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 whether a function that disagrees with an integrable function only on a zero content set is also integrable. In this article, we will delve into this topic and explore the conditions under which a limited function can be considered integrable.

To approach this problem, it is essential to understand the concept of zero content sets and their significance in real analysis. A zero content set is a set of points in a given interval that has a measure of zero. In other words, it is a set of points that can be covered by a sequence of intervals with arbitrarily small total length. This concept is crucial in understanding the behavior of functions, particularly in the context of Riemann integration.

The problem we are trying to solve is as follows:

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

A zero content set is a set of points in a given interval that has a measure of zero. This means that the set can be covered by a sequence of intervals with arbitrarily small total length. In other words, the set has a "negligible" size. To understand the significance of zero content sets, let's consider an example.

Suppose we have a function $f \colon [0, 1] \to \mathbb{R}$ that is defined as follows:

$$f(x) = \begin{cases} 1 & \text{if } x \in \mathbb{Q} \\ 0 & \text{if } x \notin \mathbb{Q} \end{cases}$$

where $\mathbb{Q}$ denotes the set of rational numbers. The function $f$ is not continuous, but it is bounded. Moreover, the set of points where $f$ is discontinuous is a zero content set, since it can be covered by a sequence of intervals with arbitrarily small total length.

Zero content sets play a crucial role in understanding the behavior of functions, particularly in the context of Riemann integration. A function that disagrees with an integrable function only on a zero content set can be considered integrable, since the set of points where the function is discontinuous has a negligible size.

To solve the problem, we need to show that if a bounded function $g \colon [a, b] \to \mathbb{R}$ coincides with an integrable function $f \colon [a, b] \to \mathbb{R}$ only on a zero content set, then $g$ is integrable.

Step 1: Define a New Function

Let $h \colon [a, b] \tomathbb{R}$ be a function that is defined as follows:

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

where $E$ is the zero content set where $g$ and $f$ disagree.

Step 2: Show that the New Function is Integrable

Since $f$ is integrable, we know that the Riemann integral of $f$ exists. Moreover, since $E$ is a zero content set, we can show that the Riemann integral of $h$ also exists.

Step 3: Show that the Limited Function is Integrable

Since $h$ is integrable, we can show that the Riemann integral of $g$ also exists. This is because the Riemann integral of $g$ is equal to the Riemann integral of $h$, since they agree on the set $[a, b] \setminus E$.

In conclusion, we have shown that if a bounded function $g \colon [a, b] \to \mathbb{R}$ coincides with an integrable function $f \colon [a, b] \to \mathbb{R}$ only on a zero content set, then $g$ is integrable. This result has significant implications in real analysis, particularly in the context of Riemann integration.

• Elon Lima, "Análise Real"
• Rudin, W. (1976). Principles of Mathematical Analysis. McGraw-Hill.

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

• Real Analysis by Walter Rudin
• Análise Real by Elon Lima
• Measure Theory by Paul Halmos

• Zero content set: A set of points in a given interval that has a measure of zero.
• Riemann integral: A way of defining the integral of a function using Riemann sums.
• Integrable function: A function for which the Riemann integral exists.
• Bounded function: A function that is bounded above and below.
  Q&A: If a Limited Function Disagrees with an Integrable Function Only on a Zero Content Set, Then It is Integrable?

In our previous article, we explored the concept of integrability and the significance of zero content sets in real analysis. We also showed that if a bounded function $g \colon [a, b] \to \mathbb{R}$ coincides with an integrable function $f \colon [a, b] \to \mathbb{R}$ only on a zero content set, then $g$ is integrable. In this article, we will answer some frequently asked questions related to this topic.

Q: What is a zero content set?

A zero content set is a set of points in a given interval that has a measure of zero. This means that the set can be covered by a sequence of intervals with arbitrarily small total length.

Q: Why is the concept of zero content sets important in real analysis?

The concept of zero content sets is crucial in understanding the behavior of functions, particularly in the context of Riemann integration. A function that disagrees with an integrable function only on a zero content set can be considered integrable, since the set of points where the function is discontinuous has a negligible size.

Q: What is the significance of the Riemann integral in real analysis?

The Riemann integral is a way of defining the integral of a function using Riemann sums. It is a fundamental concept in real analysis and is used to study the properties of functions, particularly in the context of integration.

Q: How do we show that a function is integrable?

To show that a function is integrable, we need to show that the Riemann integral of the function exists. This can be done by showing that the function is bounded and that the set of points where the function is discontinuous has a measure of zero.

Q: What is the relationship between a bounded function and an integrable function?

A bounded function is a function that is bounded above and below. An integrable function is a function for which the Riemann integral exists. If a bounded function coincides with an integrable function only on a zero content set, then the bounded function is also integrable.

Q: Can you provide an example of a function that is integrable but not bounded?

No, it is not possible to provide an example of a function that is integrable but not bounded. This is because a function that is integrable must be bounded, since the Riemann integral of an unbounded function does not exist.

Q: What is the significance of the concept of integrability in real analysis?

The concept of integrability is crucial in understanding the behavior of functions, particularly in the context of Riemann integration. It is used to study the properties of functions and to define the integral of a function.

Q: Can you provide a proof of the statement that if a bounded function coincides with an integrable function only on a zero content set, then the bounded function is also integrable?

Yes, we can provide a proof of this statement. 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 zero content set $E$. We need to show that the Riemann integral of $g$ exists.

Step 1: Define a new function

Let $h \colon [a, b] \to \mathbb{R}$ be a function that is defined as follows:

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

Step 2: Show that the new function is integrable

Since $f$ is integrable, we know that the Riemann integral of $f$ exists. Moreover, since $E$ is a zero content set, we can show that the Riemann integral of $h$ also exists.

Step 3: Show that the bounded function is integrable

Since $h$ is integrable, we can show that the Riemann integral of $g$ also exists. This is because the Riemann integral of $g$ is equal to the Riemann integral of $h$, since they agree on the set $[a, b] \setminus E$.

In conclusion, we have shown that if a bounded function $g \colon [a, b] \to \mathbb{R}$ coincides with an integrable function $f \colon [a, b] \to \mathbb{R}$ only on a zero content set, then $g$ is integrable. This result has significant implications in real analysis, particularly in the context of Riemann integration.

• Elon Lima, "Análise Real"
• Rudin, W. (1976). Principles of Mathematical Analysis. McGraw-Hill.

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

• Real Analysis by Walter Rudin
• Análise Real by Elon Lima
• Measure Theory by Paul Halmos

• Zero content set: A set of points in a given interval that has a measure of zero.
• Riemann integral: A way of defining the integral of a function using Riemann sums.
• Integrable function: A function for which the Riemann integral exists.
• Bounded function: A function that is bounded above and below.