How to show that Since $\sigma\text{-fields}$ are monotone classes, we have that $\sigma[\mathcal{C}]\supset m[\mathcal{C}]$ I am looking at the proposition from Probability theory, the proposition stated that:
Assume that $\mathcal{C}$ is a field. Then $m[\mathcal{C}]=\sigma[\mathcal{C}]$,where  $m\mathcal{[C]}$ is the minimal monotone class containing the field $\mathcal{C}$ and  $\sigma[\mathcal{C}]$ is the minimal $\sigma$-field generated by the same field $\mathcal{C}$.
In my teacher's lecture note, he said Since $\sigma\text{-fields}$ are monotone classes, we have that $\sigma[\mathcal{C}]\supset m[\mathcal{C}]$. However, I don't find this statement very obvious and try to prove this one myself. I tried to let $A\in m\mathcal{[C]}$ and show $A\in \sigma[\mathcal{C}]$.
Here's the definition for $\sigma[\mathcal{C}]$: $\sigma[\mathcal{C}]\equiv\cap\{\mathcal{F} :\mathcal{F}_\alpha\text{ is a }\sigma\text{-field of subsets of }\Omega$ which satisfies $\mathcal{C}\subset\mathcal{F_\alpha}\}$. The definition for $m[\mathcal{C}]$ should be similar. However, when I suppose $A\in m[\mathcal{C}]$ I don't know what should I say next. If I go on saying $A$ must satisfies $\cup^\infty A_n\in\cap\mathcal{F_\alpha}$ it makes no sense. Let along in $\sigma[\mathcal{C}]$ we may expect if $A\in\sigma[\mathcal{C}]$ then $A^c\in\sigma[\mathcal{C}]$, which may not be true for elements from the $m[\mathcal{C}]$.
Later on, my teacher explains his reasoning to me, where I found his logic is very straight forward, he argues: because $\sigma$-field are monotone classes, $\sigma[\mathcal{C}]$ is a monotone class containing $\mathcal{C}$, therefore it must in turn at least be as big as $m[\mathcal{C}]$ which contains $\mathcal{C}$ (because $m[\mathcal{C}]$ is the minimal monotone class containing $\mathcal{C}$). I can understand what he illustrated but I still don't get what I have wronged (so I cannot finish the proof). Could someone please point out 1) if my approach would work, what should it be like? 2) if my logic is completely wrong, where's the misconception?
Thank you so much!
 A: This result can be found in Halmos's measure theory:
If $\mathcal{A}$ is an algebra of sets, then the intersection of all monotone classes that contain $\mathcal{A}$ is $\sigma(\mathcal{A})$.
Here is a sketch of the proof:
The intersection $\mathcal{M}$ the intersection of all monotone classes that contain $\mathcal{A}$ is also a monotone class. Clearly $\mathcal{M}\subset\sigma(\mathcal{A})$. Define
$$
\mathcal{M}_0=\{B\in\mathcal{M}:X\setminus B\in\mathcal{M}\}
$$
Clearly $\mathcal{A}\subset\mathcal{M}_0$. If $\{B_n:n\in\mathbb{N}\}\subset\mathcal{M}_0$ is a monotone sequence, then  $\{X\setminus B_n:n\in\mathbb{N}\}\subset\mathcal{M}$ is also a monotone sequence. Thus  $\lim_n B_n\in\mathcal{M}$, and $X\setminus\lim_nB_n=\lim_n(X\setminus B_n)\in\mathcal{M}$. It follows that $\mathcal{M}_0$ is a monotone class, and so  $\mathcal{M}=\mathcal{M}_0$.
Define
$$
\mathcal{M}_1=\{B\in\mathcal{M}:A\in\mathcal{A}\,\text{implies}\,A\cup B\in\mathcal{M}\}
$$
Clearly $\mathcal{A}\subset\mathcal{M}_1$. If $\{B_n:n\in\mathbb{N}\}\subset\mathcal{M}_1$ is a monotone sequence and $A\in\mathcal{A}$ then, $\{B_n\cup A:n\in\mathbb{N}\}$ is a monotone sequence in $\mathcal{M}$. Thus $\lim_nB_n\in\mathcal{M}$, and  $A\cup\lim_nB_n=\lim_n(A\cup B_n)\in\mathcal{M}$.
It follows that $\mathcal{M}_1$ is a monotone class, and so  $\mathcal{M}_1=\mathcal{M}$.
Finally, define
$$
\mathcal{M}_2=\{B\in\mathcal{M}: A\in\mathcal{M}\,\text{implies}\, A\cup B\in\mathcal{M}\}
$$
As $\mathcal{M}_1=\mathcal{M}$, we have that $\mathcal{A}\subset\mathcal{M}_2$. If $\{B_n:n\in\mathbb{N}\}\subset\mathcal{M}_2$ is a monotone sequence, and $A\in\mathcal{M}$, then  $\{A\cup B_n:n\in\mathbb{N}\}$ is a monotone sequence in $\mathcal{M}$. Thus $\lim_nB_n\in\mathcal{M}$,  and $A\cup\lim_nB_n=\lim_n(A\cup B_n)\in\mathcal{M}$. It follows that $\mathcal{M}_2$ is a monotone class, and so $\mathcal{M}_2=\mathcal{M}$.
So far we have proven that  $\mathcal{M}$ is  an algebra of sets. Now let $\{B_n:n\in\mathbb{N}\}\subset\mathcal{M}$. Then $\Big\{D_n=\bigcup^n_{j=1}B_j: n\in\mathbb{N}\Big\}\subset\mathcal{M}$ is a monotone sequence, and so
$\lim_nD_n=\bigcup_nB_n\in\mathcal{M}$. Therefore $\mathcal{M}$ is a $\sigma$--algebra.
