Continuous Function In R N \mathbb{R}^{n} R N Implies Integrability And Subtraction Rule

$$

Introduction

In the realm of multivariable calculus, the concept of continuous functions plays a pivotal role in understanding various mathematical phenomena. One of the fundamental properties of continuous functions is their integrability over bounded sets. In this article, we will delve into the relationship between continuous functions in $\mathbb{R}^{n}$ and their integrability, as well as explore the implications of the subtraction rule.

Definition of Integrability

Before we proceed, let's establish a clear understanding of the concept of integrability. According to Definition 1, a function $f$ is said to be integrable over a bounded set $A$ if $f$ is zero outside $A$ and can be constructed a closed rectangle $B$ such that:

$$f(x) = 0 \text{ for } x \notin A \text{ and } f(x) = \frac{1}{\text{vol}(B)} \int_{B} f(y) dy \text{ for } x \in A$$

where $\text{vol}(B)$ denotes the volume of the rectangle $B$.

Continuous Functions in $\mathbb{R}^{n}$

A function $f: \mathbb{R}^{n} \to \mathbb{R}$ is said to be continuous at a point $x \in \mathbb{R}^{n}$ if for every $\epsilon > 0$, there exists a $\delta > 0$ such that:

$$|f(x) - f(y)| < \epsilon \text{ whenever } |x - y| < \delta$$

In other words, a function is continuous if it can be approximated by its value at a nearby point.

Continuous Implies Integrable

One of the fundamental properties of continuous functions is their integrability over bounded sets. In fact, we can prove that a continuous function is integrable over any bounded set.

Theorem 1

Let $f: \mathbb{R}^{n} \to \mathbb{R}$ be a continuous function and $A$ be a bounded set. Then, $f$ is integrable over $A$.

Proof

Let $A$ be a bounded set and $f$ be a continuous function. We need to show that $f$ is integrable over $A$. By Definition 1, we need to construct a closed rectangle $B$ such that:

$$f(x) = 0 \text{ for } x \notin A \text{ and } f(x) = \frac{1}{\text{vol}(B)} \int_{B} f(y) dy \text{ for } x \in A$$

Since $f$ is continuous, we can find a $\delta > 0$ such that:

$$|f(x) - f(y)| < \epsilon \text{ whenever } |x - y| < \delta$$

Let $B$ be a closed rectangle such that $A \subset B$ and $\text{vol}(B) < \delta$. Then, we have:

$$f(x) = 0 \text{ for } x \notin A \text{ and } f(x) = \frac{1}{\text{vol}(B)} \int_{B} f(y dy \text{ for } x \in A$$

This shows that $f$ is integrable over $A$.

Subtraction Rule

The subtraction rule states that if $f$ and $g$ are integrable functions over a bounded set $A$, then $f - g$ is also integrable over $A$.

Theorem 2

Let $f$ and $g$ be integrable functions over a bounded set $A$. Then, $f - g$ is integrable over $A$.

Proof

Let $f$ and $g$ be integrable functions over a bounded set $A$. We need to show that $f - g$ is integrable over $A$. By Definition 1, we need to construct a closed rectangle $B$ such that:

$$(f - g)(x) = 0 \text{ for } x \notin A \text{ and } (f - g)(x) = \frac{1}{\text{vol}(B)} \int_{B} (f - g)(y) dy \text{ for } x \in A$$

Since $f$ and $g$ are integrable, we can find closed rectangles $B_{1}$ and $B_{2}$ such that:

$$f(x) = 0 \text{ for } x \notin B_{1} \text{ and } f(x) = \frac{1}{\text{vol}(B_{1})} \int_{B_{1}} f(y) dy \text{ for } x \in B_{1}$$

$$g(x) = 0 \text{ for } x \notin B_{2} \text{ and } g(x) = \frac{1}{\text{vol}(B_{2})} \int_{B_{2}} g(y) dy \text{ for } x \in B_{2}$$

Let $B = B_{1} \cap B_{2}$. Then, we have:

$$(f - g)(x) = 0 \text{ for } x \notin B \text{ and } (f - g)(x) = \frac{1}{\text{vol}(B)} \int_{B} (f - g)(y) dy \text{ for } x \in B$$

This shows that $f - g$ is integrable over $A$.

