Measurable w.r.t a set but not w.r.t another set Let $f:\mathbb{N}\rightarrow\mathbb{R}$ be  a function defined by $f(x)=\sqrt{x}$, for $x\in\mathbb{N}$.
Show that $f$ is $\mathcal{P}(\mathbb{N})/\mathcal{B}(\mathbb{R})$-measurable but not $\sigma(A)/\mathcal{B}(\mathbb{R})$ for a set $A=\left \{ \left \{ 1,2,3 \right \}, \left \{ 4,5,6 \right \}, \left \{ 7,8,9 \right \}, \left \{ 10,11,12 \right \},... \right \}$

I know a set is measurable if the function between the underlying sets of two measurable spaces that preserves the structure of the space. The preimage of any measurable set is measurable.
I also see that my set $A$ is of the form $\left \{ n,n+1,n+2 \right \}$, where $n$ is a natual number and $3|(n+2)$. However, it seems undoable to show that it is measurable (the function) in these 2 cases.
 A: Let $f:\mathbb{N}\rightarrow\mathbb{R}$ be  a function defined by $f(x)=\sqrt{x}$, for $x\in\mathbb{N}$.
Item 1. Let us prove that  $f$ is $\mathcal{P}(\mathbb{N})/\mathcal{B}(\mathbb{R})$-measurable.
This is immediate, since for any $C \in \mathcal{B}(\mathbb{R})$, $f^{-1}(C)\in \mathcal{P}(\mathbb{N})$. So, $f$ is $\mathcal{P}(\mathbb{N})/\mathcal{B}(\mathbb{R})$-measurable.
Item 2. Let $A=\left \{ \left \{ 1,2,3 \right \}, \left \{ 4,5,6 \right \}, \left \{ 7,8,9 \right \}, \left \{ 10,11,12 \right \},... \right \}$
For a proof that $\{1\} \notin \sigma(A)$, see Remark below.
Now note that $\{1\} \in \mathcal{B}(\mathbb{R})$, and $f^{-1}(\{1\}) = \{1\} \notin \sigma(A)$. So, $f$ is not $\sigma(A)/\mathcal{B}(\mathbb{R})$.
Remark: The fact that $\{1\} \notin \sigma(A)$ can be proved in a easy way from a broader result.

Lemma. Let $X$ be a set and $\mathcal{S}=\{A_i\}_{i\in I}$ a countable partition of $X$. Then $\sigma(\mathcal{S})= \{ \bigcup_{i\in K} A_i : K \subseteq I \}$

Proof:
It is easy to check that $\{ \bigcup_{i\in K} A_i : K \subseteq I \}$  is a $\sigma$-algebra. In fact,

*

*$\emptyset = \bigcup_{i\in \emptyset} A_i$;

*Since $\mathcal{S}$ is a partition of $X$, $X \setminus (\bigcup_{i\in K} A_i)=\bigcup_{i\in I \setminus K} A_i $;

*Given a countable family $\{\bigcup_{i\in K_j} A_i\}_{j \in J}$, then
$\bigcup_{j \in J}(\bigcup_{i\in K_j} A_i)= \bigcup_{i\in \bigcup_{j \in J} K_j} A_i$;

*Clealy $X=  \bigcup_{i\in I} A_i $
So $\{ \bigcup_{i\in K} A_i : K \subseteq I \}$  is a $\sigma$-algebra. Since $\mathcal{S} \subseteq \{ \bigcup_{i\in K} A_i : K \subseteq I \}$, we have that
$\sigma(\mathcal{S}) \subseteq \{ \bigcup_{i\in K} A_i : K \subseteq I \}$.
Now, since $\mathcal{S}=\{A_i\}_{i\in I}$ a countable partition of $X$, we have that $I$ is countable and so, every $K \subseteq I$ is countable. So, for every $K \subseteq I$, $\bigcup_{i\in K} A_i \in \sigma(\mathcal{S})$. So, $ \{ \bigcup_{i\in K} A_i : K \subseteq I \} \subseteq \sigma(\mathcal{S})$.
Thus we proved that $\sigma(\mathcal{S}) = \{ \bigcup_{i\in K} A_i : K \subseteq I \}$. $\square$
Now let us consider the second item of the question. Assuming $\Bbb N =\{1, 2, \dots \}$. We have that
$A=\left\{\{1,2,3\},\{4,5,6\},\{7,8,9\},\{10,11,12\}\dots\right\}$ is a countable partition of $\Bbb N$. By the lemma above, any $D \in \sigma(A)$ is the union of elements of $A$. So, it is easy to see that $1 \in D$ if and only if $2, 3 \in D$. So $\{1\} \notin \sigma(A)$
If we assume that $\Bbb N =\{0, 1, 2, \dots \}$, then consider the countable partition
$C=\left\{\{0\},\{1,2,3\},\{4,5,6\},\{7,8,9\},\{10,11,12\}\dots\right\}$. Then, repeating the previous argument, we have $\{1\} \notin \sigma(C)$. Since $A \subseteq C$, it is immediate that  $\sigma(A) \subseteq \sigma(C)$. So $\{1\} \notin \sigma(A)$.
