Trying to prove Theorem $3.9$ from $I^K$ *Convergence* by Sleziak & Macaj From $I^K$ Convergence by Sleziak & Macaj


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*Theorem $3.9$ says The equivalence of (iii), (iv), (v) can be easily shown by the standard methods from the measure theory.
I know very little Measure Theory. So can I prove them with other methods? I've tried and done a few the rest are stuck. Any help is greatly appreciated.
For example,$iii)\implies iv)$ seemed pretty trivial.Here is what I did to prove $iv)\implies iii)$:
$(A_n)_{n\in \mathbb N}$ is a sequence of sets in $I.$ We define a new sequence $(A'_n)$ by $A'_1=A_1 \text{ and } A'_n=A_n\backslash \left(\cup_{i=1}^{n-1}A_i\right).$ Now for each $n\in \mathbb N,$ $A'_n\subset A_n\in I$ hence $(A'_n)$ is also a sequence of sets from $I$ and they are mutually disjoint too. 
So by $iv$, we know that  $\exists$ a sequence of sets $(B_n)_{n\in \mathbb N}$  of sets in $I$ s.t. $$\cup_{i=1}^{\infty}B_n\in I \text{ and }B_n \sim_{K} A'_n.$$ Next I try to prove that $A_n\sim_K B_n$ also holds ,i.e. both $A_n\backslash B_n\in K$ and $B_n\backslash A_n \in K$ are true at the same time.
$$B_n\backslash A_n \in K\\\implies B_n\backslash \left(A_n\backslash \bigcup_{n=1}^{n-1}A_i\right)\in K\\\implies B_n \cap \left(A_n\backslash \bigcup_{n=1}^{n-1}A_i\right)^c \in K \\\implies B_n \cap \left(A_n^c\cup \left(\bigcup _{i=1}^{n-1}A_i\right)\right)\in K\\\implies \left(B_n\cap A^c_n\right)\cup \left(B_n\cap \left(\bigcup_{i=1}^{n-1}A_i\right)\right)\in K\\\implies B_n\cap A_n^c \in K \\\implies B_n\backslash A_n \in K.$$
For the other part $$A'_n\backslash B_n \in K\\\implies \left(A_n\backslash \left(\bigcup_{i=1}^{n-1}A_i\right)\right)\cap B_n^c \in K\\\implies \left(A_n\cap \left(\bigcap_{i=1}^{n-1}A_i^c\right)\right)\cap B_n^c\in K\\\implies \left(A_n\bigcap B_b^c\right)\cap\left(\bigcap_{i=1}^{n-1}A_i^c\right)\in K\\\implies (A_n\backslash B_n)\cap \left(\bigcap_{i=1}^{n-1}A_i^c\right)\in K$$ If only I could show $(A_n\backslash B_n)\subseteq \left(\bigcap_{i=1}^{n-1}A_i^c\right)$ my proof would be done.
Help kindly or is it impossible to do it without Measure Theory $?$
 A: Probably a reasonable way to start is to notice some properties of the relation$\newcommand{\I}{\mathcal{I}}\newcommand{\J}{\mathcal{K}}\newcommand{\simJ}{\sim_{\J}}\newcommand{\N}{\mathbb N}$
$$A\simJ B \Leftrightarrow A\triangle B\in\J$$
on the set $\mathcal P(S)$.
For any ideal $\J$ on the set $S$, this relation is congruence on the Boolean algebra $\mathcal P(S)$. Without using big words, this simply says that the relation $\simJ$ "behaves nicely" w.r.t. set operations, such as union, intersection, complement. For example, we can notice that for $X_1\simJ Y_1$, $X_2\simJ Y_2$ we have


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*$X_1\cup Y_1\simJ Y_1\cup Y_2$;

*$X_1\cap Y_1\simJ Y_1\cap Y_2$;

*$S\setminus X_1 \simJ S\setminus Y_1$;

*$X_1\setminus X_1\simJ Y_1\setminus Y_2$.


Proofs of these properties should be relatively straightforward. It is also to see that the first two properties can be generalized to union and intersection of finitely many sets.

We also define $A\subseteq_{\J} B$ by $A\setminus B\in\J$.
Now the above properties can be used to show that these conditions are equivalent:


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*(i) For every sequence $(A_n)_{n\in\N}$ of sets from $\I$ there is $A\in\I$ such that $A_n\subseteq_{\J} A$ for all $n$'s.

