In the Dominated Convergence Theorem, we usually assume that |fn|≤g for some integrable function g. However, what is a counter-example where fn are not dominated by an integrable function?
And I found this example but there is one thing I don't understand and I would like your help.
So I considered the sequence of functions on $(0,1)$
$$ f_n(x) = \begin{cases}
n & \text{ if } x \in (0,1/n)\\
0 & \text{ otherwise}
\end{cases}$$
We have $\lim_{n \to \infty} f_n(x) = 0 = f(x)$. However,
$$\lim_{n \to \infty} \int_0^1f_n(x)dx = 1 \neq 0 = \int_0^1 f(x) dx$$
Question:
(1) I'm having a problem with this limit here: $\lim_{n \to \infty} f_n(x) = 0 $
because there will be always point between $0$ and $1/n$ not matter how big $n$ is. Therefore, this limit will approach infinity.
(2) Why is this integral 1? $\lim_{n \to \infty} \int_0^1f_n(x)dx = 1$
Best Answer
Here, $\lim_{n \to \infty}f_n$ is the pointwise limit of $f_n$. So, fix $x$ & let $n \to \infty$. Does this $x$ work for all $n$?
$\int_{(0,1)}f_n d\mu = \int_{(0,1)}n 1_{(0, 1/n)} d \mu= n\int_{(0,1/n)} 1 d \mu = n \mu(0, 1/n) = n. 1/n = 1$. Letting $n \to \infty$, you have $\int_{(0,1)}f_n d\mu \to 1$. (Note: $f_n = 1_{(0,1/n))}$ is another way to write $f_n$)