Is the set of probabilities with bounded density dense? Let $X$ be a Polish space and $P(X)$ be the set of all the probability measures on $X$. Fix $\mu\in P(X)$ and define
$$P_\mu(X) = \{\nu\in P(X):\mathrm{Supp}(\nu)\subseteq\mathrm{Supp}(\mu)\}\,.$$
Clearly, all $\nu\ll\mu$ are in $P_\mu(X)$. Let $\nu\in P_\mu(X)$. I wonder wether we can always find a sequence $\{\nu_n\}\subset P_\mu(X)$ such that:

*

*$\nu_n\ll\mu$ (for all $n$) and its density $\tfrac{d\nu_n}{d\mu}$ is bounded;

*$\nu_n\rightharpoonup\nu$ as $n\to\infty$, where $\rightharpoonup$ denotes the standard weak convergence for probability measures.

Note that the boundedness of the sequence does not have to be uniform (that is I do not assume that there is a $K$ such that all the densities are bounded by $K$).
Edit.
I believe that the answer is yes. I guess that to show it one can prove that $\{\nu\in P_\mu(X):\nu\ll\mu\}$ is dense (wrt the weak convergence) in $P_\mu(X)$. If this result is true, than I believe that the probabilities with bounded densities are dense in the set of probabilities absolutely continuous wrt $\mu$. Indeed, let $\nu\ll\mu$ and $r=\frac{d\nu}{d\mu}$. Then we can consider the sequence of functions $\{r_n\}$ defined as
$$r_n(x) = \min(r(x),n)\,.$$
We have that for all bounded and continuous $f$
$$\int_X r_n(x) f(x) d\mu(x) \to \int_X r(x) f(x) d\mu(x) = \mathbb E_\nu[f]\,.$$
(First we assume that $f\geq 0$ and use monotone convergence, then we write $f$ as the difference of its positive and negative parts.)
Now in particular, if $f\equiv 1$, we have that $Z_n = \int_X r_n(x)d\mu(x)\to 1$, and so if we define $\nu_n$ as the measure with density $r_n/Z_n$ wrt $\mu$ we conclude that
$$\mathbb E_{\nu_n}[f]\to\mathbb E_\nu[f]$$
for all bounded and continuous $f$, that is $\nu_n\rightharpoonup\nu$.
So the question can be restated as:
Is $\{\nu\in P_\mu(X): \nu\ll\mu\}$ dense in $P_\mu(X)$?
 A: Firstly, let us simplify the statement. Since $\mathrm{supp} \mu$ is closed, it is a Polish space. Thus, the assertion is equivalent to the following:

Let $X$ be a Polish space, and $\mu$ be a probability on $X$ with full support. Show that $L^\infty(\mu)\cap \mathscr{P}^{\ll\mu}$ is weakly dense in $\mathscr{P}(X)$.

Here, $\mathscr{P}^{\ll\mu}=\{f\in L^1(\mu)^+: \|f\|_{L^1(\mu)}=1\}$.
We show a stronger statement:

Let $\mathcal{F}\subset \mathscr{P}^{\ll\mu}$ be any family so that $\overline{\mathcal F}^{\mathrm{weak}\, L^1(\mu)}\supset\mathscr{P}^{\ll\mu}$. Then $\mathcal{F}$ is weakly dense in $\mathscr{P}(X)$.

