# Thread: Lebesgue Integral

1. ## Lebesgue Integral

Let $\displaystyle (f_n)$ be a sequence in $\displaystyle L^2 (\Re)$.Suppose that $\displaystyle \|f_n\|_2 < M$ and $\displaystyle f_n \rightarrow f$ a.e. on $\displaystyle \Re$.
Show that $\displaystyle \int f_ng\rightarrow \int fg$ for all g $\displaystyle \in L^2(\Re)$.

$\displaystyle L^2$ is a Hilbert space,hence it is complete under the norm $\displaystyle \|.\|_2$.So, if $\displaystyle f_n\rightarrow f$,then $\displaystyle f\in L^2(\Re)$.But I am not sure whether we can claim the same thing on almost everywhere converges sequence in Hilbert space.
Am I in the right way to prove the question?If I am wrong,can anyone please help me

2. Hello,

The claim should be okay for the almost everywhere. There's some thing dealing with equivalence classes.

Can't you use Lebesgue dominated theorem for your problem ?

3. Originally Posted by problem
Let $\displaystyle (f_n)$ be a sequence in $\displaystyle L^2 (\Re)$.Suppose that $\displaystyle \|f_n\|_2 < M$ and $\displaystyle f_n \rightarrow f$ a.e. on $\displaystyle \Re$.
Show that $\displaystyle \int f_ng\rightarrow \int fg$ for all g $\displaystyle \in L^2(\Re)$.
Your question has been puzzling me...

Fatou's lemma gives $\displaystyle \int f(x)^2 dx\leq \liminf_n \int f_n(x)^2 dx\leq M^2$, hence $\displaystyle f\in L^2(\mathbb{R})$. (and not because of completeness)
I think the conclusion (what we want to prove) is equivalent to $\displaystyle \|f_n-f\|_2\to_n 0$.

It would be tempting to apply the bounded convergence theorem, like Moo suggests, but there is no obvious dominating function. Sure we can write $\displaystyle \int |f_ng|\leq M \|g\|_2$ by Cauchy-Schwarz, but what we would need is $\displaystyle |f_n(x) g(x)|\leq \phi(x)$ for some integrable function $\displaystyle \phi$.

If nobody can provide an answer on the forum, please tell us your teacher's solution when you get it.

4. I see this question randomly from a book which I had forgotten the title of the book.

5. I don't know if this is useful (I personally think I'm in the right direction, but there are some things missing) so I'm posting it here to see if anyone can do something with it:

Since $\displaystyle \Vert f_n \Vert \leq M$ it follows that there exists $\displaystyle (f_{n_k}) \subset (f_n)$ such that $\displaystyle f_{n_k} \rightarrow h \in L^2(\mathbb{R} )$ weakly i.e. $\displaystyle \int \ f_{n_k}g \rightarrow \int \ hg$ for all $\displaystyle g \in L^2 (\mathbb{R} )$. I'm stuck trying to prove $\displaystyle f=h$ and that in fact $\displaystyle f_n \rightarrow h$ weakly.

I know it's not much, but at least it looks like what we're trying to prove. Hope it helps, I'll keep trying to figure it out.

6. Originally Posted by Moo
Hello,

The claim should be okay for the almost everywhere. There's some thing dealing with equivalence classes.

Can't you use Lebesgue dominated theorem for your problem ?
I agree, wouldn't DCT be enough? Since $\displaystyle |f_n|<M$ so $\displaystyle |f_ng|<|Mg|\in L^2(\Re)$?

7. Originally Posted by putnam120
I agree, wouldn't DCT be enough? Since $\displaystyle |f_n|<M$ so $\displaystyle |f_ng|<|Mg|\in L^2(\Re)$?
Yeah, you're right. Hölder's inequality and DCT is all that's needed.

8. Originally Posted by putnam120
I agree, wouldn't DCT be enough? Since $\displaystyle |f_n|<M$ so $\displaystyle |f_ng|<|Mg|\in L^2(\Re)$?
Originally Posted by Jose27
Yeah, you're right. Hölder's inequality and DCT is all that's needed.

wait, I just noticed that this is not valid, because what is bounded is the norm in $\displaystyle L^2$ of $\displaystyle f_n$, and to use DCT you need $\displaystyle \vert f_n(x)g(x) \vert$ to be bounded by an integrable function which doesn't follow from Hölder, the best you would get is $\displaystyle \Vert f_ng \Vert _1 \leq M \Vert g \Vert _2$ (where $\displaystyle \Vert . \Vert _1$ is the norm in $\displaystyle L^1$ ). so DCT is not enough, at least not like that.

9. Originally Posted by Jose27
I don't know if this is useful (I personally think I'm in the right direction, but there are some things missing) so I'm posting it here to see if anyone can do something with it:

Since $\displaystyle \Vert f_n \Vert \leq M$ it follows that there exists $\displaystyle (f_{n_k}) \subset (f_n)$ such that $\displaystyle f_{n_k} \rightarrow h \in L^2(\mathbb{R} )$ weakly i.e. $\displaystyle \int \ f_{n_k}g \rightarrow \int \ hg$ for all $\displaystyle g \in L^2 (\mathbb{R} )$. I'm stuck trying to prove $\displaystyle f=h$ and that in fact $\displaystyle f_n \rightarrow h$ weakly.

