## $\sigma$-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

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

I tried the following:

Let $$1 \le p and $$\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 $$1/q-1$$.

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 $$L^q$$)?