Why Must A Translation-Invariant, Finitely Additive Function Be The Lebesgue Measure?

by ADMIN 86 views

Introduction

In the realm of real analysis and measure theory, the concept of a translation-invariant, finitely additive function is crucial in understanding the properties of measures. A measure is a function that assigns a non-negative real number to each subset of a given set, in such a way that it satisfies certain properties. In this article, we will explore the relationship between translation-invariant, finitely additive functions and the Lebesgue measure.

Translation-Invariant, Finitely Additive Functions

A function $ \nu: \mathcal{I}{\mathbb{R}} \to [0, \infty) $ is said to be translation-invariant if for any $ (a,b] \in \mathcal{I}{\mathbb{R}} $ and any $ c \in \mathbb{R} $, we have $ \nu((a,b]) = \nu((a+c,b+c]) $. This means that the measure of an interval does not change under a translation.

On the other hand, a function $ \nu: \mathcal{I}{\mathbb{R}} \to [0, \infty) $ is said to be finitely additive if for any finite collection of disjoint intervals $ (a_1,b_1], (a_2,b_2], ..., (a_n,b_n] $ in $ \mathcal{I}{\mathbb{R}} $, we have $ \nu\left(\bigcup_{i=1}^{n} (a_i,b_i]\right) = \sum_{i=1}^{n} \nu((a_i,b_i]) $. This means that the measure of a finite union of disjoint intervals is equal to the sum of the measures of the individual intervals.

The Lebesgue Measure

The Lebesgue measure is a measure that assigns a non-negative real number to each subset of $ \mathbb{R} $, in such a way that it satisfies the following properties:

  • Non-negativity: $ \mu(A) \geq 0 $ for any $ A \subseteq \mathbb{R} $.
  • Countable additivity: For any countable collection of disjoint sets $ A_1, A_2, ... $ in $ \mathbb{R} $, we have $ \mu\left(\bigcup_{i=1}^{\infty} A_i\right) = \sum_{i=1}^{\infty} \mu(A_i) $.
  • Translation invariance: For any $ A \subseteq \mathbb{R} $ and any $ c \in \mathbb{R} $, we have $ \mu(A) = \mu(A+c) $.

Why Must a Translation-Invariant, Finitely Additive Function Be the Lebesgue Measure?

To show that a translation-invariant, finitely additive function must be the Lebesgue measure, we need to prove that it satisfies the properties of the Lebesgue measure.

Step 1: Non-Negativity

Let $ \nu: \mathcal{I}{\mathbb{R}} \to [0, \infty) $ be a translation-invariant, finitely additive function. We need to show that $ \nu(A) \geq 0 $ for any $ A \subseteq \mathbb{R} $. Let $ A = (a,b] $ be an interval in $ \mathcal{I}{\mathbb{R}} $. Since $ \nu $ is finitely additive, we have $ \nu(A) = \nu((a,b]) \geq 0 $. Therefore, $ \nu(A) \geq 0 $ for any $ A \subseteq \mathbb{R} $.

Step 2: Countable Additivity

Let $ \nu: \mathcal{I}{\mathbb{R}} \to [0, \infty) $ be a translation-invariant, finitely additive function. We need to show that for any countable collection of disjoint sets $ A_1, A_2, ... $ in $ \mathbb{R} $, we have $ \nu\left(\bigcup{i=1}^{\infty} A_i\right) = \sum_{i=1}^{\infty} \nu(A_i) $. Let $ A_i = (a_i,b_i] $ be an interval in $ \mathcal{I}{\mathbb{R}} $ for each $ i $. Since $ \nu $ is finitely additive, we have $ \nu\left(\bigcup{i=1}^{n} A_i\right) = \sum_{i=1}^{n} \nu(A_i) $ for any finite $ n $. Therefore, we can write $ \nu\left(\bigcup_{i=1}^{\infty} A_i\right) = \lim_{n \to \infty} \nu\left(\bigcup_{i=1}^{n} A_i\right) = \lim_{n \to \infty} \sum_{i=1}^{n} \nu(A_i) = \sum_{i=1}^{\infty} \nu(A_i) $.

Step 3: Translation Invariance

Let $ \nu: \mathcal{I}{\mathbb{R}} \to [0, \infty) $ be a translation-invariant, finitely additive function. We need to show that for any $ A \subseteq \mathbb{R} $ and any $ c \in \mathbb{R} $, we have $ \nu(A) = \nu(A+c) $. Let $ A = (a,b] $ be an interval in $ \mathcal{I}{\mathbb{R}} $. Since $ \nu $ is translation-invariant, we have $ \nu(A) = \nu((a+c,b+c]) = \nu(A+c) $.

Conclusion

In conclusion, we have shown that a translation-invariant, finitely additive function must be the Lebesgue measure. This is because it satisfies the properties of the Lebesgue measure, namely non-negativity, countable additivity, and translation invariance.

References

  • Lebesgue, H. (1902). "Intégrale, longueur, aire." Annali di Matematica Pura ed Applicata, 7(1), 231-359.
  • Halmos, P. R. (1950). Measure Theory. Van Nostrand.

Further Reading

  • Real Analysis by Royden, H. L. and Fitzpatrick, P. M.
  • Measure Theory by Cohn, D. L.
  • Lebesgue Measure and Integration by Wheeden, R. L. and Zygmund, A.
    Q&A: Why Must a Translation-Invariant, Finitely Additive Function Be the Lebesgue Measure? =====================================================================================

Q: What is the Lebesgue measure?

