Q:

Problematic deductions in measure theory proof,

In G.de Barra's text "Measure Theory and Integration" I'm struggling on a line in the proof of the fact that $\mathcal{M}$ (class of Lebesgue measurable sets) is a $\sigma$-algebra over $\mathbb{R}$. The line I'm confused with occurs proving closure under countable unions (pg.  31 if you have the text.) 
If $\{E_j\}$ is a sequence of sets in $\mathcal{M}$ we have established that for an arbitrary set $A$ and any positive integer $n$: 
$$m(A)\geq m(A\cap E_1) + \sum_{i=2}^{n}m \bigg(A \cap E_i\cap \big (\bigcup_{j<i}{E_j} \big)^c\bigg)+m\bigg(A\cap \big(\bigcup_{j=1}^{\infty}E_j\big)^c\bigg)$$
He then says that it therefore follows that: 
$$m(A)\geq m(A\cap E_1) + \sum_{i=2}^{\infty}m \bigg(A \cap E_i\cap \big (\bigcup_{j<i}{E_j} \big)^c\bigg)+m\bigg(A\cap \big(\bigcup_{j=1}^{\infty}E_j\big)^c\bigg)$$
I cannot follow this jump as because the measure is always $\geq 0$: going from a finite summation of measure to an infinite summation of measure can only increase the value correct?  There must be some nuanced aspect to the sets involved or something that I'm not seeing.  I'd appreciate being shown why we can make this deduction.  

A:

Take the limit of both sides of your first inequality as $ n \to \infty$ to get the final inequality. 
Note if $x_i > 0$ and $\sum_{i=1}^nx_i \leqslant x < \infty$ for all $n$ then the partial sums form a bounded monotonically increasing sequence.  Hence, the series converges and $\sum_{i=1}^\infty x_n \leqslant x.$
Apply this to $x_i = m \bigg(A \cap E_i\cap \big (\bigcup_{j<i}{E_j} \big)^c\bigg).$

