Proved Lemma $3.5$ from the paper $I^K$-convergence by Macaj & Sleziak ; need it checked $I^K$ Convergence by Sleziak & Macaj
Lemma $3.5$. If $I$ and $K$ are ideals on a set $S$ and $f : S → X$ is a function such that   $K\lim f = x$,then $I^K \lim f = x$.
I've done the proof I need it checked for correctness::
proof: Take any $M\in \mathcal F(I)$. Define $$ g(s) =
\begin{cases}
f(s),  & s\in M \\
x, & s\notin M
\end{cases}$$
To show $K \lim g=x$ we have to show $$g^{-1}(U)\in \mathcal F(K)$$
Now $$g^{-1}(U)=\{s\in S:g(s)\in U\}\\ \subset \left(\left(\{s\in S: f(s)\in U\}\right)\cap M \right)\cup \{s\in S: s\notin S\}\\ \subset\{s\in S:f(s)\in U\}\cup (S\backslash M)\\ \in \mathcal f(K) $$ because $\{s\in S : f(s)\in U\}\in \mathcal F(K)$ and $\mathcal F(K)$ being a filter, contains the supersets of its members. 
Hence we have $K\lim g=x$ implying $I^k \lim f=x$(proved)
Thank you. 

The necessary Definitions :

$I$-Convergence :Let $I$ be an ideal on a set $S$ and $X$ be a topological space. A function $f : S → X$  is said to be $I$-convergent to $x ∈ X$ if
$f^{−1}(U)=\{s ∈ S; f(s) ∈ U\} ∈ \mathcal F(I)$ holds for every neighborhood $U$ of the point $x$.
We use the notation $I- \lim f = x$.
$I^K$-Convergence: Let $K$ and $I$ be ideals on a set $S$, let $X$ be a topological space and let $x$ be an element of $X$. The function $f : S → X$ is said to be $I^K$-convergent to $x$ if there exists a set $M ∈ \mathcal F(I)$ such that the function $g : S → X$ given by
$$ g(s) =
\begin{cases}
f(s),  & s\in M \\
x, & s\notin M
\end{cases}$$
is $K$-convergent to $x$. If $f$ is $I^K$-convergent to $x$, then we write $I^K \lim f = x$.
 A: Definition of $\mathcal I^{\mathcal K}$ convergence says that there exists a set $M\in\mathcal F(\mathcal I)$ such that $f$ is $\mathcal K$-convergent "along" $M$.
To prove the above lemma, you can simply take $M=S$. In this case you get $g=f$ and thus you immediately have that $g$ is $\mathcal K$-convergent. 

You made things slightly more difficult in that you are trying to prove the same thing for every $M\in\mathcal F(\mathcal I)$. (However, this is still true - if it is $\mathcal K$-convergent on $S$, the same will be true for $\mathcal K$-convergence "along" subsets.)
Your proof seems to contain some typos and minor mistakes, but if you realize that the inclusions you wrote are actually equalities and you have
\begin{align*}
g^{-1}(U) &= \{s\in S; g(s)\in U\}\\
&= (\{s\in S; f(s)\in U\}\cap M) \cup (S\setminus M)\\
&= \{s\in S; f(s)\in U\} \cup (S\setminus M)\\
&= f^{-1}(U) \cup (S\setminus M)
\end{align*}
then you get that $g^{-1}(U)\in \mathcal F(\mathcal K)$.
The way your proof is currently presented it contains some typos and minor problems. Here is quote from the current revision of your post:

Now $$g^{-1}(U)=\{s\in S:g(s)\in U\}\\ \subset \left(\left(\{s\in S: f(s)\in U\}\right)\cap M \right)\cup \{s\in S: s\notin S\}\\ \subset\{s\in S:f(s)\in U\}\cup (S\backslash M)\\ \in \mathcal f(K) $$ because $\{s\in S : f(s)\in U\}\in \mathcal F(K)$ and $\mathcal F(K)$ being a filter, contains the supersets of its members.

Some things seem to be obvious typos (but I decided not to correct them myself - since you asked for critique of your proof, maybe it is better not to edit it in places where this could change the intended meaning):


*

*You probably wanted to write $\{s\in S: s\notin M\}$ instead of $\{s\in S: s\notin S\}$.

*You probably wanted to write $\mathcal F(K)$ instead of $\mathcal f(K)$.


The main problem is that you want to show that $g^{-1}(U)$ is a superset of some set from the filter. Instead what you wrote is that it is a subset of a set from the filter. (But since the inclusions you wrote are in fact equalities, this can be easily corrected.) 