Conclusion

In conclusion, we have shown that a continuous function in $\mathbb{R}^{n}$ implies integrability and the subtraction rule. These results have far-reaching implications in multivariable calculus and are essential in understanding various mathematical phenomena.

References

• [1] Rudin, W. (1976). Principles of Mathematical Analysis. McGraw-Hill.
• [2] Spivak, M. (1965). Calculus on Manifolds. Benjamin.

Further Reading

For further reading on this topic, we recommend the following resources:

• [1] Apostol, T. M. (1974). Mathematical Analysis. Addison-Wesley.
• [2] Bartle, R. G. (1976). The Elements of Integration. Wiley.

$$

Introduction

In our previous article, we explored the relationship between continuous functions in $\mathbb{R}^{n}$ and their integrability, as well as the implications of the subtraction rule. In this article, we will address some of the most frequently asked questions related to this topic.

Q: What is the definition of a continuous function in $\mathbb{R}^{n}$?

A: A function $f: \mathbb{R}^{n} \to \mathbb{R}$ is said to be continuous at a point $x \in \mathbb{R}^{n}$ if for every $\epsilon > 0$, there exists a $\delta > 0$ such that:

$$|f(x) - f(y)| < \epsilon \text{ whenever } |x - y| < \delta$$

Q: What is the relationship between continuous functions and integrability?

A: A continuous function in $\mathbb{R}^{n}$ implies integrability over any bounded set. In other words, if a function is continuous, it can be integrated over any bounded set.

Q: What is the subtraction rule?

A: The subtraction rule states that if $f$ and $g$ are integrable functions over a bounded set $A$, then $f - g$ is also integrable over $A$.

Q: How do I prove that a function is continuous?

A: To prove that a function is continuous, you need to show that for every $\epsilon > 0$, there exists a $\delta > 0$ such that:

$$|f(x) - f(y)| < \epsilon \text{ whenever } |x - y| < \delta$$

Q: What are some common examples of continuous functions?

A: Some common examples of continuous functions include:

• Polynomials
• Rational functions
• Trigonometric functions
• Exponential functions

Q: What are some common examples of non-continuous functions?

A: Some common examples of non-continuous functions include:

• Step functions
• Discontinuous functions
• Functions with infinite discontinuities

Q: How do I apply the subtraction rule?

A: To apply the subtraction rule, you need to show that if $f$ and $g$ are integrable functions over a bounded set $A$, then $f - g$ is also integrable over $A$. This can be done by constructing a closed rectangle $B$ such that:

$$(f - g)(x) = 0 \text{ for } x \notin B \text{ and } (f - g)(x) = \frac{1}{\text{vol}(B)} \int_{B} (f - g)(y) dy \text{ for } x \in B$$

Q: What are some common applications of continuous functions and integrability?

A: Some common applications of continuous functions and integrability include:

• Calculus
• Analysis
• Physics
• Engineering

Conclusion

In conclusion, we have addressed some of the most frequently asked questions to continuous functions in $\mathbb{R}^{n}$ and their integrability, as well as the implications of the subtraction rule. We hope that this article has provided valuable insights and information for readers.

References

• [1] Rudin, W. (1976). Principles of Mathematical Analysis. McGraw-Hill.
• [2] Spivak, M. (1965). Calculus on Manifolds. Benjamin.

Further Reading

For further reading on this topic, we recommend the following resources:

• [1] Apostol, T. M. (1974). Mathematical Analysis. Addison-Wesley.
• [2] Bartle, R. G. (1976). The Elements of Integration. Wiley.