Construction of a measure space from some weird functional Here is the complete problem I am trying to solve, but currently, I am just interested in proving that $\Sigma$ is a $\sigma$-algebra.

Let $X$ be a set and $\phi: 2^X \to [0, \infty]$ be a functional such that:

*

*$\phi \varnothing = 0$.


*$\phi(A \cup B) \ge \phi A + \phi B$, for all $A,B \subseteq X$ disjoint.


*If $(A_n)$ is a non-increasing sequence of subsets of $X$ and $\phi A_0 < \infty$, then:
$$\phi\left( \bigcap_{n \in \Bbb N} A_n \right) = \lim_{n \to \infty} \phi A_n$$

*

*If $\phi A = \infty$, then for all $a \in \Bbb R$, there exists $B \subseteq A$ such that $a \le \phi B < \infty$
Put:
$$\Sigma = \{ E \subseteq X \mid \phi(A \cap E) + \phi(A \setminus E) = \phi A \text{ for every $A \subseteq X$} \}$$
Show that $(X, \Sigma, \phi\mid_{\Sigma})$ is a measure space.

It is straightforward to prove that $\varnothing \in \Sigma$ and that for any $E \in \Sigma$, we have $X \setminus E \in \Sigma$.
So, basically my question is how to establish that $\Sigma$ is closed under countable unions. Here is what I did:
If $E, F \in \Sigma$, then for any $A \subseteq X$, we have:
$$\phi(A \cap (E \cup F)) = \phi( (A \cap (E \cup F)) \cap E) + \phi((A \cap (E \cup F) \setminus E) \\ = \phi(A \cap E) + \phi((A \setminus E) \cap F)$$
Hence,
$$\phi(A \cap (E \cup F)) + \phi(A \setminus (E \cup F)) = \phi(A \cap E) + \phi((A \setminus E) \cap F) + \phi((A \setminus E) \setminus F) \\ = \phi(A \cap E) + \phi(A \setminus E) = \phi A$$
and so $E \cup F \in \Sigma$.
Hence $\Sigma$ is closed under finite unions. Here's a bunch of possibly useful things as well:

*

*$\Sigma$ is also closed under set difference: if $E, F \in \Sigma$, then $X \setminus (E \setminus F) = (X \setminus E) \cup F \in \Sigma \implies E\setminus F \in \Sigma$.


*To prove $E \in \Sigma$, it is sufficient to verify that for any $A \subseteq X$, $\phi A \le \phi(A \cap E) + \phi(A \setminus E)$ because of the second property of $\phi$.


*$\phi$ is increasing (in the sense that if $A \subseteq B$, then $\phi A \le \phi B$).
For each $n \ge 0$, put $G_n = \cup_{m \le n} E_m$. We have that $(G_n) \subset \Sigma$ and for $A \subseteq X$, if it is true that for some $n_0 \in \Bbb N$, $\phi(A \setminus G_{n_0}) < \infty$, then we are done: using the third property of $\phi$ and that $(G_n)$ is increasing and that $\cup E_n = \cup G_n$, we have
$$\phi \left( A \setminus \bigcup_n E_n \right) = \phi \left( \bigcap_n (A \setminus G_n) \right) = \lim \phi(A \setminus G_n) \ \ \ \ \ (1)$$
also we have:
$$\phi \left( A \cap \bigcup_n E_n \right) \ge \lim \phi(A \cap G_n) \ \ \ \ \ (2)$$
Thus, the sum of the LHS's in $(1)$ and $(2)$ is $\ge \lim \{ \phi(A \setminus G_n) + \phi(A \cap G_n) \} = \phi A$.

The other case is: for all $n \ge 0$, $\phi(A \setminus G_n) = \infty$.

Is there a way to finish the argument from here? Or would it be better to take a different approach?
P.S. For reference, this is problem $113Y(g)$ in Measure Theory by Fremlin.
 A: Notation: $Y^c= X$ \ $Y$ for $Y\subset X.$
To show that $\Sigma$ is closed under countable unions, once we know that $\Sigma$ is closed under finite unions and complements, it suffices to show that $\Sigma$ is closed under countable unions of pairwise disjoint collections.
Let $F=\{E_n\}_{n\in N}\subset \Sigma $ where $m\ne n \implies E_m\cap E_n=\emptyset.$  
For $n \in N$ let $D_n=\cup_{j=1}^nE_j.$
For $A\subset X $ we have $\phi (A)=\phi (A\cap D_n)+\phi (A\cap D_n^c).$ 
For $A\subset X$ we have  $\phi (A\cap (\cup F) )\geq \phi (A\cap D_n)+\phi (A\cap \cup_{j>n}E_j).$ 
For $A\subset X$ we have $ A\cap (\cup F)^c=\cap_{n\in N}(A\cap D_n^c) . $ And since $A\cap D_{n+1}^c\subset A\cap D_n^c$ we have $\phi (B\cap (\cup F)^c)=\lim_{n\to \infty}\phi (B\cap D_n^c)$ whenever  $B\subset A$ and $\phi (B)<\infty.$
Therefore. with $\delta_n$ denoting a quantity that $\to 0$ as $n\to \infty,$ we have, for any $B\subset A$ such that $\phi (B)<\infty $ :$$\phi (B)\geq \phi (B\cap (\cup F))+\phi (B\cap (\cup F)^c)\geq$$ $$\geq \phi (B\cap D_n)+\phi (B\cap \cup_{j>n}E_n)+\phi (B\cap D_n^c)+\delta_n=$$ $$=[ \;\phi (B\cap D_n)+\phi (B\cap D_n^c)\;]+\phi (B\cap \cup_{j>n}E_j)+\delta_n=$$ $$=\phi (B)+\phi (B\cap \cup_{j>n}E_j)+\delta_n\geq$$ $$\geq  \phi (B)+\delta_n.$$ Letting $n\to \infty$ we have $$\phi(B)\geq \phi (B\cap (\cup  F))+\phi (B\cap (\cup F)^c)\geq \phi (B).$$  Now if $\phi (A)<\infty$ we can choose $B=A$ and we are done.
But if $\phi (A)=\infty$ then we can choose $B\subset A$ with $\phi (B)$ finite but as large as we like, and we have $$\phi (A)\geq \phi (A\cap (\cup F))+\phi (A\cap (\cup F)^c)\geq$$ $$\geq \sup \{\phi (B\cap (\cup F))+\phi (B\cap (\cup F)^c) :B\subset A \land \phi (B)<\infty \}=$$ $$=\sup \{\phi (B): B\subset A \land  \phi (B)<\infty \}=\infty =\phi (A).$$
