Necessary and sufficient condition for weak convergence and convergence of density Let $(\mu_n)_n$ and $\mu$ be two probability measure, having respectively density $(f_n)_n$ and $f$ for the measure $\lambda$ on $(\mathbb{R},B(\mathbb{R})).$

*

*Prove that the following statement are equivalent:
a) $(\mu_n)_n$ converges weakly to $\mu$ and $$\forall \epsilon>0,\exists \delta>0;\forall n \in \mathbb{N}, \forall E \in B(\mathbb{R}),\lambda(E)\leq \delta\implies\int_Ef_n(x)dx \leq \epsilon$$
b) $(\mu_n)_n$ converges weakly to $\mu$ and $$\lim_{k\to+\infty}\sup_{n \in \mathbb{N}}\int_{\left\{f_n>k \right\}}f_n(x)dx=0.$$
c) $\forall E \in B(\mathbb{R}),\lim_{n\to+\infty}\mu_n(E)=\mu(E).$


*If $(\mu_n)_n$ converges weakly to a probability measure $\sigma$ and for all $\epsilon>0,$ there exist $\delta>0$ such that for all $n \in \mathbb{N},$ for all $E \in B(\mathbb{R})$ such that $\lambda(E)\leq \delta,\int_Ef_n(x)dx \leq \epsilon.$ Is it true that $\sigma$ have a probability density ? (There exist $\phi:\mathbb{R}\to\mathbb{R}^+,$ such that $\int_{\mathbb{R}}\phi(x)dx=1, \sigma(U)=\int_U\phi(x)dx,$ for all $U \in B(\mathbb{R})$)
This is the attempt so far.
a) $\implies$ b). Take $\epsilon>0.$ there exist $\delta>0$ such that $$\forall n \in \mathbb{N},\forall E \in B(\mathbb{R}),\lambda(E) \leq \delta \implies \int_Ef_ndx \leq \epsilon.$$ Let $k \geq \frac{1}{\delta}.$ So $$\forall n \in \mathbb{N},\lambda(\left\{f_n>k \right\}) \leq \frac{1}{k} \leq \delta$$ which means that $$\forall n \in \mathbb{N},\int_{\left\{f_n>k \right\}}f_n \leq\epsilon,$$ Then $\sup_n\int_{\left\{f_n>k \right\}}f_n(x)dx \leq \epsilon.$
b) $\implies$ a). Let $\epsilon>0.$ there exist $k>0$ such that $$\sup_n \int_{\left\{f_n>k \right\}}f_n(x)dx \leq \epsilon/2.$$
Let $n \in \mathbb{N},E \in B(\mathbb{R})$ such that $\lambda(E) \leq \frac{\epsilon}{2(k+1)}.$
$$\int_E f_n(x)dx \leq k\lambda(E)+\int_{\left\{f_n>k \right\}}f_n(x)dx \leq \epsilon.$$
How can we proceed with c) $\implies$ a)? 2) is the statement correct?
 A: Your proof for (a) $\Leftrightarrow$ (b) is correct in my opinion.
(a) implies (c):
I do not know how to use the hint, but I think I found an alternative approach.
Let $\varepsilon>0$ and a set $E\in B(\Bbb R)$ be given.
Let $\delta>0$ be given according to (a).
Then there is an open set $E_o\supset E$ such that
$\lambda(E_o\setminus E)\leq \delta$.
It follows that $\mu_n(E_o\setminus E)\leq \varepsilon$.
For open sets we know that we have $\mu(E_o)\leq \liminf \mu_n(E_o)$.
Thus we have
$$
\mu(E)\leq \mu(E_o) \leq \liminf \mu_n(E_o) \leq \liminf \mu_n(E) +\varepsilon.
$$
Since $\varepsilon>0$ was arbitrary, this implies
$\mu(E)\leq \liminf \mu_n(E)$
for all Borel sets $E$.
Applying this observation to the complement of $E$
will lead to
$\mu(E)=\lim \mu_n(E)$ after a couple of rearrangements.
an observation regarding densities:
If a measure $\nu$ has a density $g$, then
we have
$$
\forall \varepsilon>0 \exists \delta>0 \forall E\in B(\Bbb R):
\lambda(E)\leq\delta \implies \nu(E)=\int_E g(x)\mathrm dx \leq \varepsilon.
$$
In particular, this holds for $\mu$ and each single $\mu_n$.
One funny way to see this is to apply
"(b) implies (a)" to the constant sequence of measures $\nu$.
The difference between this property and the property from (a) is that the $\delta>0$
can be chosen independently of $n$.
Note that for finitely many $n$, one can always find a common $\delta>0$
by choosing the minimum of the respective $\delta$'s for each $n$.
It also follows that
$$
\label{equiv1}\tag{1}
\forall \varepsilon>0 \exists \delta>0 \exists n_0\in \Bbb N\forall n\geq n_0,
\forall E\in B(\Bbb R):
\lambda(E)\leq\delta \implies \int_E f_n(x)\mathrm dx \leq \varepsilon
$$
is equivalent to
$$
\label{equiv2}
\tag{2}
\forall \varepsilon>0 \exists \delta>0 \forall n\in\Bbb N,\forall E\in B(\Bbb R):
\lambda(E)\leq\delta \implies \mu_n(E)=\int_E f_n(x)\mathrm dx \leq \varepsilon.
$$
(c) implies (a):
The weak convergence is probably clear for you.
Suppose that the rest of (a) is not true.
Then there exists $\varepsilon>0$
and a sequence $n_k\in\Bbb N,E_k\in B(\Bbb R)$ such that
$\lambda(E_k)\leq \hat\delta 2^{-k}$ and
$\mu_{n_k}(E_k)>\varepsilon$ hold.
Here, $\hat\delta>0$ is given such that
$
\forall E\in B(\Bbb R): \lambda(E)\leq\hat\delta \implies \mu(E)\leq \varepsilon/2
$.
Such an $\hat\delta$ exists by the observation regarding density above.
Suppose that $M:=\sup_k n_k<\infty$,
i.e. $n_k$ takes only finitely many values.
Then, by the observation regarding density above,
one could find $\delta>0$ independent of $k$ such that
$\lambda(E_k)\leq \delta$ implies $\mu_{n_k}(E_k)\leq\varepsilon$,
which would be a contradiction to our assumption.
Thus, we have $M:=\sup_k n_k=\infty$,
and we can (without loss of generality) assume that $n_k\to\infty$
as $k\to\infty$.
Alternatively, one can also obtain $n_k\to\infty$ by using the
negation of (\ref{equiv1}) instead of the negation of (\ref{equiv2})
as the assumption at the beginning of this proof.
We now define the set $E:=\cup_{k\in\Bbb N} E_k$.
Then we have $ \lambda(E)\leq\hat\delta $
and therefore
$$
\mu(E)\leq \varepsilon/2
< \varepsilon < \mu_{n_k}(E_k) \leq \mu_{n_k}(E).
$$
This is a contradiction to $\mu_{n}\to\mu(E)$.
2.:
I think this is true.
The above proof for (a) implies (c) does not rely on the assumption that
$\mu$ has a density.
Thus we know that
$\mu_n(E)\to \sigma(E)$ holds for all $E\in B(\Bbb R)$.
Since the $\mu_n$ measures are absolutely continuous,
it follows that $\sigma$ must be absolutely continuous, too.
Therefore $\sigma$ has a density.
Remark regarding $\Bbb R^n$:
As far as I can see, all these arguments for 1. and 2. also work for $\Bbb R^n$ instead
of $\Bbb R$.
