Let $X$ be a compact metric space. Let $C_{+}(X)$ be the set of all continuous non negative functions on $X$. Let $\lambda : C_{+}(X) \to [0,\infty)$ such that $$\lambda(f+g)=\lambda(f)+\lambda(g)$$ Let $\phi_i \in C_{+}(X)$ for all $i$ such that $\sum\limits^\infty \phi_i \ge \phi$ where $\phi \in C_+(X)$. Show that $$\sum^{\infty} \lambda(\phi_i) \ge \lambda(\phi)$$
If $\sum\limits^\infty \phi_i \in C_{+}(X)$, then we are done by MCT. But $\sum\limits^\infty \phi_i$ may not be in $C_+(X)$.
It seems the following.
Assume the converse. Then
$$\varepsilon=\lambda(\phi)-\sum^{\infty}_{i=1}\lambda(\phi_i)>0.$$
Let $1_X:X\to \{1\}$ be a constant function. There exists a natural number $m$ such that $$\lambda((3/m)\cdot 1_X)= 3\lambda(1_X)/m<\varepsilon.$$
Let $x\in X$ be an arbitrary point. Since $$\sum\limits^\infty \phi_i \ge \phi,$$ there exists a number $n_x$ such that
$$\sum\limits_{i=1}^{n_x} \phi_i(x) \ge \phi(x)-1/m.$$ Since each of the functions $\phi$, $\phi_1,\dots, \phi_{n(x)}$ is continuous, there exists an open neighborhood $O_x$ of the point $x$ such that
$$|\phi(y)-\phi(x)|<1/m$$
and
$$|\phi_i(y)-\phi_i(x)|<1/(mn_x)$$ for each $i=1,\dots, n_x$ and each point $y\in O_x$.
Let $y\in O_x$ be an arbitrary point. Then
$$\sum\limits_{i=1}^{n_x} \phi_i (y)-\phi(y)=$$ $$\sum\limits_{i=1}^{n_x} \phi_i(x)-\phi(x)+\sum\limits_{i=1}^{n_x}(\phi_i (y)-\phi_i(x))-(\phi(y)-\phi(x))\ge $$ $$\sum\limits_{i=1}^{n_x} \phi_i(x)-\phi(x)- \sum\limits_{i=1}^{n_x}|\phi_i (y)-\phi_i(x)|-|\phi(y)-\phi(x)|\ge $$ $$-1/m-n_x\sum\limits_{i=1}^{n_x} 1/(mn_x)-1/m=-3/m.$$
The family $\{O_x:x\in X\}$ is an open cover of the space $X$. Since the space $X$ is compact, there exists a finite subset $F$ of the set $X$ such that $X=\bigcup\{O_x: x\in F\}$. Put $n=\max\{n_x:x\in F\}$. The construction of a neighborhood $O_x$ implies that $$\sum\limits_{i=1}^{n} \phi_i (y)-\phi(y)\ge -3/m$$ for each point $y\in O_x$ for each point $x\in F$. Since $X=\bigcup\{O_x: x\in F\}$, we have
$$\sum\limits_{i=1}^{n} \phi_i (y)-\phi(y)\ge -3/m$$ for each point $y\in X$. Then
$$\sum\limits_{i=1}^{n} \phi_i+(3/m)\cdot 1_X\ge\phi, $$ that is a function
$$\eta=\sum\limits_{i=1}^{n} \phi_i+(3/m)\cdot 1_X -\phi$$ belongs to $C_+(X)$.
Thus
$$\sum\limits_{i=1}^{n}\lambda(\phi_i)+ \lambda((3/m)\cdot 1_X)=\lambda(\phi)+ \lambda(\eta)\ge \lambda(\phi),$$
$$\lambda(\phi)>\sum^{\infty}_{i=1}\lambda(\phi_i)+\varepsilon\ge$$
$$\sum\limits_{i=1}^{n}\lambda(\phi_i)+ \lambda((3/m)\cdot 1_X)\ge \lambda(\phi),$$
a contradiction.