Let $(X,d)$ be a separable metric space, let $\mu$ be a Borel locally finite measure on $(X,d)$, let $(g_n)\subseteq L_1(X,\mu)$ be a sequence with $g_n$ non-negative such that $\sup\limits_{n}\int\limits_X g_n d\mu <+\infty$. For every $n$ let $\mu_n$ denote the Borel measure generated by $g_n$, i.e. $\mu_n(A)=\int\limits_A g_nd\mu$ for every Borel set $A$. My question: is there exists a subsequence $(\mu_{n_k})$ that weakly converges (in the sence of probability measures) to some finite Borel measure $\nu$ on $(X,d)$, i.e. for every continuous function $\phi\colon X\to [-1,1]$ it holds that $\lim\limits_{k\to +\infty} \int\limits_X \phi\, d\mu_{n_k}=\int\limits_X \phi \,d\nu$.
The analogous statement holds for any $p\in (1,+\infty)$, since every $L_p(X,\mu)$-bounded family in $L_p(X,\mu)$ contains a weakly converging (in $L_p(X,\mu)$ sence) sequence. However, if $p=1$, there exist $L_1(X,\mu)$-bounded families in $L_1(X,\mu)$ that do not have convergent sequences. But may be it is possible to extract a sequence that weakly converges (in the sence of probability measures) to a finite Borel measure. I tried to find some information about it, but there was no an explicit answer.
Edit: this is definitely not true, @geetha290krm provided a counter-example, a sequence $(\chi_{[n,n+1]})$. The questions below, however, are still relevant.
I found some information about Prokhorov's theorem that gives a criteria for a family of Borel probability measures on a separable metric space to be relatively weakly sequentially compact in the space of all Borel probability measures in terms of tightness of this family. The formulation of it in Wikipedia (https://en.wikipedia.org/wiki/Prokhorov%27s_theorem) confused me a bit, since, as I understood, if $\mu(X)$ is not equal to the supremum of the mesures of compact subsets of $X$, then every family of Borel probability measure on $(X,d)$ is not weakly sequentially compact. This seems to be absurd, since every measure when viewed as a constant sequence clearly converges to itself. Or may be the closure of every measure is more than this measure itself? Correct me please if I misunderstood something with this theorem.
If the answer on my question is "no" in general, what sufficient conditions, on $(X,d)$, $\mu$ or $(g_n)$, do exist to guarantee the existence of weakly converging subsequence?
Thanks in advance.