One problem in theorem $2$ from "Some further results in Ideal Convergence in Topological Spaces" by P. Das It is Theorem $2$ in this paper by P.Das. I will quote the entire proof here to help discussion by not having to go to the paper repeatedly.:

Theorem $2$. Let $X$ be a first countable space. For any sequence $(x_n)_{n∈N}$ in $X$ the set $I(L_x)$ is an $F_σ$-set provided $I$ is an analytic  $P$-ideal.
Proof:  Since $I$ is an analytic $P$-ideal, there exists a lower semicontinuous submeasure $ϕ$ satisfying (∗). For any $r ∈ \mathbb N$ let
  $$F_r =\left\{p ∈ X: ∃A = \{n_1 < n_2 < n_3 < ···\} ⊂ \mathbb N, \lim_k x_{n_k}=p \text{ and} \lim_{n→∞} ϕ(A\backslash[1,n])\ge {1\over r}\right\}$$
We shall now show that each $F_r$ is a closed subset of $X$. Let $α ∈ \bar F_r$ and let $U$ be a neighborhood of $α$. Since $X$ is first countable, there is a sequence $(α_j)_{j∈\mathbb N}$ in $F_r$ converging to $α$. For each $α_j$, we can find $A_j ⊂ \mathbb N$ with ${\lim_{n→∞}}_{n∈ {A_j}} x_n=α_j$
  and $\lim_{n→∞}ϕ(A_j\backslash [1,n])\ge {1\over r}$. Let $(\epsilon_j)_{j∈\mathbb N}$ be a monotonically decreasing sequence of positive real numbers converging to $0$.
  We now proceed as follows: 
    First choose $n_1 ∈ \mathbb N$ such that $ϕ(A_1\backslash [1,n_1])\ge {1\over r}−{{\epsilon_1}\over 2}$. Now lower semicontinuity of $ϕ$ implies
  that $ϕ(A_1\backslash [1,n_1]) = \lim_{n→∞} ϕ[(A_1\backslash [1,n_1]) ∩ [1,n]]$. Choose $m ∈ \mathbb N$ such that $ϕ[(A_1\backslash [1,n_1]) ∩ [1,n]]\ge ϕ(A_1\backslash [1,n_1]) −{\epsilon_1\over 2}∀n\ge m$. Again there exists $m_1 ∈ \mathbb N$ such that $ϕ(A_2\backslash [1,n]) \ge {1\over r} −{\epsilon_2 \over 2}
∀n\ge m_1$. Now choose $n_2 > n_1,m,m_1$. Then clearly
  we have simultaneously $ϕ(A_1 ∩ (n_1,n_2])\ge {1\over r}−\epsilon_1$ and also $ϕ(A_2\backslash [1,n_2])\ge {1\over r}−{\epsilon_2\over 2}$.  Proceeding as above we now choose
  $n_3 > n_2$ such that $ϕ(A_2 ∩ (n_2,n_3])\ge {1\over r}−\epsilon_2$ and $ϕ(A_3\backslash [1,n_3])\ge {1\over r}-{\epsilon_3 \over 2}$ and so on. Thus we can construct a sequence
  $n_1 < n_2 < n_3 < ···$ of positive integers such that
  $$ϕ(A_j∩(n_j,n_{j+1}])\ge {1\over r}−\epsilon_j, j ∈ \mathbb N.$$
  Let us define $$A=\bigcup_j\{A_j∩(n_j,n_{j+1}]\}$$. Then clearly ${\lim_{n→∞} ϕ(A\backslash [1,n])\ge {1\over r}}$ and so ${A \not∈ I}$. Let $A=\{l_1<l_2<l_3< ···\}$. Since
  $\lim_j α_j = α$ and $α ∈ U$ so $α_j ∈ U ∀ j\ge j_0$ for some $j_0∈\mathbb N$. $\color{blue}{\text{ This implies that }x_n∈U\text{ for all but a finite number of indices }n\text{ of the set }A}$. Therefore $α∈F_r$. Hence $F_r$ is a closed subset of $X$. The assertion now immediately follows from the fact that $I(L_x)=\bigcup _{r=1}^∞Fr$.

