Prove that $\mathfrak{R}(f^{-1}(\mathfrak{N}))=f^{-1}(\mathfrak{R}(\mathfrak{N}))$. (a famous book by Kolmogorov and Fomin, a set ring) 
Definition:
Let $\mathfrak{R}$ be a non-empty set of sets.
$\mathfrak{R}$ is called a ring when $\mathfrak{R}$ satisfies the following conditions:
If $A\in\mathfrak{R}, B\in\mathfrak{R}$, then $A\triangle B\in\mathfrak{R}, A\cap B\in\mathfrak{R}$.


Definition:
Let $\mathfrak{S}$ be a non-empty set of sets.
We define $\mathfrak{R}(\mathfrak{S})$ as the smallest ring $\mathfrak{R}$ such that $\mathfrak{R} \supset\mathfrak{S}$.


Definition:
Let $f$ be a mapping from $M$ to $N$.
Let $\mathfrak{N}\subset 2^N$.
$f^{-1}(\mathfrak{N}):=\{A\in2^M | A = f^{-1}(B)\text{ for some } B\in\mathfrak{N}\}$.

I am reading a famous book by Kolmogorov and Fomin.
The authors wrote the following equation holds without a proof:

$\mathfrak{R}(f^{-1}(\mathfrak{N}))=f^{-1}(\mathfrak{R}(\mathfrak{N}))$.

I guess it is easy for many people to prove the above equation holds, but I cannot prove that.
I can prove that $f^{-1}(\mathfrak{R}(\mathfrak{N}))$ is a ring and $f^{-1}(\mathfrak{N})$ is a subset of  $f^{-1}(\mathfrak{R}(\mathfrak{N}))$, so $\mathfrak{R}(f^{-1}(\mathfrak{N}))\subset f^{-1}(\mathfrak{R}(\mathfrak{N}))$.
 A: Let $N_0=f[M]$, and let $\mathfrak{N}_0=\{A\cap N_0:A\in\mathfrak{N}\}$; then
$$\mathfrak{R}(\mathfrak{N}_0)=\{A\cap N_0:A\in\mathfrak{R}(\mathfrak{N})\}\,.$$
Clearly $f^{-1}(\mathfrak{N}_0)=f^{-1}(\mathfrak{N})$, so $f^{-1}(\mathfrak{R}(\mathfrak{N}_0))=f^{-1}(\mathfrak{R}(\mathfrak{N}))$.
Say that a subset $A$ of $M$ is saturated if it is a union of fibres of $f$, i.e., if $A=f^{-1}[f[A]]$. If $A$ and $B$ are saturated subsets of $M$, then so are $A\cap B$ and $A\mathop{\triangle}B$, and every member of $f^{-1}(\mathfrak{N})$ is saturated, so every member of $\mathfrak{R}(f^{-1}(\mathfrak{N}))=\mathfrak{R}(f^{-1}(\mathfrak{N}_0))$ must be saturated as well.
Now note that $f[A\cap B]=f[A]\cap f[B]$ and $f[A\mathop{\triangle}B]=f[A]\mathop{\triangle}f[B]$ for any saturated sets $A,B\subseteq M$. And $f[f^{-1}[A]]=A$ for each $A\in\mathfrak{N}_0$, so $\{f[R]:R\in\mathfrak{R}(f^{-1}(\mathfrak{N_0}))\}$ is a ring containing $\mathfrak{N}_0$, and therefore
$$\{f[R]:R\in\mathfrak{R}(f^{-1}(\mathfrak{N}_0))\}\supseteq\mathfrak{R}(\mathfrak{N}_0)\,.$$
Finally, then,
$$f^{-1}(\mathfrak{R}(\mathfrak{N}))=f^{-1}(\mathfrak{R}(\mathfrak{N}_0))\subseteq\mathfrak{R}(f^{-1}(\mathfrak{N}_0))=\mathfrak{R}(f^{-1}(\mathfrak{N}))\,,$$
completing the proof.
A: My answer is not rigorous. This is an idea.
We can write $\mathfrak{R}(\mathfrak{N})$ explicitly like the image below:

Let $A\in f^{-1}(\mathfrak{R}(\mathfrak{N}))$.
Then $A=f^{-1}(B)$ for some $B\in\mathfrak{R}(\mathfrak{N})$.
For example, we can write $B$ like this: $$B=(N_1\cap(N_2\triangle N_3))\triangle(N_4\cap N_5)$$ for some $N_1,N_2,N_3,N_4,N_5\in\mathfrak{N}$.
In general, $$f^{-1}(N_1\cap N_2)=f^{-1}(N_1)\cap f^{-1}(N_2)$$ and $$f^{-1}(N_1\triangle N_2)=f^{-1}(N_1)\triangle f^{-1}(N_2)$$ for any $N_1, N_2\in 2^N$.
So, $A=f^{-1}(B)=(f^{-1}(N_1)\cap (f^{-1}(N_2)\triangle f^{-1}(N_3)))\triangle(f^{-1}(N_4)\cap f^{-1}(N_5))\in\mathfrak{R}(f^{-1}(\mathfrak{N}))$.
So, $f^{-1}(\mathfrak{R}(\mathfrak{N}))\subset\mathfrak{R}(f^{-1}(\mathfrak{N}))$.
