# Fatou's Lemma - Example

Let $$f_n(x)=I_{[n,\infty)}(x)=\begin{cases} 1, x\geq n\\ 0 , x We can see that for all $$n$$, $$f_n:\mathbb{R}\to [0,\infty]$$ is lebesgue measurable, $$f_n$$ is decreasing as a function of $$n$$, $$f_1\geq > f_2\geq...$$

1.why $$[o,\infty]$$ and not $$[0,1]$$?

and

$$lim_{n\to\infty}f_n(x)=0=f(x)$$

But $$\int _{\mathbb{R}}f(x)dm=0\neq \infty=\lim_{n\to \infty}\int _{\mathbb{R}}f_n(x)dm$$

1. $$\lim_{n\to \infty}\int _{\mathbb{R}}f_n(x)dm=\infty$$ is due to $$\lim_{n\to \infty}\int _{\mathbb{R}}f_n(x)dm=\lim_{n\to \infty}\int _{\mathbb{R}}I_{[n,\infty)}dm=m([n,\infty))=\infty?$$

2. In general how do we evaluate $$\lim_{n\to \infty}\int _{\mathbb{R}}f_n(x)dm$$ for a non negative $$f_n$$

And due to Fatou's Lemma:

$$\int _{\mathbb{R}}f(x)dm=0< \infty=\liminf\int_{\mathbb{R}}f_n(x)dm$$

1. $$\infty=\liminf\int _{\mathbb{R}}f_n(x)dm$$ because if there is a limit so it is equal to the liminf and limsup?

$$(1)$$ Look up the difference between codomain and image.
$$(2)$$ Almost. We have $$\int_{\Bbb R}f_n(x) \,dx = \int_n^\infty 1\, dx = +\infty$$. Because $$\int_{\Bbb R}f_n(x) \,dx = +\infty$$ for all $$n$$, the limit as $$n\to\infty$$ is also $$+\infty$$.
$$(3)$$ Like above, evaluate each $$\int_{\Bbb R}f_n(x) \,dx$$. This is a (extended) real number, so you're just taking the limit of a sequence of real numbers.
$$(4)$$ $$+\infty$$ is the $$\liminf$$ because each term in the sequence is $$+\infty$$. Your reasoning also works in the context of extended real numbers.
• 1. What I thought, thanks for the conformation 2 and 3. I am trying to use the definition using simple function integration so it is $\int_{\mathbb{R}}f_n(x)dm=1\cdot I_{[n,\infty)}=1\cdot \mu([n,\infty))=1\cdot\infty=\infty$? – newhere Jan 16 '20 at 18:19