If A Limited Function G G G Disagrees With An Integrable Function F F F Only On A Zero Content Set, Then G G G Is Integrable?

If a Limited Function Disagrees with an Integrable Function Only on a Zero Content Set, Then It is Integrable?

In real analysis, the concept of integrability is crucial in understanding the behavior of functions, particularly in the context of Riemann integration. A function is said to be integrable if its Riemann integral exists. However, the question arises: what happens when a function, which is not necessarily integrable, agrees with an integrable function on a set of zero content? In this article, we will delve into this topic and explore the conditions under which a limited function, which disagrees with an integrable function only on a zero content set, is integrable.

Before we proceed, let's establish some basic definitions and concepts.

• A function $f \colon [a, b] \to \mathbb{R}$ is said to be integrable if its Riemann integral exists, i.e., if $\int_{a}^{b} f(x) dx$ is defined.
• A set $E \subset [a, b]$ is said to have zero content if for every $\epsilon > 0$, there exists a finite collection of intervals $\{I_1, I_2, \ldots, I_n\}$ such that $E \subset \bigcup_{i=1}^{n} I_i$ and $\sum_{i=1}^{n} |I_i| < \epsilon$.
• A function $g \colon [a, b] \to \mathbb{R}$ is said to be limited if it is bounded, i.e., if there exists a constant $M > 0$ such that $|g(x)| \leq M$ for all $x \in [a, b]$.

Given a bounded function $g \colon [a, b] \to \mathbb{R}$ that coincides with an integrable function $f \colon [a, b] \to \mathbb{R}$ only on a set of zero content, we want to determine whether $g$ is integrable.

Let $g \colon [a, b] \to \mathbb{R}$ be a bounded function that coincides with an integrable function $f \colon [a, b] \to \mathbb{R}$ only on a set of zero content. Then, $g$ is integrable.

Let $E \subset [a, b]$ be the set where $g$ and $f$ disagree. By assumption, $E$ has zero content. We need to show that $\int_{a}^{b} g(x) dx$ exists.

Since $f$ is integrable, we know that $\int_{a}^{b} f(x) dx$ exists. Let $\epsilon > 0$ be given. By the definition of integrability, there exists a partition $P$ of $[a, b]$ such that $U(P, f) - L(P, f) < \epsilon$.

Now, let $P'$ be a refinement of $P$ such that $E \subset \bigcup_{i=1}^{n} I_i$, where $I_i$ are the intervals in $P'$. Since $E$ has zero content, we can choose $P such that $\sum_{i=1}^{n} |I_i| < \epsilon$.

We can now define a new function $h \colon [a, b] \to \mathbb{R}$ by

$$h(x) = \begin{cases} g(x) & \text{if } x \in [a, b] \setminus E \\ f(x) & \text{if } x \in E \end{cases}$$

Since $g$ and $f$ coincide on $E$, we have $h = g$ on $[a, b] \setminus E$ and $h = f$ on $E$. Therefore, $h$ is integrable, and we have

$$\int_{a}^{b} h(x) dx = \int_{a}^{b} g(x) dx$$

Since $h$ is integrable, we know that $\int_{a}^{b} h(x) dx$ exists. Therefore, $\int_{a}^{b} g(x) dx$ also exists, and $g$ is integrable.

In this article, we have shown that if a bounded function $g$ disagrees with an integrable function $f$ only on a set of zero content, then $g$ is integrable. This result has important implications in real analysis, particularly in the context of Riemann integration.

• Elon Lima, "Análise Real"
• Rudin, W. (1976). Principles of mathematical analysis. McGraw-Hill.
• Bartle, R. G. (1964). The elements of real analysis. John Wiley & Sons.

For further reading on real analysis and Riemann integration, we recommend the following resources:

• Rudin, W. (1976). Principles of mathematical analysis. McGraw-Hill.
• Bartle, R. G. (1964). The elements of real analysis. John Wiley & Sons.
• Folland, G. B. (1999). Real analysis: modern techniques and their applications. John Wiley & Sons.

We hope this article has provided a clear understanding of the concept of integrability and its relationship with bounded functions that disagree with integrable functions only on a set of zero content.
Q&A: Integrability and Bounded Functions

In our previous article, we explored the concept of integrability and its relationship with bounded functions that disagree with integrable functions only on a set of zero content. In this article, we will answer some frequently asked questions related to this topic.

A set having zero content is a crucial concept in real analysis. It means that the set can be covered by a collection of intervals whose total length is arbitrarily small. In the context of integrability, a set of zero content is used to define the behavior of a function on a set where it is not integrable.

The concept of zero content is closely related to integrability. If a function is integrable, then it must be bounded on the set where it is integrable. A set of zero content is used to define the behavior of a function on a set where it is not integrable. If a function disagrees with an integrable function only on a set of zero content, then it is integrable.

Bounded functions play a crucial role in the study of integrability. If a function is bounded, then it is integrable if and only if it is bounded on the set where it is integrable. In other words, a bounded function is integrable if and only if it is bounded on the set where it is not integrable.

No, a function cannot be integrable if it is not bounded. Integrability requires that the function be bounded on the set where it is integrable. If a function is not bounded, then it is not integrable.

The theorem that states that a bounded function that disagrees with an integrable function only on a set of zero content is integrable is a fundamental result in real analysis. It provides a necessary and sufficient condition for a function to be integrable. In other words, it provides a way to determine whether a function is integrable or not.

Yes, consider the function $f(x) = \begin{cases} 0 & \text{if } x \in [0, 1] \\ 1 & \text{if } x \in (1, 2] \end{cases}$. This function is bounded and disagrees with the integrable function $g(x) = 0$ only on the set $\{1\}$, which has zero content.

Yes, consider the function $f(x) = \begin{cases} x & \text{if } x \in [0, 1] x^2 & \text{if } x \in (1, 2] \end{cases}$. This function is not bounded and is not integrable.

In this article, we have answered some frequently asked questions related to the concept of integrability and bounded functions. We hope that this article has provided a clear understanding of the relationship between bounded functions and integrability.

• Elon Lima, "Análise Real"
• Rudin, W. (1976). Principles of mathematical analysis. McGraw-Hill.
• Bartle, R. G. (1964). The elements of real analysis. John Wiley & Sons.

For further reading on real analysis and Riemann integration, we recommend the following resources:

• Rudin, W. (1976). Principles of mathematical analysis. McGraw-Hill.
• Bartle, R. G. (1964). The elements of real analysis. John Wiley & Sons.
• Folland, G. B. (1999). Real analysis: modern techniques and their applications. John Wiley & Sons.