Monotone convergence theorem in a special case 
Suppose $C^{+}[0,1]$ be the set of all continuous functions with
  domain $[0,1]$ taking non-negative values only. Let $\lambda : C^{+}[0,1] \to [0,\infty)$ be a map that satisfies
  $$\lambda(f+g)=\lambda(f)+\lambda(g)$$ Now suppose $\{g_k\}$ be a
  sequence of functions in $C^{+}[0,1]$ that satisfies $$0\le g_1\le g_2\le \cdots, \ \ \ \  \ \ \ \ \ g_k \to g \in C^{+}[0,1]$$ Show that
  $$\lim_{k\to\infty} \lambda(g_k)=\lambda(g)$$

My progress:
I was able to show that $\lim\limits_{k\to\infty} \lambda(g_k)\le \lambda(g)$. So, I am trying to show $\lambda(g) \le \lim\limits_{k\to\infty} \lambda(g_k)$. I am stuck here.
I have knowledge of the proof of actual monotone convergence theorem. There it uses the notion of simple function which are easy to handle and then they prove the general case by taking the supremum over simple functions. But note that simple functions do not belong to $C^{+}[0,1]$. So I think we cannot use the same technique. 
Any idea to tackle this one? Hints, links, solutions all are welcome.

We consider $f_n = g-g_n \in C^+([0,1])$. Then $f_n\ge f_{n+1}\ge \cdots$ and $\{f_n\}$ converges uniformly to the zero function. Consider the constant function $I(x) = 1$ and let $C = \lambda (I)$. One can show by induction that 
$$\lambda \left(\frac 1k I\right) = \frac{C}{k}.$$
Now for each $k$, there is $n_k$ so that $f_m \le \frac 1k I$ for all $m \ge n_k$. Thus 
$$\lambda (f_m) \le \frac Ck$$
for all $m\ge n_k$. Hence $\lim_{n\to \infty} \lambda (f_n) = 0 = \lambda (0)$. Thus $\lim_{n\to \infty} \lambda(g_n) = \lambda (g)$. 
