About an integral over measurable sets Let $(X, \Sigma, \mu)$ a measurable space and $f$ an integrable function. Show that if $(F_n)_{n\in\mathbb N}$ is a decreasing sequence of measurable sets and $F=\bigcap_{n} F_n$, then
$$\int_{F}fd\mu = \lim_{n \to \infty} \int_{F_n}fd\mu$$
 A: Fix $\varepsilon>0$. By definition of Lebesgue integral, we can find a simple function $g=g_{\varepsilon}$ such that $\int_X(|f|-g)d\mu\leq \varepsilon$. Hence we have
$$\left|\int_Ffd\mu-\int_{F_n}fd\mu\right|\leq \int_X|f|(\chi_{F_n}-\chi_F)d\mu\leq \varepsilon+\int_Xg(\chi_{F_n}-\chi_F)d\mu,$$
so we have to prove the result when $g$ has the form $\sum_{k=1}^Na_k\chi_{A_k}$, with $A_k\in \Sigma$ have finite measure and $a_k\geq 0$. So we get 
$$\limsup_{n\to +\infty}\left|\int_Ffd\mu-\int_{F_n}fd\mu\right|\leq\varepsilon+
\sum_{k=1}^Na_k\limsup_{n\to +\infty}\left[\mu(A_k\cap F_n)-\mu(A_k\cap F)\right].$$
Now we will be able to conclude after having showed the following result:

If $(X,\Sigma,\mu)$ is a measure space with $\mu$ positive, and $\{B_n\}\subset\Sigma$ is a decreasing sequence with $\mu(B_0)<\infty$, then $\mu(B_n)\to \mu\left(\bigcap_{j=1}^{+\infty}B_j\right)$. To see that, we work with the sequence of pairwise disjoint sets $C_k:=B_k\setminus B_{k+1}$, and use $\sigma$-additivity of $\mu$. 

A: Well as per the question ${F_n}$ is a descending countable collection of measurable subsets of $\Sigma$ such that $F =\cap F_{n} $ 
Intuition Thoughts
Observe that when we calculate the above intersection we are excluding space from F with each intersection. This is done while iterating over a countably infinte collection of measurable sets so hence we should ask for the limit of this process.
Which would be $F =\cap  _{n=1}^{\infty }F_n = \lim_{n \to \infty} F_n$
This justifies us in stating
$$\int _{F}fd\mu = \int _{\cap  _{n=1}^{\infty }F_{n}}fd\mu =  \lim_{n \to \infty} \int_{F_n}fd\mu$$
Edit:
We could use Lebesgue dominated convergence Theorem to justify the limit. If you want more details i can improve my post.
A more rigorous Proof attempt 
Let $n$ be a natural number. Define $f_n = f.\chi_n$ where $\chi_n$ is the characteristic function of the measurable set $F_n$. Then $f_n$ is a measurable function on $F$ and $|f_n|\leq |f|$ on $F$ for all $n$.
We observe that under the above construction ${\{f_n\}}\rightarrow f$ pointwise a.e on $F$. Thus, by the Lebesgue Dominated Convergence Theorem.
$\int_F f =\lim_{n \to \infty} \int_{F}f_n = \lim_{n \to \infty} \int_{F_n}f_n =  \lim_{n \to \infty} \int_{F_n}f$.
Please suggest if firstly there is something wrong with the proof and if so how could i improve upon it. I am new to measure theory and relatively to proofs themselves. So this would give me an opportunity to improve and i would be very great full. :-)
