Let $$f_n(x)=I_{[n,\infty)}(x)=\begin{cases} 1, x\geq n\\ 0 , x<n\\ \end{cases}$$ 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 $[0,1]$?



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.

  • $\begingroup$ 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 $\int_{\mathbb{R}}f_n(x)dm=1\cdot I_{[n,\infty)}=1\cdot \mu([n,\infty))=1\cdot\infty=\infty$? $\endgroup$ – newhere Jan 16 '20 at 18:19
  • 1
    $\begingroup$ Yes, that would be correct. $\endgroup$ – Fimpellizieri Jan 16 '20 at 18:27

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