How to Prove the Inequality for a Limit of Functions and Integrals?

  • Thread starter Thread starter Chris L T521
  • Start date Start date
Click For Summary
SUMMARY

The discussion centers on proving the inequality for the limit of functions and integrals using the sequence of functions defined by \(f_n\). Specifically, it demonstrates that \(\int_{\mathbb{R}}\liminf_{n\to\infty}f_n\,dm < \liminf_{n\to\infty}\int_{\mathbb{R}}f_n\,dm\). Ackbach provided a correct solution to this problem, confirming the validity of the inequality through rigorous mathematical reasoning.

PREREQUISITES
  • Understanding of Lebesgue integration
  • Familiarity with the concept of limit inferior in sequences
  • Knowledge of measure theory
  • Basic proficiency in mathematical analysis
NEXT STEPS
  • Study Lebesgue Dominated Convergence Theorem
  • Explore the properties of limit inferior in sequences of functions
  • Investigate examples of functions with discontinuities in measure theory
  • Learn about the implications of the Fatou's Lemma in integration
USEFUL FOR

Mathematicians, students of analysis, and anyone interested in advanced topics in measure theory and integration techniques.

Chris L T521
Gold Member
MHB
Messages
913
Reaction score
0
Here's this week's problem.

-----

Problem: Consider\[f_n = \begin{cases}1 & \forall\,x\in\left[n,n+1\right)\\ 0 & \forall\,x\in\mathbb{R}\backslash\left[n,n+1\right)\end{cases}\]
Show that
\[\int_{\mathbb{R}}\liminf_{n\to\infty}f_n\,dm < \liminf_{n\to\infty}\int_{\mathbb{R}}f_n\,dm.\]

-----

 
Physics news on Phys.org
This week's problem was correctly answered by Ackbach. You can find his solution below.

This follows directly from Fatou's Lemma, but you can also simply compute both sides directly:

Note that
$$ \liminf_{n \to \infty} \int_{ \mathbb{R}}f_{n} \, dm= \liminf_{n \to \infty} \left[1 \cdot m([n,n+1)) + 0 \cdot m( \mathbb{R} \setminus [n,n+1)) \right]
= \liminf_{n \to \infty}1=1.$$
But
$$ \int_{ \mathbb{R}} \left[ \liminf_{n \to \infty} f_{n} \right] \, dm= \int_{ \mathbb{R}} 0 \, dm = 0.$$

Since $0<1$, we are done.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
7K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K