OMG I'm holding a bachelor degree in EE and I worked in the industry for 2 years. Now I'm back studying for my master and I doing a master level course in measure theoretic probability, which is quite a math shock as you can imagine. Those videos came in just a perfect time, thank you.
I really love how measure theory is so strongly connected to real analysis. The limit of f_n might not exist but the limit inferior must exist. Since the limit inferior is the smallest(infimum) accumulation value of the sequence. And because the interval [0, inf] is compact, there is at least one accumulation value on the interval, then we just find the smallest (infimum).
When you have an increasing series, X for example, we have the equality X_n = min(X_k, k ≥ n) = Inf(X_k, k ≥ n) since by definition of an increasing series, X_n is lesser than every value that comes after it. Since all g_n are point-wise monotonically increasing we know that the integrals of g_n are increasing and therefore you can substitute the integral of g_n by the infimum of g_k over k ≥ n. and finally you can turn the limit into limit inf.
I was confused by the inequality sign in the Fatou's lemma. It led me thinking: is there another result that would be "stronger", with equality sign, but that's just harder to prove and this happens to be enough for the further proofs, OR, is there some deeper reason *why* there is an inequality there. After googling for it a bit, turns out that there are cases where the inequality is strict! First of all, the general understanding: The left side, the function converges "pointwise" whereas on the right side the integral converges as a whole. Then, consider the function series 𝑓𝑛(𝑥) = {n for 0 ≤𝑥≤1/𝑛; 0 otherwise }. That is a function that on the limit n→∞, converges to 0 pointwise, because the "width" of the non-zero part gets smaller and smaller. However, the height of that part gets higher and higher; it's carefully defined for the area under the curve to stay as 1. That means that the integral as whole doesn't converge to zero, it stays bigger than the integral of the pointwise limit function f. I guess a similar argument is also the reason we talk about the limit infinum, not limit. (I don't know why though.) So here's the missing example / piece of intuition for those who were mystified by the statement of Fatou's lemma!
I guess for the "lim inf" part is cause its always defined in the extended real number line, and its monotonically increasing, which makes the monotonic convergence theorem works, as opposed to the "lim sup"
I think the main point of the proof is the construction of the function gn(x), since gn(x) = inf(f_k(x) n>=k), as n increase， the size of the set {f_k(x)| k>=n} is decreasing so inf{f_k|k>=n} will also increase because {f_k|k>=2} is always a fraction of {f_k|k>=1}. I don't know whether my understanding is correct.
6:55 I'm trying to understand the interpretation of inf(f_n). How can inf(f_n) be smaller than simply f_n? Is it possible for inf(f_n) to be smaller than f_n for a deterministic function f_n, or can inf(f_n) be smaller only if f_n is a stochastic function?
I am still loving this video series. You can be sure I will check out more of your channel. I am confused how you infer at 5:00 that g_n is monotonic? Is it perhaps you have restrictions on f_n that you did not include? As far as I understand from this point, f_n could be any function at all in no particular order so long as it is measurable and its codomain is nonnegative. This is the first serious snag I have encountered in the whole video series. I did consider that the f_n could be monotone as in previous videos but I have no right to assume that. As you must know as a math teacher, one can quickly end up recycling notation in a different context. I would appreciate any clarification on the properties of f_n. I know I bought this issue up before but, at 4:00 you say you need the "closed on infinity" statement (which I still don't understand-I thought infinity was always an open boundary) to have the statement in the integral you point out. That does not make sense to me because you are using a limit statement, which by definition means one can get as close as one likes to infinity; one does not need to reach it. That is at the heart of the idea of open sets. I know there might be a language barrier with my questions even though your English in the videos is perfect. As you can see, I did not understand your earlier answer to me, but the issue was more of a technicality. However, the new issue I raised has left me with a real logical gap.
Okay, that is a long question. Therefore I split my answer into several parts: (1) (g_n(x)) is monotonic simply by definition. It is the infimum of f_k(x) starting with n. (2) The function is allowed to take the symbolic value "inf". For example, you could define: f : [0,1] -> [0,inf] with f(x) = 1/x for x >0 and f(0) = inf. Why not? :) (3) By adding inf as a symbol in the set, a limit process that goes to infinity now has a limit point: The symbol inf. This makes your life just way easier. However, if you really want to, you could do everything without this neat trick but then you need to write a lot more at some points. (4) I don't think that there is a language barrier here. It is just that this topic is just not easy at all.
