$\phi$ is a lower semicontinous submeasure on $\mathbb N.$ $K\subset \mathbb N$ such that $$\lim_{n\rightarrow \infty}\phi(K\backslash [1,n])=\beta$$ is given.
Then the $K$ is decomposed into disjoint $K_i's$ such that $$K=\bigcup_{i=1}^{\infty}K_i.$$ and $$K_i=\bigcup_{j\in D_i}C_j;\left(\mathbb N= \bigcup_{j=1}^{\infty}D_j \text{ being a decomposition of }\mathbb N \text{ into disjoint subsets }\right)$$ where the $C_j's$ are finite disjoint subsets of $\mathbb N$ such that $C_j\subset K$ and $\lim_{j\rightarrow \infty}\phi(C_j)=\beta$
I have to show that $$\lim_{n\rightarrow \infty}\phi(K_i\backslash [1,n])=\beta.$$
Now I know that $$K_i\backslash [1,n]\subset K\backslash [1,n]\\ \implies \phi(K_i\backslash [1,n])\le \phi(K\backslash [1,n])\\ \implies \lim_{n\rightarrow \infty}\phi(K_i\backslash [1,n])\le \beta$$
On the other hand $$\lim_{n\rightarrow \infty }\phi(C_i\backslash [1,n])\le \lim_{n\rightarrow \infty}\phi(K_i\backslash [1,n])$$ If only I could show that the limit on the LHS is $\beta$ then I would get $$\beta \le \lim_{n\rightarrow \infty}\phi(K_i\backslash [1,n])\le \beta$$ implying $$\lim_{n\rightarrow \infty}\phi(K_i\backslash [1,n])=\beta.$$ But that is I cannot do.
Thanks for any help.
It is found in the paper Some further results on Ideal Convergence in Topological Spaces by P. Das.