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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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Hint: assume $f \geq 0$ and consider the finite measure $d\nu = f d\mu$. – t.b. Sep 16 '12 at 19:47
Hi LFRC, welcome to Math.SE. This looks like a homework question, and for such questions we ask that you follow certain guidelines; see the FAQ at…;. In particular, simply stating a problem without any indication of what you tried, what you know, or which part is causing you trouble, is not acceptable. – Nate Eldredge Sep 16 '12 at 20:04
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? – LFRC Sep 16 '12 at 21:31

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. – Comic Book Guy Sep 16 '12 at 23:21

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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@t.b. Thank you for your feedback. I am relatively new to measure theory so please bear with me here. Choosing $F_n$ to be what you stated in your comment the intersection would result into a set with measure zero as it would be an empty set right. What about if i choose to formulate a proof based on $f_n = f\chi_n$ so that there is point wise convergence from the simple functions above to f on the subsets. Could I then justify using the Lebesgue dominated Convergence theorem to state the result ? – Comic Book Guy Sep 16 '12 at 20:48
Yes, that's absolutely right, Lebesgue dominated convergence leads to the result, as you intend. In my example I tried to point out that you could get $0 = \infty$ in the last displayed equation if you're not careful. The exercise is however much more elementary to solve than by using dominated convergence. Among the first facts one proves in measure theory is: if $F_n$ is a decreasing sequence of sets, $F = \bigcap F_n$ and for some $k$ we have $\mu(F_k) \lt \infty$ then $\mu(F) = \lim\limits_{n\to\infty} \mu(F_n)$. (this follows from $\sigma$-subadditivity of $\mu$) – t.b. Sep 16 '12 at 21:01
Sorry i was editing, so did not see your comments before. i am reading now. – Comic Book Guy Sep 16 '12 at 21:10
+1 now. Just a small suggestion for your more rigorous argument: I would define $g = f \chi_{F}$ and $g_n = f\chi_{F_n}$ then $|g_n|,|g| \leq |f|$ and $g_n \to g$ pointwise a.e. Thus, dominated convergence gives $$\int_{F} f = \int g = \lim_{n} \int g_n = \lim_{n} \int_{F_n} f$$ which is the desired conclusion. – t.b. Sep 16 '12 at 21:27
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 :-) – Comic Book Guy Sep 16 '12 at 21:39

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