A: The Lebesgue measure is a measure that assigns a non-negative real number to each subset of $ \mathbb{R} $, in such a way that it satisfies the following properties:

  • Non-negativity: $ \mu(A) \geq 0 $ for any $ A \subseteq \mathbb{R} $.
  • Countable additivity: For any countable collection of disjoint sets $ A_1, A_2, ... $ in $ \mathbb{R} $, we have $ \mu\left(\bigcup_{i=1}^{\infty} A_i\right) = \sum_{i=1}^{\infty} \mu(A_i) $.
  • Translation invariance: For any $ A \subseteq \mathbb{R} $ and any $ c \in \mathbb{R} $, we have $ \mu(A) = \mu(A+c) $.

Q: What is a translation-invariant, finitely additive function?

A: A function $ \nu: \mathcal{I}{\mathbb{R}} \to [0, \infty) $ is said to be translation-invariant if for any $ (a,b] \in \mathcal{I}{\mathbb{R}} $ and any $ c \in \mathbb{R} $, we have $ \nu((a,b]) = \nu((a+c,b+c]) $. This means that the measure of an interval does not change under a translation.

On the other hand, a function $ \nu: \mathcal{I}{\mathbb{R}} \to [0, \infty) $ is said to be finitely additive if for any finite collection of disjoint intervals $ (a_1,b_1], (a_2,b_2], ..., (a_n,b_n] $ in $ \mathcal{I}{\mathbb{R}} $, we have $ \nu\left(\bigcup_{i=1}^{n} (a_i,b_i]\right) = \sum_{i=1}^{n} \nu((a_i,b_i]) $. This means that the measure of a finite union of disjoint intervals is equal to the sum of the measures of the individual intervals.

Q: Why must a translation-invariant, finitely additive function be the Lebesgue measure?

A: To show that a translation-invariant, finitely additive function must be the Lebesgue measure, we need to prove that it satisfies the properties of the Lebesgue measure. Specifically, we need to show that it is non-negative, countably additive, and translation-invariant.

Q: How do we show that a translation-invariant, finitely additive function is non-negative?

A: Let $ \nu: \mathcal{I}{\mathbb{R}} \to [0, \infty) $ be a translation-invariant, finitely additive function. We need to show that $ \nu(A) \geq 0 $ for any $ A \subseteq \mathbb{R} $. Let $ A = (a,b] $ be an interval in $ \mathcal{I}{\mathbb{R}} $. Since $ \nu $ is finitely additive, we have $ \nu(A) = \nu((a,b]) \geq 0 $. Therefore, $ \nu(A) \geq 0 $ for any $ A \subseteq \mathbb{R} $.

Q: How do we show that a translation-invariant, finitely additive function is countably additive?

A: Let $ \nu: \mathcal{I}{\mathbb{R}} \to [0, \infty) $ be a translation-invariant, finitely additive function. We need to show that for any countable collection of disjoint sets $ A_1, A_2, ... $ in $ \mathbb{R} $, we have $ \nu\left(\bigcup{i=1}^{\infty} A_i\right) = \sum_{i=1}^{\infty} \nu(A_i) $. Let $ A_i = (a_i,b_i] $ be an interval in $ \mathcal{I}{\mathbb{R}} $ for each $ i $. Since $ \nu $ is finitely additive, we have $ \nu\left(\bigcup{i=1}^{n} A_i\right) = \sum_{i=1}^{n} \nu(A_i) $ for any finite $ n $. Therefore, we can write $ \nu\left(\bigcup_{i=1}^{\infty} A_i\right) = \lim_{n \to \infty} \nu\left(\bigcup_{i=1}^{n} A_i\right) = \lim_{n \to \infty} \sum_{i=1}^{n} \nu(A_i) = \sum_{i=1}^{\infty} \nu(A_i) $.

Q: How do we show that a translation-invariant, finitely additive function is translation-invariant?

A: Let $ \nu: \mathcal{I}{\mathbb{R}} \to [0, \infty) $ be a translation-invariant, finitely additive function. We need to show that for any $ A \subseteq \mathbb{R} $ and any $ c \in \mathbb{R} $, we have $ \nu(A) = \nu(A+c) $. Let $ A = (a,b] $ be an interval in $ \mathcal{I}{\mathbb{R}} $. Since $ \nu $ is translation-invariant, we have $ \nu(A) = \nu((a+c,b+c]) = \nu(A+c) $.

Q: What are some examples of translation-invariant, finitely additive functions?

A: Some examples of translation-invariant, finitely additive functions include:

  • The Lebesgue measure
  • The counting measure
  • The Dirac measure

Q: What are some applications of translation-invariant, finitely additive functions?

A: Some applications of translation-invariant, finitely additive functions include:

  • Measure theory
  • Real analysis
  • Functional analysis
  • Probability theory

Conclusion

In conclusion, we have shown that a translation-invariant, finitely additive function must be the Lebesgue measure. This is because it satisfies the properties of the Lebesgue measure, namely non-negativity, countable additivity, and translation invariance. We have also discussed some examples of translation-invariant, finitely additive functions and some applications of these functions.

References

  • Lebesgue, H. (1902). "Intégrale, longueur, aire." Annali di Matematica Pura ed Applicata, 7(1), 231-359.
  • Halmos, P. R. (1950). Measure Theory. Van Nostrand.

Further Reading

  • Real Analysis by Royden, H. L. and Fitzpatrick, P. M.
  • Measure Theory by Cohn, D. L.
  • Lebesgue Measure and Integration by Wheeden, R. L. and Zygmund, A.