*(iii) For every sequence $(A_n)_{n\in\mathbb N}$ of sets belonging to $\I$ there exists a sequence $(B_n)_{n\in\mathbb N}$ of sets from $\I$  such that $A_j \simJ B_j$ for $j\in\mathbb N$ and $B=\bigcup_{j\in\N} B_j\in\I$.

*(iv) For every sequence of mutually disjoint sets
$(A_n)_{n\in\mathbb N}$ belonging to $\I$ there exists a sequence
$(B_n)_{n\in\mathbb N}$ of sets belonging to $\I$ such that $A_j \simJ B_j$
for $j\in\N$ and $B=\bigcup_{j\in\N} B_j\in\I$.

*(v) For every non-decreasing sequence $A_1\subseteq A_2 \subseteq \dots \subseteq A_n \subseteq \dots$
of sets from $\I$ there exists a sequence
$(B_n)_{n\in\N}$  of sets belonging to $\I$ such that $A_j \simJ B_j$
for $j\in\N$ and $B=\bigcup_{j\in\N} B_j\in\I$.


We immediately have that (iii) implies both (iv) and (v).
EDIT: As was pointed out in a comment, the following proof only shows that (iv) implies (v) and not, as I originally claimed, that (iv) implies (iii). To get the equivalence of these conditions, I had to include also the condition (i) (see below). At the moment, I do not have a direct proof that either (iv) or (v) implies (iii) (without going through the condition (i).)
Proof of (iv)$\Rightarrow$(v). Let $(A_n)$ be a non-decreasing sequence of sets from $\I$. Then we define $A'_n=A_n\setminus\left(\bigcup\limits_{i=1}^{n-1}A_i\right)$. Then $A'_n$ is a sequence of mutually disjoint sets from $\I$. By (iv) there exists a sequence of sets $B'_i\in\I$ such that $A'_i\simJ B'_i$ for each $i$ and also $\bigcup B'_i\in\I$.
If we put $B_n=\bigcup\limits_{i=1}^n B'_i$, then we get
$$A_n = \bigcup_{i=1}^n A'_i \simJ \bigcup_{i=1}^n B'_i = B_n$$
and each $B_n$ belongs to $\I$. Moreover, $\bigcup_{i\in\N} B_n=\bigcup_{i\in\N} B'_i \in \I$. $\square$
Proof of(v)$\Rightarrow$(iv). Let $(A_n)$ be a sequence of mutually disjoint sets from $\I$. We put $A'_n=\bigcup\limits_{i=1}^n A_n$. Then $(A'_n)$ is a non-decreasing sequence, so (v) gives us a sequence of sets $B'_n\in\I$ such that $B'_n\simJ A'_n$ for each $n$ and $\bigcup B'_n\in\I$.
Now if we put $B_n=B'_n\setminus B_{n-1}$ then
$$ A_n = A'_n \setminus A_{n-1} \simJ B'_n \setminus B'_{n-1} = B_n$$
and $\bigcup\limits_{n\in\N} B_n = \bigcup\limits_{n\in\N} B'_n \in\I$. And we also have that $B_n\in\I$ for each $n$. $\square$ 
Now in the paper you linked, the implications (i)$\Rightarrow$(v) and (iii)$\Rightarrow$(i) are shown. However, the proof of (i)$\Rightarrow$(v) without any substantial change also gives:
Proof of (i)$\Rightarrow$(iii). Let $A_n\in\I$ for $n\in\N$. Let
$A'_n=\bigcup\limits_{i\leq n}A_i$. Since each $A'_n\in\I$, the condition
(i) yields existence of $A\in\I$ with $A'_n\subseteq_{\J} A$ for
$n\in\N$. Now we put $B_n:=A\cap A_n$. Since $B_n\subseteq A$, we
have $B_n\in\I$. Moreover, $B_n\triangle A_n =A_n\setminus A \subseteq
A'_n\setminus A \in \J$, thus $B_n\simJ A_n$. Finally, $\bigcup_{j\in\N}
B_j\subseteq A\in\I$. $\square$
It is worth mentioning that this is basically the same proof as the proof of the corresponding part of Proposition 1 in Balcerzak, Dems, Komisarski: Statistical convergence and ideal convergence for sequences of functions, Journal of Mathematical Analysis and Applications, Vol. 328, No. 1, p. 715-72,
doi: 10.1016/j.jmaa.2006.05.040. (The only differnce is that in that paper the authors work with $\J=\mathrm{Fin}=$the ideal consisting of all finite subsets of $\mathbb N$).
