Density Lp spaces

Question

If $ 1 \leq p \leq r \leq q < \infty$ and $f \in L^r(\mathbb{R})$. Why is $f \chi _{\lbrace x : \frac{1}{n} \leq |f| \leq n \rbrace} \in L^p \cap L^q $ for all n? I've tried to show that the set ${\lbrace x : \frac{1}{n} \leq |f| \leq n \rbrace}$ is of finite measure by using chebychev-ish bounds but this hasn't gotten me anywhere. (This question refers to Is $L^p \cap L^q$ dense in $L^r$?).

Answer 1

Use the inequalities $|f|^p \le n^{r-p} |f|^r$ and $|f|^q \le n^{q-r} |f|^r$ which are both valid on the set $\{1/n \le |f| \le n\}$.

Answer 2

In fact any bounded measurable function on a set $E$ of finite measure belongs to $L^s(E), 0 < s\le \infty.$ That has a very simple proof. Now in your problem, set $E= \{1/n\le |f| \le n\}.$ To show $E$ has finite measure, note

$$\infty > \int_E |f|^r \ge (1/n)^r\mu(E).$$