Lower-semicontinuous submeasure on $\mathbb N$ vs. function on Cantor space On the page 7 of Farah's Analytic Quotients, immediately after the definition of lower semicontinuous  submeasure (see below) there is a brief remark saying that "This obviously corresponds to $\phi$ being lower semicontinuous in the Cantor-set
topology on $\mathcal P(\mathbb N)$." I hope I correctly interpreted this remark as:

Let $\phi$ be a submeasure on $\mathbb N$. Then $\phi$ is lower semicontinuous (as a submeasure) if and only if the corresponding function $\{0,1\}^{\mathbb N}\to{[0,{\infty}]}$ is lower semicontinuous w.r.t. the product topology. (Here $\{0,1\}$ is endowed with the discrete topology, so $\{0,1\}^{\mathbb N}$ is product of countably many two-point discrete spaces.)

This observation seems to be useful when dealing with lsc submeasures. (For example, we immediately get the fact that supremum of lsc submeasures is again a lsc submeasure - since supremum of any family of lower semicontinuous function is again lower semicontinuous.)
I would be grateful for any comments on my proof posted as an answer, alternative proofs, further insights into this.
Relevant definitions
A submeasure on $\mathbb N$ is a function ${\phi}\colon{\mathcal P(\mathbb N)}\to{[0,{+\infty}]}$ such that
\begin{gather*}
\phi(\emptyset)=0\\
\phi(A) \le \phi(A\cup B)\leq \phi(A)+\phi(B)
\end{gather*}
Notice that the second property also means that $A\subseteq B$ implies $\phi(A)\le\phi(B)$.
In the other words, a submeasure is a function ${\phi}\colon{\mathcal P(\mathbb N)}\to{[0,{+\infty}]}$ which is monotone and subadditive.
A submeasure $\phi$ is called lower semicontinuous if
$$\phi(A)=\lim_{n\to\infty} \phi(A\cap [1,n])$$
for every $A\subseteq\mathbb N$.
References


*

*Farah I. Analytic Quotients. Theory of Liftings for Quotients over Analytic Ideals on the Integers, Memoirs of AMS, 2000.

 A: $\newcommand{\powerset}[1]{\mathcal P({#1})}\newcommand{\intrv}[2]{[#1,#2]}$Recall that a function $f\colon X \to \mathbb R$ is lower semicontinuous iff
$$f^{-1}(a,\infty)=\{x\in X; f(x)>a\}$$
is open for every $a\in\mathbb R$.
This is equivalent to validity of
$$f(p) \le \liminf\limits_{x\to p} f(x)$$
for every $p\in X$. If $X$ is a metric space, it suffices to require
$$f(p) \le \liminf\limits_{n\to\infty} f(x_n)$$
for every sequence $x_n\to p$.

Proof. In the proof we are identifying $\powerset{\mathbb N}$ with $\{0,1\}^{\mathbb N}$ anyway; we will also use $\phi$ both for the submeasure (=function $\powerset{\mathbb N}\to{\intrv0\infty}$) and the corresponding function $\phi\colon{\{0,1\}^{\mathbb N}}\to{\intrv0\infty}$.
$\boxed{\Rightarrow}$ Let $a$ be a real number and $\phi(A)>a$.
From the semicontinuity of the submeasure $\phi$ we get that there exists $n_0$ such that
$$\phi(A\cap[1,n])>a$$
for each $n \ge n_0$. Therefore the set of all sets $B\subseteq\mathbb N$ such that
$$B\cap[1,n_0]=A\cap[1,n_0]$$
is a neighborhood $\mathcal U$ of $A$ (in the product topology) such that $\phi(B)>a$ for each $B\in\mathcal U$.
$\boxed{\Leftarrow}$ The sequence $A\cap[1,n]$ converges to $A$ in the product topology. So we have
$$\phi(A) \le \liminf_{n\to\infty} \phi(A\cap[1,n])$$
from lower semicontinuity. But since $A\cap[1,n]\subseteq A$, we also get
$$\limsup_{n\to\infty} \phi(A\cap[1,n]) \le \phi(A)$$
from monotonicity. $\square$
