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The definition of Monotone classes says it should be closed under countable increasing unions.
That is, if the sequence of sets, {$E_n$}$_{n\in \mathbb{N}}\subset C$, and $E_1\subset E_2\subset E_3\subset ....... $ , then $\cup_{n=1}^\infty E_n\in C$
But if the sequence is increasing, then isn't it true that the last set, will contain all the other sets and since it's already in C, what's the whole point of the countable increasing union being closed. I think I'm confused.
Can anyone help?

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Beware that according to the definition, a monotone class is a certain set of subsets. Let's fix some notation. We call $X$ the "environment". The $E_i$ are subsets of $X$. Then $C$ is a collection of subsets of $X$. Then:

$C$ is a monotone class if for all sequence $(E_n)_{n \ge 1}$ of elements of $C$ (i.e. $E_i \in C$) s.t. $E_i \subset E_{i+1}$ , then $\bigcup_{n \ge 1}E_n \in C$.

Note that the relation between the $E_i$ and $C$ is a membership relation $\in$, not an inclusion $\subset$ (which does not really make sense here because $C$ and $E_i$ are different "types" of objects).

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    Damn. I just saw that it's a membership relation, not an inclusion.But still, each $E_i$ will contain some elements, right? And will the last set will contain each of those elements, and it still is a member of C, so isn't the Union always closed? and I'll edit the question to make the containment correct. – 2017-02-06
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    @Sum-Meister: I'm not sure to understand. $\bigcup_{n \ge 1} E_n$ is not necessarily in $C$. If it is the case for all $(E_n)$, then $C$ is called a monotone class (or, equivalently "$C$ is closed by countable monotone union"). – 2017-02-06
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    A counterexample will be the best thing right now....I don't get how can the union not be in C. – 2017-02-06
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    Consider $\mathbb{Z},$ and the collection of subsets $C=\{\varnothing,\mathbb{Z}\}\cup\{S\subset\mathbb{Z}:|S|<\infty\}.$ Then $C$ is closed under finite unions and intersections, but not under countable increasing unions: $\bigcup_{k=1}^{\infty}\{-2k,2-2k,\ldots,2k-2,2k\}=2\mathbb{Z}\not\in C.$ – 2017-02-06