-finite.

$x^\alpha$ is integrable on $(1,\infty) \Leftrightarrow \alpha <-1$.

Choose $b$ so that $1/q<b<1/p \Leftrightarrow -bq<-1, -bp>-1$.

Then $x^{-b}\chi_{(1,\infty)}$$\in L^q$ but $\notin L^p$ because $x^{-bq}$ is integrable because the exponent $-bq<-1$ and $x^{-bp}$ isn't integrable because the exponent $-bp>-1$. Now I found a function that is in $L^p$ but not in $L^q$. But that doesn't really show that $L^p \subsetneq L^q$, meaning $L^p$ is a proper subset of $L^q$, right (because I don't know if every element of $L^p$ is also an element of $L^q$)?

Thanks in advance!

Answer 1

The inclusion $L^p(\mu) \subset L^q(\mu)$ for $1\leq p < q \leq +\infty$ holds if and only if $$ \inf\{\mu(A):\ A\in\mathcal{M},\ \mu(A)>0\} > 0. $$ Hence your $\mu$ must necessarily be an atomic measure.

For example, you can consider the spaces $\ell^p$. More precisely, you equip $\mathbb{N}$ with the counting measure, so that $$ \ell^p := \{x = (x_1, x_2, \ldots):\ \sum_{j=1}^\infty |x_j|^p < \infty\}. $$