The theorem states
Let $f$ be a function on a cell $A$ and let $f_n$ be a sequence of McShane (Lebesgue) integrable functions such that, $f_n\le f$ and $f_n\to f$. Than, if $\lim_{n\to\infty} \int_A f_n d\alpha$ is finite, than $$\int_A fd\alpha =\lim_{n\to\infty} \int_A f_nd\alpha$$
Now, let $$I=\lim_{n\to\infty}\int_A f_nd\alpha$$ and $$A=\cup_{i=0}^kA_i$$
The proof states
The theorem is proved by showing that $$|\sum_{i=0}^k f(x_i)\alpha(A_i) - I|<\epsilon|2-\alpha(A)|\tag{1}\label{1}$$
Question
As $\alpha(A_i)=\alpha(b)-\alpha(a)$ where $b$ and $a$ are the endpoints of the interval $A_i$, how can this be so? Suppose the range of $\alpha$ is $\mathbb{R}$. Does that not make the right hand side of \eqref{1} infinite? Surely the $\alpha(A)$ must be finite?
Please help me understand, I do not know why $\eqref{1}$ implies the theorem. As a self taught amateur, I have no one but stackexchange to ask...
Edit
To clarify, I know how to prove the theorem, if $I$ assume that the stateent \eqref{1} implies the theorem, as I know how to prove \eqref{1}. My problem is, that I do not understand why \eqref{1} implies the theorem, if, lets say, $\alpha(A)=\alpha(b)-\alpha(a)$ is not finite, because lets say, $a$ or $b$ are not finite. Than It would seem that the right hand side of \eqref{1} is not finite. So how does \eqref{1} imply the theorem even when $\alpha(A)$ is not finite?
Let $A = \cup_{i=1}^n A_i$ with $\alpha(A_i) < \infty$ and $A_i$'s are disjoint. WLOG, say $\alpha(A_i) = 1$. Then suppose $\int_A f_n d\alpha < \infty$. Then one can write $$ \int_A f_n d\alpha = \sum_{i=1}^\infty \int_{A_i} f_nd\alpha.$$ Thus let $I = \sum_{i=1}^\infty I_i = \int_A f d\alpha$ where $I_i = \int_{A_i} fd\alpha$. Thus it suffices to show that for each $i$, $$ \left|I_i - \int_{A_i} f_n d\alpha \right| < \frac{\epsilon}{2^i}.$$ Because this implies $$ \left|I - \int_A f_n d\alpha \right| \le \sum_{i=1}^\infty\left|I_i - \int_{A_i} f_n d\alpha \right| < \sum_{i=1}^\infty \frac{\epsilon}{2^i} \le \epsilon. $$