If $\mu$ has a density with respect to the Lebesgue measure, is $C_c(\mathbb R)$ dense in $L^p(\mu)$?

Question

Let $\mu$ be a probability measure on $(\mathbb R,\mathcal B(\mathbb R))$.

Is $C_c^\infty(\mathbb R)$ dense in $L^p(\mu)$ for all $p\ge1$?

Let $\lambda$ denote the Lebesgue measure on $(\mathbb R,\mathcal B(\mathbb R))$. We know that $C_c(\mathbb R)$ is dense in $L^p(\lambda)$ for all $p\ge1$. Since, $C_c^\infty(\mathbb R)$ is dense in $C_c(\mathbb R)$, we can conclude that $C_c^\infty(\mathbb R)$ is dense in $L^p(\lambda)$ for all $p\ge1$.

Now, I'm especially interested in the case where $\mu$ has a density $f$ with respect to $\lambda$. It would be even fine for me to assume that $f\in C^2(\mathbb R)$ and that $f>0$. Moreover, it would be sufficient for me to obtain the desired claim for $p=2$?

Is there any chance to use the known result for the Lebesgue measure?

Answer 1

There is the following general statement which you can find in Measures, Integals and Martingales by R. Schilling (Corollary 17.9 in the 2nd edition).

Theorem: Let $\mu$ be a measure on $(\mathbb{R}^n,\mathcal{B}(\mathbb{R}^n))$ which assigns finite measure to compact sets. Then the compactly supported smooth functions $C_c^{\infty}(\mathbb{R}^n)$ are dense in $L^p(\mu)$ for any $p \geq 1$.

If $\mu$ is a probability measure on $(\mathbb{R}^n,\mathcal{B}(\mathbb{R}^n))$, then the assumption that $\mu$ assigns finite measure to compact sets is trivially satisfied, and hence $C_c^{\infty}(\mathbb{R}^n)$ is dense in $L^p(\mu)$ for all $p \geq 1$.