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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

$\int_{F}fd\mu = \lim_{n \to \infty} \int_{F_n}fd\mu$

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    I'm so sorry, but I'm new in the website, I will follow the guidelines since now. What I've tried is to redefine the sequence in this way: $E_1 = F_1, E_2 = F_2 - F_1, ...E_n =F_n - F_{n-1}$, later I tried to make the integral over the characteristic function for use the convergence dominated theorem but I'm a few confused about this way, I'm ok?2012-09-16

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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 $\mu$.

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    +(1) Nicely done.2012-09-16
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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. :-)

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    That is quite clever and definitely neater. Thank you for your feedback. I really appreciate it. In the absence of an instructor/mentor hopefully in near future i can get some more measure theory proofs reviewed from you :-)2012-09-16