Approximating $L^1(\mu)$ functions by smooth functions Let $\mu$ be a Borel measure on $\mathbb R^{d+1}$ having continuous positive density with respect to the Lebesgue measure. Is $C^\infty(\mathbb R^{d+1})$ dense in $L^1(\mu)$ ? What is an approximating sequence?
Let $\psi\in L^1(\mu)\,$. I would like to find a sequence $\psi_n\in C^\infty(\mathbb R^{d+1})$ such that
$$\int_{\mathbb R^{d+1}}|\psi_n-\psi|\,d\mu\to0 \;.$$
Moreover I am interested in preserving the support in the following sense: if $\psi$ is continuous and
$$\text{supp}\psi\subseteq K\times\mathbb R$$
for some $K$ compact subset of $\mathbb R^d$, then I would like $\psi_n$ to have the same property (possibly for a larger compact set $K_n$).
Can I choose
$$ \psi_n(x) \,=\, (\psi*f_n)(x) \,=\, \int_{\mathbb R^{d+1}}\psi(x-y)\,f_n(y)\,d y$$
where $f_n$ is a sequence of mollifiers?
 A: 
Step 1: $\mu$ is a Radon measure.

Theorem 1.28 in these notes shows that any locally finite Borel measure on $\mathbb{R}^d$ is Radon. It is clear that either assumption mentioned above is sufficient for this hypothesis.

Step 2: If we define the boxes $B_{x,n} = \{y \in \mathbb{R}^d: \|x - y\|_\infty \leq 2^{-n}\}$ and let $\chi_{x,n}$ be the characteristic function of $B_{x,n}$ then $F = \operatorname{span}\{\chi_{x,n}: x \in \mathbb{Q}^d, n \geq 1\}$ is dense in $L^1(\mu)$.

It suffices to show that $\chi_E \in \overline{F}$ for an arbitrary measurable set $E$ of finite measure.
First suppose that $E$ is open. Then there exists a countable set of dyadic rationals $x_n$ and integers $k_n \geq 1$ such that $E$ can be written as a disjoint union
$$E = \bigsqcup_{n} B_{x_n, k_n}.$$
Let $E_j = \bigsqcup_{n = 1}^j B_{x_n,k_n}$. Then $\|\chi_E - \chi_{E_j}\|_1 = \mu(E\setminus E_j) \to 0$ as $j \to 0$ since $E$ has finite measure. Hence $\chi_E \in \overline{F}$.
For an arbitrary measurable set $E$ of finite measure, by outer regularity of $\mu$, there exist open sets $E_k$ such that $E \subseteq E_k$ and $\mu(E_k \setminus E) < k^{-1}$. Since $\chi_{E_k} \in \overline{F}$ by the above, it follows that $\chi_E \in \overline{F}$ also.

Step 3: $\mu(B_{x+y,n} \setminus B_{x,n}), \mu(B_{x,n} \setminus B_{x+y,n}) \to 0$ as $y \to 0$.

We deal with the first case. We have that $$B_{x+y,n} \subseteq B_{x,n}^y := \{z \in \mathbb{R}^d: \|x - z\|_\infty \leq 2^{-n} + \|y\|_\infty\}.$$ Then $$\mu(B_{x+y,n} \setminus B_{x,n}) \leq \mu(B_{x,n}^y \setminus B_{x,n}) \to \mu(\bigcap_{y > 0} B_{x,n}^y \setminus B_{x,n}) = \mu(\emptyset) = 0$$ as $y \to 0$.
For the second case, a similar argument yields
$$\lim_{y \to 0} \mu(B_{x,n} \setminus B_{x+y,n}) \leq \mu(\partial B_{x,n}) = 0$$
by absolute continuity with respect to the Lebesgue measure.

Step 4: For $g \in L^1(\mu)$, $\|g(\cdot + y) - g(\cdot)\|_1 \to 0$ as $y \to 0$.

Here I only sketch the argument. One checks that set of functions $g$ such that the claim is true is a closed vector subspace of $L^1(\mu)$. Step 3 shows that it contains the dyadic cubes introduced in Step 2 and so the result follows from Step 2.

Step 5: For $\psi \in L^1(\mu)$ and $f$ a smooth probability density function with support in the unit ball, define $f_n(x) = 2^{nd} f(2^n x)$. Then $\psi_n := \psi \ast f_n$ converges to $\psi$ in $L^1(\mu)$ as $n \to \infty$.

One has that $$|(\psi_n - \psi)(x)| \leq \int_{\mathbb{R}^d} |\psi(x-y) - \psi(x)| f_n(y) dy. $$
Hence, by Fubini,
$$\|\psi_n - \psi\|_1 \leq \int_{\mathbb{R}^d} \delta(y) f_n(y) dy$$
where $\delta(y) := \int_{\mathbb{R}^d} |\psi(x-y) - \psi(x)| \mu(dx)$.
Therefore we have that
$$\|\psi_n - \psi\|_1 \leq \left(\sup_{y: |y| < 2^{-n}} |\delta(y)|\right) \|f\|_{L^1(\mathbb{R}^d)} = \sup_{y: |y| < 2^{-n}} |\delta(y)| \to 0$$
as $n \to \infty$ by Step 4.