The sequence f(n) of functions is not monotonic, but the sequence g(n) IS monotonic, because of how g(n) is defined. You are correct: each f(n) is just an arbitrary measurable nonnegative integrable function, but this has no bearing on the definition of g(n), and it does not change the fact that g(n) is monotonic. Remember that g(n) is defined as inf({f(k) | k >= n}), and this particular combination of operations is necessarily monotonic.
Remarkable and very helpful to whoever would grasp hard concepts of Higher Mathematics, i'm a little bit uncertain on what stated about second equality in the proof, as lim shifts to lim inf.
5:15 Shouldn't it be "the inf can only get smaller not bigger ? " That would change the G function to monotonically decreasing , but the proof remains valid Thank u very much , Keep it up👌
If we consider less elements in the set, the infimum of this set can get bigger. If you drop the smallest elements, you get a bigger infimum. That is what happened here.
As you see, the fn measurable functions defined from the beginning are non negative, so as our Sir has clarified, as we move the index k in the set of positive integers, the infimum in the set of those measurable functions gets increased.
At 6:13 I'm not sure that I get how you substitute the limit with the limit inferior, since the limit refers to an integral and the limit inf has been defined as a function... Unless we consider the constant function which maps some variable to the integral of gn over X ?
Please help me to check whether my understanding is correct lim(inf{I(f_n)}) means inf{I(f_1),I(f_2),....I(f_n)}? Sorry for that since this notation is a little bit confusing to me.
![Measure Theory 9 | Fatou's Lemma [dark version]](http://i.ytimg.com/vi/dy0naGNogs4/mqdefault.jpg)
OMG I'm holding a bachelor degree in EE and I worked in the industry for 2 years. Now I'm back studying for my master and I doing a master level course in measure theoretic probability, which is quite a math shock as you can imagine. Those videos came in just a perfect time, thank you.
How have your studies been going?
I really love how measure theory is so strongly connected to real analysis. The limit of f_n might not exist but the limit inferior must exist. Since the limit inferior is the smallest(infimum) accumulation value of the sequence. And because the interval [0, inf] is compact, there is at least one accumulation value on the interval, then we just find the smallest (infimum).
many thanks for putting this series together, it is outstanding!!!
Fatou? More like "Thank you!" These videos are all excellent.
In 6:10, why can we substitute the limit with the limit inf?
When you have an increasing series, X for example, we have the equality X_n = min(X_k, k ≥ n) = Inf(X_k, k ≥ n) since by definition of an increasing series, X_n is lesser than every value that comes after it.
Since all g_n are point-wise monotonically increasing we know that the integrals of g_n are increasing and therefore you can substitute the integral of g_n by the infimum of g_k over k ≥ n. and finally you can turn the limit into limit inf.
@@sghaiermohamed2905 Thank you
just stopping by to say i love you
I was confused by the inequality sign in the Fatou's lemma. It led me thinking: is there another result that would be "stronger", with equality sign, but that's just harder to prove and this happens to be enough for the further proofs, OR, is there some deeper reason *why* there is an inequality there.
After googling for it a bit, turns out that there are cases where the inequality is strict! First of all, the general understanding: The left side, the function converges "pointwise" whereas on the right side the integral converges as a whole.
Then, consider the function series 𝑓𝑛(𝑥) = {n for 0 ≤𝑥≤1/𝑛; 0 otherwise }. That is a function that on the limit n→∞, converges to 0 pointwise, because the "width" of the non-zero part gets smaller and smaller. However, the height of that part gets higher and higher; it's carefully defined for the area under the curve to stay as 1. That means that the integral as whole doesn't converge to zero, it stays bigger than the integral of the pointwise limit function f. I guess a similar argument is also the reason we talk about the limit infinum, not limit. (I don't know why though.)
So here's the missing example / piece of intuition for those who were mystified by the statement of Fatou's lemma!
Wow
I guess for the "lim inf" part is cause its always defined in the extended real number line, and its monotonically increasing, which makes the monotonic convergence theorem works, as opposed to the "lim sup"
I think the main point of the proof is the construction of the function gn(x), since gn(x) = inf(f_k(x) n>=k), as n increase， the size of the set {f_k(x)| k>=n} is decreasing so inf{f_k|k>=n} will also increase because {f_k|k>=2} is always a fraction of {f_k|k>=1}. I don't know whether my understanding is correct.
6:55 I'm trying to understand the interpretation of inf(f_n). How can inf(f_n) be smaller than simply f_n? Is it possible for inf(f_n) to be smaller than f_n for a deterministic function f_n, or can inf(f_n) be smaller only if f_n is a stochastic function?
A lower bound of a set need not be in this set. The same is true for the infimum. In fact, the infimum of an open set is not an element of the set.
