Csdrhrt

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!

The inclusion $L^p(mu) subset L^q(mu)$ for $1leq p < q leq +infty$ holds if and only if
$$
inf{mu(A): Ainmathcal{M}, mu(A)>0} > 0.
$$
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
$$
ell^p := {x = (x_1, x_2, ldots): sum_{j=1}^infty |x_j|^p < infty}.
$$