I know it's not much, but at least it looks like what we're trying to prove. Hope it helps, I'll keep trying to figure it out.
I also believe you are in the correct direction. The ball with radius M in $\displaystyle L^2$ is weakly compact, hence any "weaker" (less fine than the weak) Hausdorff topology agree with the weak topology in this ball. The point is showing that almost everywhere convergence implies a convergence in a topology with this characteristiscs, but unfortunately I don't see how to define such a topology. Perhaps it is trivial but at the moment I don't see.

Perhaps it is possible to answer the first question in your given argument. $\displaystyle (f_{n_k}-h)$ tends weakly to 0. Then I believe that taking $\displaystyle g\in L^2$ charecteristic functions of bounded intervals implies almost everywhere convergence to zero of $\displaystyle (f_{n_k}-h)$ (using DCT in bounded intervals because the functions are essentially bounded in each interval by $\displaystyle 2\sqrt{M}$ divided the lenghth of the interval) and then $\displaystyle f=h$ in $\displaystyle L^2$. But at the moment I don't know how to deduce from this fact the convergence of the whole sequence.

10. Originally Posted by Enrique2
The ball with radius M in $\displaystyle L^2$ is weakly compact
This seems to be the right way indeed. More precisely, the fact that bounded sequences in $\displaystyle L^2$ have weakly convergent subsequences.

Let us prove that any weakly convergent subsequence of $\displaystyle (f_n)_n$ converges weakly toward $\displaystyle f$ (i.e. that $\displaystyle f$ is the only cluster point of the sequence $\displaystyle (f_n)_n$ for the weak topology). Then we will be able to conclude that $\displaystyle (f_n)_n$ converges weakly toward $\displaystyle f$. Indeed, if not, then there would be $\displaystyle \epsilon>0$ and $\displaystyle g\in L^2$ such that, for all $\displaystyle n$, $\displaystyle |\int f_n g -\int f g|>\epsilon$, which contradicts the existence of a subsequence of $\displaystyle (\int f_n g)_n$ converging toward $\displaystyle \int fg$. This is a general feature: in a sequentially compact space (here, a closed $\displaystyle L^2$-ball for weak topology), if a sequence has only one cluster point, then it converges (to the unique cluster point).

So, let's assume that a subsequence $\displaystyle (f_{\varphi(n)})_n$ converges weakly toward some limit $\displaystyle h\in L^2$. In order to ease the writing, I shall take $\displaystyle \varphi(n)=n$ (or I could write $\displaystyle f'_n=f_{\varphi(n)}$ and study $\displaystyle (f'_n)_n$). We know that $\displaystyle \int f_n g\to \int h g$ for all $\displaystyle g\in L^2$, and $\displaystyle f_n(x)\to f(x)$ for almost-all $\displaystyle x\in\mathbb{R}$.

Let us first reduce to a finite measure space (maybe there's a quicker way, I am biased by my acquaintance with probability). Let $\displaystyle I$ be a bounded interval of $\displaystyle \mathbb{R}$. We have of course $\displaystyle \int_I f_n g\to\int_I h g$ for all measurable function $\displaystyle g$ on $\displaystyle I$ (taking $\displaystyle g(x)=0$ outside $\displaystyle I$). Let $\displaystyle \epsilon>0$. By Egoroff's theorem, there is a measurable subset $\displaystyle A_\epsilon\subset I$ such that $\displaystyle \lambda(A_\epsilon^c)<\epsilon$ (Lebesgue measure) and $\displaystyle (f_n)_n$ converges uniformly toward $\displaystyle f$ on $\displaystyle A_\epsilon$. Then $\displaystyle \int_{A_\epsilon} f_n g\to \int_{A_\epsilon} f g$ because of Lebesgue's bounded convergence theorem: $\displaystyle |(f_n(x)-f(x))g(x)|\leq \|(f_n-f) 1_{A_\epsilon}\|_\infty |g(x)|\leq C |g(x)|$ (by uniform convergence) and $\displaystyle g\in L^2(I)\subset L^1(I)\subset L^1(A_\epsilon)$ (because the measure of $\displaystyle I$ is finite). Thus, we have both $\displaystyle \int_{A_\epsilon} f_n g\to \int_{A_\epsilon} f g$ and $\displaystyle \int_{A_\epsilon} f_n g \to \int_{A_\epsilon} h g$ (by applying our hypothesis to the function $\displaystyle g 1_{A_\epsilon}$), hence $\displaystyle \int_{A_\epsilon} f g= \int_{A_\epsilon} h g$ for all $\displaystyle g\in L^2(A_\epsilon)$, and finally $\displaystyle f=h$ almost-everywhere in $\displaystyle A_\epsilon$. This gives $\displaystyle \lambda(\{x\in I| f(x)\neq h(x)\})\leq \lambda(A_\epsilon^c)\leq \epsilon$. Therefore, $\displaystyle \lambda(\{x\in I| f(x)\neq h(x)\})=0$. In other words, $\displaystyle f(x)=h(x)$ for almost all $\displaystyle x$ in $\displaystyle I$. Since $\displaystyle \mathbb{R}$ is $\displaystyle \sigma$-finite (a big word for just writing $\displaystyle \mathbb{R}=\bigcup_{n\in \mathbb{Z}} [n,n+1]$), we deduce finally from the finite measure case that $\displaystyle f(x)=h(x)$ for almost all $\displaystyle x\in\mathbb{R}$. QED. If there is no mistake.