I have difficulty understanding the $\color{blue}{blue}$ part.
Since $\alpha\in U$ is the limit of the sequence $\alpha_j$ so from above we have $\alpha_s\in U$ for some $s>j_0.$ Now,$\lim_{n\rightarrow \infty;n\in A_s}x_n=\alpha_s.$ So again there is a $p\in \mathbb N$ such that $x_n\in U\forall n\ge p.$ Let $p=l_t\in A$ and $x_p\in U.$ Now how can I show that $x_{l_{t+1}}\in U$ too? Is not that $x_n$ necessary to show thatall but finite indices go inside $U$? or how else? Please advice. Thank you.
EDIT: This question is cross posted in both MSE and MathOverFlow.This one has an answer that I accepted although the one in OverFlow doesn't have any answers or comments.the link to the same question in MathOverFlow.
 A: In the proof,$\newcommand{\intrvr}[2]{(#1,#2]}\newcommand{\ve}{\epsilon}\newcommand{\I}{\mathcal I}$
a sequence $(n_j)$ is constructed in such way that $$\phi(A_j\cap\intrvr{n_j}{n_{j+1}})\ge \frac1r-\ve_j.\tag{1}$$ The purpose of this is to get in the end the set $A$ such that $A\notin\I$.
However, we also need that $x|_A$ converges to $\alpha$. Here I see the problem, because we only know that the subsequence $x|_{A_j}$ converges to $\alpha_j$, but it is still possible that this subsequence only gets close to $\alpha_j$ for indices much bigger than $n_{j+1}$. 
So I do not see how the argument can be finished in the way described in the paper. (I hope that if I am simply missing something, somebody will be able to explain how it can be proved this way.) 

However, I can still suggest slight modification of the proof which leads to the desired result. We simply add one more condition on the sequence $(n_j)$ which will help us to get the desired limit.
Since we are working in a first countable space, there is a countable neighborhood basis of $\alpha$. Let us choose such a basis $\{U_n; n\in\mathbb N\}$. Moreover, we will require the sequence $U_n$ to be non-increasing, i.e., $U_n\supseteq U_{n+1}$.
We can assume w.l.o.g. that $\alpha_n\in U_n$. (If needed, we can replace the original sequence $(\alpha_n)$ by a subsequence, simply by omitting some terms. Of course, the subsequence converges to $\alpha$ as well.)
In this way, we know that since the subsequence $x|_{A_j}$ converges to $\alpha_j$, there is some $n\in\mathbb N$ such that 
$$k\ge n, k\in A_j \qquad\implies\qquad x_k\in U_j.\tag{2}$$
(Here we used only the fact that $U_j$ is a neighborhood of $\alpha_j$.)
So we will modify the above proof in such way that we will require from the sequence $n_j$ that:


*

*$x_k\in U_j$ for each $k\in A_j$ such that $k>n_j$;

*$\phi(A_j\cap\intrvr{n_j}{n_{j+1}})\ge \frac1r-\ve_j$.


The first condition is satisfied for every large enough $n_j$ because of $(2)$. And we can get also $(1)$ will hold at the same time - in the inductive step, if we want to choose $n_{j+1}$, changing $n_{j+1}$ to a larger value cannot change validity of $(1)$. And if we choose it to be large enough, we get that also $(1)$ holds.
After this modification we can really get that $x|_A$ converges to $\alpha$.
It suffices to show that each $U_j$ contains all but finitely many terms of the subsequence $x|_A$. And from the above conditions we see that for $k\in A$ such that $k>n_j$ we have $x_k\in U_j$ 

The proof of Theorem 2 from the mentioned paper P. Das: Some further results on ideal convergence in topological spaces, doi: 10.1016/j.topol.2012.04.007, seems to be following the proof of Theorem 1.1 from 
P. Kostyrko, M. Mačaj, T. Šalát, O. Strauch: On statistical limit points, doi: 10.1090/S0002-9939-00-05891-3. This is the same result for the ideal of sets having zero asymptotic density.
As far as I can say, there is the same problem in this proof. And I think that similar modification can be done in this proof too. So I am also curious whether somebody can find a way how the proof works in the original version (without additional condition on the choice of indices determining the set $A$.) But personally I have doubts whether it can be done. 
