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

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!