TITLE: Is $L^p \cap L^q$ dense in $L^r$?
QUESTION [5 upvotes]: It is known that $L^p \cap L^q \subset L^r$, where $1 \le p \le r \le q \le \infty$. Are all of these inclusions dense? I.e., do we have 
\begin{equation*}
\overline{L^p \cap L^q} = L^r
\end{equation*}
(closure w.r.t. $\Vert \cdot \Vert_r$)?
To prove this, one would have to show the existence of a sequence of functions $f_n \in L^p \cap L^q$ such that $f_n \stackrel{L^r}{\longrightarrow} f$, given arbitrary $f \in L^r$. I know how this works in the case $p=1$, $q=\infty$. But is it even true in the most general case?
If it is true, as a consequence, any function in $L^p$ could be approximated by functions $f_n \in L^q$ for any $1 \le q \le \infty$. This sounds quite like a utopia to me, as it would make a lot of things I am currently working on easier. I don't see why it would be wrong, however. Many thanks in advance!

REPLY [3 votes]: The result is false for $p<r=q=\infty$, which would read
$$ \overline{L^p\cap L^\infty} \overset{?}{=} L^\infty$$
For instance take $1\in L^\infty(\mathbb R)$ with Lebesgue measure; if $\|f-1\|_{L^\infty}\le 1/2$ then say $|f|\ge 1/4$ a.e. so $f\notin L^p$ as $\int_{\mathbb R} |f|^p = \int_{|f|\ge 1/4} |f|^p \ge \frac1{4^p}\int_\mathbb R 1 = \infty. $ So no collection of $f\in L^p$, in particular $f\in L^p\cap L^\infty$ can approximate $1$.