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 $\color{blue}{\lim_{n→∞} ϕ(A\backslash [1,n])\ge {1\over r}}$ and so $\color{blue}{A \not∈ I}$. 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 $j_0∈\mathbb N$. This implies that $x_n∈U$ for all but a finite number of indices $n$ 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 $I(L_x)=\bigcup _{r=1}^∞ Fr$.
My problem is at the $\color{blue}{blue\ part}$
$$A\backslash [1,n]\\=\bigcup_j\{A_j∩(n_j,n_{j+1}]\}\backslash [1,n]\\=\cup_{n\gt n_j}(A_j\cap (n,n_{j+1}])\bigcup (\cup_{n\le n_j}(A_j\cap (n_j,n_{j+1}]))$$
Now it follows that
$$\phi(A\backslash [1,n])\\=\phi (\cup_{n\gt n_j}(A_j\cap (n,n_{j+1}])\bigcup (\cup_{n\le n_j}(A_j\cap (n_j,n_{j+1}])))\\ \le \sum_{n\gt n_j}\phi(A_j\cap (n,n_{j+1}])+\sum_{n\le n_j}\phi (A_j\cap (n_j,n_{j+1}])))$$ Now from here, how do I say that $\phi(A\backslash [1,n]) \ge {1\over r}?$. Please help. Thanks.
Since the sequence $n_j$ is increasing, we have $\lim\limits_{j\to\infty} n_j=+\infty$.
So for any $n$ there exists $j$ such that $n_j>n$. This means that $(A_j\cap(n_j,n_{j+1}])\setminus[1,n]=A_j\cap(n_j,n_{j+1}]$ and \begin{align*} A\setminus [1,n] &\supseteq A_j\cap(n_j,n_{j+1}]\\ \phi(A\setminus [1,n]) &\ge \phi(A_j\cap(n_j,n_{j+1}]) \ge \frac1r-\epsilon_j \end{align*} (Notice that $j$ in the above inequality depends on $n$, but the notation like $n_{j_n}$ or $n_{j(n)}$ would be rather cumbersome. So I simply denoted it $n_j$, but we should keep in mind that $j$ and $n_j$ depend on $n$, we have chosen $j$ in such way that $n_j>n$.)
If $n\to\infty$, then $n_j\to\infty$, $j\to\infty$ and $\epsilon_j\to 0$. Thus we get $$ \lim\limits_{n\to\infty}\phi(A\setminus [1,n]) \ge \lim\limits_{j\to\infty}\frac1r-\epsilon_j = \frac1r. $$