Problem: Let $ f_n $ be a sequence of integrable functions on $ [0, +\infty ) $.
Prove/Disprove the following statements:
(A) If $ f_n $ converges pointwise to $ f(x) = 0 $ on $ [0, +\infty ) $ then $\lim _{n \rightarrow \infty} \int_{0}^{\infty} f_{n}(x) d x=0$
(B) If $ f_n $ converges uniformly to $ f(x) = 0 $ on $ [0, +\infty ) $ then $\lim _{n \rightarrow \infty} \int_{0}^{\infty} f_{n}(x) d x=0$
The answers are that both of the above statements are false and the following sequence of functions are brought for disprove:
For (A) :
$ f_n(x)=\begin{cases} n &\text{if}\;x \in [0,1/n]\\\\0 &\text{else}\; \end{cases}$
For (B):
$ f_n(x)=\begin{cases} 1/n &\text{if}\;x \in [0,n]\\\\0 &\text{else}\; \end{cases}$
When I draw these functions, for (A) I understand that the function will thin-out to be of zero thickness, and (B) will flatten-out to be of zero height, so it might make sense that both of these functions converge to zero pointwise. However, I can't seem to see it mathematically since $ x $ depends on $ n $ and when taking limit of $ n $ , $ x $ will change and so it obscures things for me.
For example, I want to show the function for $ (A) $ converges pointwise to $0 $.
How does one show the function converges to $0$ as $n$ goes to infinity? it seems as if i'll take $ x=0 $ and take the limit of $n $ then I'll get $ f_n(0) \to \infty $. Then, my second attempt is that I went by definition and did as follows:
Let $ \epsilon >0 $. There exists $ N \in \mathbb{N} $ s.t. $ x< \frac{1}{N} $, and so $ f_n(x) = N $, in particular for all $ n > N $ it will be the case that $ f_n(x) = n $ ..... ( this is mumbo-jumbo I know, I feel as I've got stuck in an infinite loop of what comes before - the chicken or the egg ) ( I think my error is that I mis-took the $ n $ that defines the function $ f_n $ to be the $ n $ of the limit definition )
In anycase, can you please explain, perhaps mathematically why the above functions converge pointwise to $ 0 $? I'm stuck on this for a while and I'll very much appreciate the help!
The first function does not converge to 0 pointwise, as we have $f_n(0) \to \infty$. The remedy could be to take it equal to $n$ on $(0,1/n]$ instead (and $f_n(0) =0$ for all $n$). Then every $x\in [0,\infty)$ will eventually fall into the set where all the next $f_n$'s are equal to 0. This implies that $f_n\to 0$ pointwise.
To get that the function in (B) converges to 0 uniformly, it suffices to note that by its definition we have $|f_n(x)|\leq \frac 1n$ for every $n\in \mathbb{N}$ and $x\in [0,\infty)$.
By the way, you can as well use the example from (B) to answer to (A).