It is clear that we can choose $\mathcal{F}=L^\infty(\mu)\cap \mathscr{P}^{\ll\mu}$, by truncation and normalization of functions in $L^1(\mu)$, but we could also replace $L^\infty(\mu)$ with some much smaller space of (e.g., continuous) functions.
Step 1 For every $x\in X$ there exists $f_n\in \mathscr{P}^{\ll\mu}$ with $f_n\mu\rightharpoonup \delta_x$.
Proof. Fix a distance $d$ metrizing $X$. Since $\mu$ has full support, for every natural $n$ we have $\mu B_{1/n}(x)>0$. Set $f_n:= \frac{1_{B_{1/n}(x)}}{\mu B_{1_n}(x)}$. It is not difficult to show that $f_n\mu\rightharpoonup \delta_x$ by the Mean Value Theorem.
Step 2 For every $\nu=\sum_{i=1}^N s_i\delta_{x_i}$ there exists $f_n\in \mathcal{C}_b(X)^+\cap \mathbb{S}^{L^1(\mu)}_1$ with $f_n\mu\rightharpoonup \nu$.
Proof. Since the weak convergence of probability measures is linear, Step 1 immediately implies Step 2.
Step 3 The set of purely atomic measures with finite support is dense in $\mathscr{P}(X)$.
Proof. This is quite classical, and there are many possible proofs. For example, let $\nu$ be the measure we want to approximate by a sequence $\nu_n$, and $\nu_{pa}$ be the purely atomic part of $\nu$.
If $\nu_{pa}=\nu$ has finitely many atoms choose $\nu_n:=\nu$ for every $n$.
If $\nu_{pa}=\nu=\sum_{i=1}^\infty a_i \delta_{x_i}$ has infinitely many atoms, it is clear that $\nu_n:=\sum_{i=1}^n a_i\delta_{x_i}$ weakly converges to $\nu$.
Since the weak convergence is homogeneous (i.e. $\nu_n\rightharpoonup \nu \implies a\nu_n\rightharpoonup a\nu$ for $a>0$) and linear (i.e. $\nu_n-\nu_{n,pa}\rightharpoonup\nu-\nu_{pa}$ and $\nu_{n,pa}\rightharpoonup \nu_{pa}$ implies $\nu_n\rightharpoonup\nu$) we may restrict to the case of atomless $\nu$. In this case, for every $n$ let $(A_{n,i})_{i\leq n}$ be a measurable partition of $X$ into $n$ pieces and assume that
$$(*)\qquad \lim_n \max_{i\leq n} \mathrm{diam}_d A_{n,i}=0.$$
Set $\nu_n:= \sum_{i:\nu A_{n,i}>0} \frac{1_{A_{n,i}}}{\nu A_{n,i}}\delta_{x_{n,i}}$ for some $x_{n,i}\in A_{n,i}$. Again it is not difficult to show that $\nu_n \rightharpoonup\nu$.
Step 4
Simple topological fact: if $B\subset\overline{A}$ then $\overline B\subset \overline{A}$.
Conclusion
Choose $\mathcal F$ with $\overline{\mathcal{F}}^{\mathrm{weak} L^1(\mu)}\supset \mathscr{P}^{\ll\mu}$.
Note that, since $\mu$ has full support, $\mathcal{C}_b(X)\hookrightarrow L^\infty(\mu)$. Therefore the weak $L^1(\mu)$-convergence of $f_n$ to $f$ implies the weak convergence $f_n\mu \rightharpoonup f\mu$. Thus, the weak closure of $\mathcal{F}$ in $\mathscr{P}(X)$ contains $\mathscr{P}^{\ll\mu}$.
By Steps 1-3, $\mathscr{P}^{\ll\mu}$ is weakly dense in $\mathscr{P}(X)$, thus $\mathcal{F}$ too is weakly dense in $\mathscr{P}(X)$ by Step 4.
Note. Any sequence of partitions as in $(*)$ is called a null-array. Whereas for simplicity we fixed a metric $d$ metrizing the topology of $X$, it can be shown that $(*)$ does not depend on $d$. We are not using this fact, but it is helpful to reconcile the fact that we are choosing $d$ with the fact that weak convergence does not depend on $d$, but only on the topology.
A: The answer to the above question is yes. However I am sure there is a much easier way to prove it.
I'll show that $\{\nu\in P_\mu(X): \nu\ll\mu\}$ is dense in $P_\mu(X)$. First, let $D$ be any metric that make $X$ a complete metric space. For each $n\geq 0$ consider a partition $\mathcal B_n$ of $X$, such that each $B\in\mathcal B_n$ is measurable and has diameter smaller than $1/n$ (measured wrt $D$). We can assume that for all $B\in\mathcal B_n$ such that $B\cap\mathrm{Supp}(\mu)\neq\emptyset$, $\mu(B)>0$. Indeed, if there is a $B$ such that $\mu(B)=0$ and there is $x\in B\cap\mathrm{Supp}(\mu)$, then $x$ must be on the boundary of $B$, as we can find arbitrarily small open neighbourhood of $x$ which has non zero measure $\mu$, but that cannot be contained in $B$. So there must be another set in $B_n$ that has non-zero $\mu$ measure and has $x$ as a limit point.
Now, since $\mathrm{Supp}(\nu)\subseteq\mathrm{Supp}(\mu)$, we have that if $\nu(B_n)>0$, then $\mu(B_n)>0$. Now, for each $x$, let $B_n$ such that $x\in B_n$.  We define $\rho_n(x)$ to be equal to $\frac{\nu(B_n)}{\mu(B_n)}$ if $\mu(B_n)>0$ and $x\in\mathrm{Supp}(\mu)$, and to $0$ otherwise.
Now define $\nu_n = \rho_n \mu$. Clearly $\nu_n\ll\mu$ and $\mathrm{Supp}(\nu_n)\subseteq\mathrm{Supp}(\mu)$.
Now, fix a function $f:X\to\mathbb R$, bounded and continuous. For all $n$, for all $B_n\in\mathcal B_n$, choose a point in $B_n$ and denote it as $x_{B_n}$. If $B_n\cap\mathrm{Supp}(\mu)\neq\emptyset$, we can suppose that $x_{B_n}\in\mathrm{Supp}(\mu)$. Define the function $f_n$ as $f(x_{B_n})$ for all $x\in B_n$.
Now, since $f$ is continuous, we have that $f_n(x)\to f(x)$, as $n\to\infty$, for all $x$. Moreover, since $f$ is bounded we have that
$$\mathbb E_\nu[f] = \lim_{n\to\infty} \mathbb E_\nu[f_n] = \lim_{n\to\infty}\sum_{B\in\mathcal B_n} f(x_n)\nu(B_n) = \lim_{n\to\infty}\sum_{B\in\mathcal B_n} f(x_n)\rho_n(x_n)\mu(B_n)= \lim_{n\to\infty}\mathbb E_{\nu_n}[f]\,.$$
Since $f$ is arbitrary we conclude that $\nu_n\rightharpoonup\nu$.
