A finite additive measure $\displaystyle \mu $ is a measure iff it is continuous from above. [sorry, I made a mistake in the title, should be "from above"]

I finished the proof from finite additive to continuous from above, but stuck on the other one...

Proof so far.

Suppose that $\displaystyle \mu : \mathbb {M} \rightarrow [0, \infty ] $ is continuous from above, that is, if $\displaystyle \{ E_j \} ^ \infty _{j=1} \subset \mathbb {M} $ with $\displaystyle E_{j+1} \subset E_j $, then I have $\displaystyle \lim _{j \rightarrow \infty } \mu (E_j)= \mu ( \bigcap ^ \infty _{j=1} E_j ) $.

I will show that $\displaystyle \mu $ is a measure.

Let $\displaystyle \{ E_j \} ^n _{j=1} \subset \mathbb {M} $ be disjoint sets.

Claim: $\displaystyle \mu ( \bigcup ^n _{j=1} E_j) = \sum _{j=1}^n \mu (E_j) $

Let $\displaystyle F_n = \bigcup ^ \infty _{j=1}E_j \ \bigcup ^n _{i=1} E_i$, note that $\displaystyle F_j \subset F_{j+1} $ and $\displaystyle \bigcup ^n_{j=1}F_j$

Now, since $\displaystyle \mu $ is continuous from above, I have $\displaystyle \lim _{n \rightarrow \infty } \mu (F_n) = \mu ( \bigcup ^ \infty _{j=1} F_j ) $

I'm defining the sequence as such base on how I work the other direction, but I can't seem to be able to break the left hand side into a sum of $\displaystyle \mu (E_1) + . . . + \mu (E_n) $.

Any hints? Thank you