# Show that for any $1\leq p<\infty$, the set $L^1\cap L^p$ is a dense subset of $L^p$

Show that for any $1\leq p<\infty$, the set $L^1\cap L^p$ is a dense subset of $L^p$.

Let $f\in L^p-L^1$. We need to find a sequence $\{\phi_n\}_n$ in $L^1\cap L^p$ converging to $f$. And I know the simple approximation theorem.

I think the following lemma is useful.

Lemma: If a simple function in a measure space $(X,\mathfrak{B},\mu)$ which belongs to $L^p(\mu)$, $1\leq p < \infty$, also belongs to $L^1(\mu)$.

Attempt: Let $g$ be simple function in $L^p$. Then we have $g=\Sigma_{i=1}^{m}a_i\chi_{E_i}$ for some $E_1,...,E_m\in\mathfrak{B}$ and some $a_1,...,a_m$ in $\mathbb{R}$. So $|g|^p=\Sigma_{i=1}^{m}|a_i|^p\chi_{E_i}$. Since $g \in L^p$, we have $\infty >\int|g|^pd\mu=\Sigma_{i=1}^{m}|a_i|^p\int \chi_{E_i}d\mu=\Sigma_{i=1}^{m}|a_i|^p\mu{E_i}$ . So $\mu(E_i)<\infty$ for all $i=1,...,m$. So $\infty>\Sigma_{i}^{m}|a_i|\mu(E_i)=\Sigma_{i=1}^{m}|a_i|\int \chi_{E_i}d\mu=\int|g|d\mu$. So $g \in L^1(\mu)$.

How is my attempt? How can we conclude the proof? Thanks!

It may be easier to work with truncations rather than simple functions. Given $f \in L^p$ define $f_n(x) = f(x)$ if $|f(x)| > \frac 1n$, and $0$ otherwise. Then $f_n(x) \to f(x)$ for all $x$. Since $|f_n| \le |f|$ you have that $f_n \in L^p$, and since $|f_n - f|^p \le 2^p |f|^p$ LDCT implies $$\lim_{n \to \infty} \int |f_n -f|^p \, d\mu = 0.$$
On the other hand, assuming $p > 1$, you have by Holder's inequality and Chebyshev's inequality $$\int |f_n| \, d\mu = \int_{\{|f| > \frac 1n\}} |f| \, d\mu \le \mu(\{|f| > \tfrac 1n\})^{1/p'} \|f\|_p < \infty$$ so that $f \in L^1$ too.
Your proof is correct! As for density, one usually argues that $C_c^{\infty}$ is dense AND contained in $L^p$ for all $p\in [1, \infty)$ and hence lies also in your intersection.
• How is $C_c^\infty$ defined on an arbitrary space $X$? – Umberto P. Jan 21 '15 at 19:54