1
$\begingroup$

How does the following equality hold? $f$ is nonnegative and integrable $$\int_{-\infty}^\infty f=\lim_{n\to\infty} \int_{-n}^nf=\lim_{n\to\infty} \int_{\{f\ge(1/n)\}}f=\lim_{n\to\infty} \int_{\{f\le n\}}f$$

I can see how all the equality hold but I am having trouble expressing it on paper.

For example $\int_{-\infty}^\infty f$=$\lim_{n\to\infty} \int_{\{f\ge(1/n)\}}f$ because as n approaches infinitely, {$ f\ge (1/n)$} eventuall become the whole domain of $f$ since $f$ is nonnegative.

For $\int_{-\infty}^\infty f$=$\lim_{n\to\infty} \int_{\{f\le n\}}f$ this holds since the set {$f \ge n$} is also eventually the whole domain of $f$.

Would this be ok? $\int_{-\infty}^\infty f$=$\int \lim_{n\to\infty}f \cdot X_{f \ge(1/n)}$=$lim_{n\to\infty} \int_{\{f\ge(1/n)\}}f$ by MCT?

  • 0
    To make a new paragraph outside of the math, just use two carriage returns. – 2017-02-19
  • 0
    @menag yes $f$ is nonnegative and lebesgue integrable, forgot to mention it – 2017-02-19
  • 2
    Use monotone convergence, so yes, this would be okay. – 2017-02-19
  • 0
    If $S_n\subseteq\mathbb R$ then $\int_{S_n}f$ is a notation for $\int f_n$ where $f_n:=f1_{S_n}$. So by nonnegative $f$ and $S_1\subseteq S_2\subseteq\cdots$ and $\bigcup_{n=1}^{\infty}S_n=\mathbb R$ you can apply MCT. – 2017-02-19
  • 0
    And in the stated circumstances, you don't even need the integrability of $f$, just the measurability. – 2017-02-19

0 Answers 0