@@angelmendez-rivera351 thanks!
I am still loving this video series. You can be sure I will check out more of your channel.
I am confused how you infer at 5:00 that g_n is monotonic? Is it perhaps you have restrictions on f_n that you did not include? As far as I understand from this point, f_n could be any function at all in no particular order so long as it is measurable and its codomain is nonnegative. This is the first serious snag I have encountered in the whole video series. I did consider that the f_n could be monotone as in previous videos but I have no right to assume that. As you must know as a math teacher, one can quickly end up recycling notation in a different context. I would appreciate any clarification on the properties of f_n.
I know I bought this issue up before but, at 4:00 you say you need the "closed on infinity" statement (which I still don't understand-I thought infinity was always an open boundary) to have the statement in the integral you point out. That does not make sense to me because you are using a limit statement, which by definition means one can get as close as one likes to infinity; one does not need to reach it. That is at the heart of the idea of open sets.
I know there might be a language barrier with my questions even though your English in the videos is perfect. As you can see, I did not understand your earlier answer to me, but the issue was more of a technicality. However, the new issue I raised has left me with a real logical gap.
Okay, that is a long question. Therefore I split my answer into several parts:
(1) (g_n(x)) is monotonic simply by definition. It is the infimum of f_k(x) starting with n.
(2) The function is allowed to take the symbolic value "inf". For example, you could define:
f : [0,1] -> [0,inf] with f(x) = 1/x for x >0 and f(0) = inf. Why not? :)
(3) By adding inf as a symbol in the set, a limit process that goes to infinity now has a limit point: The symbol inf. This makes your life just way easier. However, if you really want to, you could do everything without this neat trick but then you need to write a lot more at some points.
(4) I don't think that there is a language barrier here. It is just that this topic is just not easy at all.
The sequence f(n) of functions is not monotonic, but the sequence g(n) IS monotonic, because of how g(n) is defined. You are correct: each f(n) is just an arbitrary measurable nonnegative integrable function, but this has no bearing on the definition of g(n), and it does not change the fact that g(n) is monotonic. Remember that g(n) is defined as inf({f(k) | k >= n}), and this particular combination of operations is necessarily monotonic.
Remarkable and very helpful to whoever would grasp hard concepts of Higher Mathematics, i'm a little bit uncertain on what stated about second equality in the proof, as lim shifts to lim inf.
It shifts because there is equality between the infimum of the set in consideration and the value of the function in question
5:15
Shouldn't it be "the inf can only get smaller not bigger ? "
That would change the G function to monotonically decreasing , but the proof remains valid
Thank u very much , Keep it up👌
If we consider less elements in the set, the infimum of this set can get bigger. If you drop the smallest elements, you get a bigger infimum. That is what happened here.
As you see, the fn measurable functions defined from the beginning are non negative, so as our Sir has clarified, as we move the index k in the set of positive integers, the infimum in the set of those measurable functions gets increased.
Somehow this video slipped under my radar
At 6:13 I'm not sure that I get how you substitute the limit with the limit inferior, since the limit refers to an integral and the limit inf has been defined as a function... Unless we consider the constant function which maps some variable to the integral of gn over X ?
Take a look at the other comments, as this has already been answered.
Also, lim inf is not defined as a function here, it is just a real number the same way lim is a real number.
At 6:20 how do we know that we can swap the limit with the limit inferior and maintain equality?
If the limit exists, it's equal to the limit inferior.
@@brightsideofmaths Awesome! Thanks so much for your reply!
This is just like the property of lower semi-continuity!
Please help me to check whether my understanding is correct lim(inf{I(f_n)}) means inf{I(f_1),I(f_2),....I(f_n)}? Sorry for that since this notation is a little bit confusing to me.
why can we change lim to lim inf in 6:10?
The question would be: Why not? What is the difference between lim and lim inf?
@@brightsideofmaths I've already understood this, thanks
@@brightsideofmaths does this mean that you could also have substituted lim with lim sup?
@@harisfawad1738 In this instance, this would also be possible. However, this would weaken the claim, of course.
@@brightsideofmaths Is it because the limit converges to one function and not alternating functions? (or for a certain x, f_n(x) -> f(x)).
Sir, so in the worst case of Fatou's lemma, that could be infinity (as a symbol) in the last inequality? Now that i have seen it, cannot be unseen.
In which cases can
liminf of f_k, where k>=n, n->inf
be not equal to
lim f_n, n->inf
?
Thank you, you are the best
Could there be a similar lemma to fatou but for limit superior?
it's the fattest Lemma in the world