If a normalized sequence of finite measures converges weakly, does the same hold true for the nonnormalized one? Let $E$ be a metric space, $(\mu_n)_{n\in\mathbb N}$ be a sequence of finite nonnegative measures on $\mathcal B(E)$ and $\mu$ be a probability measure on $\mathcal B(E)$ s.t. $$\frac{\mu_n}{\mu_n(E)}\xrightarrow{n\to\infty}\mu\tag1$$ with respect to the to the topology of weak convergence of measures.
Assuming that $c:=\sup_{n\in\mathbb N}\mu_n(E)<\infty$, are we able to deduce that the nonnormalized sequence $(\mu_n)_{n\in\mathbb N}$ is convergent as well?
If not, does the other direction of this assertion hold, i.e. can we infer that the normalized sequence (i.e. the left-hand side of $(1)$) is convergent from knowing that the nonnormalized sequence is convergent? It's easy to see that $(\mu_n(E))_{n\in\mathbb N}$ is convergent in that case (since the function constantly equal to $1$ is bounded and continuous). So, this should be possible to obtain.
Remark: Without assuming $c<\infty$, the first implication is clearly wrong. We simply can consider any finite nonnegative measure $\nu$ on $\mathcal B(E)$ and set $\nu_n:=n\nu$ for $n\in\mathbb N$.
 A: Put $E = $ real line. Consider $\mu_{2n} = 2\delta_{ \{0 \} }$ and  $\mu_{2n-1} = \delta_{ \{0 \} }$, where $\delta_{ \{x \} }$  is a Dirac measure.
So the answer is negative.
The opposite is true if $lim_n \mu_n (E)$ (it exists, as you noticed) is not equal to $0$.
But if  $lim_n \mu_n (E) = 0$ we may have smth. like that: $\mu_{n} = \frac{(-1)^n + 1}{n}\delta_{ \{0 \} }$, that is counterexample, because normalized version of $\mu_{2n+1}$ doesn't exist.
Addition. Suppose that normalized sequence of measures converges to some measure $\nu$ and that $\mu_n(E) \to a \ne 0$, (hence $c < \infty$ and $a = \nu(E)$).
Statement: $\mu_n \to a \nu$ and $\nu$ is normalized (i.e. a probability measure).
Indeed, as $\lim_n \mu_n(E) \ne 0$ we have $\mu_n(E) \ne 0$ for all $n\ge n_0$. Thus $$\mu_n = \mu_n(E) \cdot \frac{\mu_n}{\mu_n(E)}.$$ Put $a_n =  \mu_n(E)$, $\nu_n = \frac{\mu_n}{\mu_n(E)}$. We want to show that $a_n \nu_n \to a \nu$. We know that  $\nu_n \to \nu$ and $a_n \to a \ne 0$.
By definition $\int f(x) d\nu_n \to \int f(x) d\nu$ for all bounded continuous $f$. Thus $$\int f(x) d(a_n \nu_n) = a_n \int f(x) d\nu_n \to a \int f(x) d\nu = \int f(x) d(a \nu)$$ and hence $a_n \nu_n \to a \nu$.
Further, $\nu(E) = \lim_n \nu_n(E) = 1$.
Moreover, under the condition $\lim_n \mu_n(E) \ne 0$ convergence of $\nu_n$ and convergence of $\mu_n$ are equivalent (because we may multiply and devide convergent sequence of measures by $ \mu_n(E)$).
Conclusion: if normalized measures converge then $\mu_n$ may not converge. If $\mu_n$ converges then in general case it doesn't mean that normalized measures converge (al least because normalized measures  may not exist). But  if $\mu_n$ converge and there is an additional condition $lim_n \mu_n(E) \ne 0$ then normalized measures converge.
A: By the same argument as presented by Botnakov N. we can even show more:
Let $\mathcal M(E)$ denote the space of finite signed measures on $\mathcal B(E)$ equipped with the total variation norm $\left\|\;\cdot\;\right\|$, $(\mu_t)_{t\in I}$ be a net in $\mathcal M(E)$ with $$c:=\lim_{t\in I}\left\|\mu_t\right\|\ne0\tag2$$ and $\mu\in\mathcal M(E)$ with $$\left(\frac{\mu_t}{\left\|\mu_t\right\|}\right)_{t\in I}\to\mu\tag3$$ with respect to the topology of weak convergence of measures. Then, $$\mu_tf=\left\|\mu_t\right\|\frac{\mu_t}{\left\|\mu_t\right\|}f\xrightarrow{t\in I}c\mu f\;\;\;\text{for all }f\in C_b(E)\tag4;$$ i.e. $$(\mu_t)_{t\in I}\to c\mu\tag5.$$
