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Prove that iff is measurable, then so is \(|f|\).

Short Answer

Expert verified
If \( f \) is measurable, then \( |f| \) is measurable because \( \{ x : |f(x)| > a \} = \{ x : f(x) > a \} \cup \{ x : f(x) < -a \} \), and both sets are measurable.

Step by step solution

01

Understand the concept of measurable functions

A function \( f: X \to \mathbb{R} \) is said to be measurable if, for every real number \( a \), the set \( \{ x \in X : f(x) > a \} \) is a measurable set in the given sigma-algebra on \( X \). Similarly, \( \{ x \in X : f(x) \ge a \} \), \( \{ x \in X : f(x) < a \} \), and \( \{ x \in X : f(x) \le a \} \) are also measurable.
02

Establish the objective for |f| measurability

We need to show that if \( f \) is measurable, then \( |f| \) is measurable as well. This means we need to show that for each real number \( a \), the set \( \{ x \in X : |f(x)| > a \} \) is measurable.
03

Express |f| in terms of f

Notice that the absolute value \(|f(x)|\) can be expressed as two cases depending on the value of \( f(x) \): \( |f(x)| = f(x) \) when \( f(x) \ge 0 \), and \( |f(x)| = -f(x) \) when \( f(x) < 0 \). This allows expressing \( |f| \) using measurable sets related to \( f \).
04

Analyze the sets for |f| using inequalities

Consider the set \( \{ x \in X : |f(x)| > a \} \), which can be rewritten using two cases: \( \{ x \in X : f(x) > a \} \cup \{ x \in X : f(x) < -a \} \). Both of these sets involve \( f \) and are measurable since \( f \) itself is measurable.
05

Demonstrate measurability of |f|

The key to measure the set \( \{ x \in X : |f(x)| > a \} \) is to use the property of measurability on sets involving \( f(x) > a \) and \( f(x) < -a \). Since \( \{ x \in X : f(x) > a \} \) and \( \{ x \in X : f(x) < -a \} \) are measurable, their union \( \{ x \in X : f(x) > a \} \cup \{ x \in X : f(x) < -a \} \), which is equal to \( \{ x \in X : |f(x)| > a \} \), is also measurable. Hence, \( |f| \) is measurable.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Measurability
Measurability is a fundamental concept in measure theory and forms the basis for understanding the behavior of functions in relation to a given measure space.
A function is said to be measurable if it maps sets in a certain structured way that respects the underlying measure, typically defined by a sigma-algebra.
Consider a function \(f: X \to \mathbb{R}\). This function is measurable if, for any real number \(a\), the set \(\{ x \in X : f(x) > a \}\) is a measurable set. Similarly, the sets \(\{ x \in X : f(x) \ge a \}\), \(\{ x \in X : f(x) < a \}\), and \(\{ x \in X : f(x) \le a \}\) must also be measurable.

Such requirements ensure that the set where the function exceeds, equals, or falls below any given real number is measurable. This concept is vital as it allows us to integrate and analyze functions over these measurable sets.
Whenever a function \(f\) is measurable, it guarantees that operations like addition, multiplication, and taking absolute values, which involve \(f\), also yield measurable functions.
Sigma-Algebra
Sigma-algebra is a key mathematical structure used in measure theory to ensure that certain operations can be systematically applied to sets.
It is a collection of subsets of a given set \(X\), satisfying specific properties crucial for defining measures.

Important properties of a sigma-algebra include:
  • It contains the empty set, meaning \(\emptyset\) is included.
  • If a set \(A\) is in the sigma-algebra, then its complement \(A^c\) is also included.
  • The sigma-algebra is closed under countable unions; that is, if \(A_1, A_2, A_3, \ldots \) are in the sigma-algebra, then the union \(\bigcup_{i=1}^{\infty} A_i\) is also in the sigma-algebra.
These properties ensure that sets and their operations are well-behaved from a measure-theoretic perspective.

In contexts where functions are defined on spaces with a sigma-algebra, such as measurable functions, the sigma-algebra serves as the framework within which the concept of measurability is applied.
By ensuring that measurable sets are compatible with the sigma-algebra, we can integrate, analyze, and manipulate functions meaningfully.
Absolute Value Function
The absolute value function, often a simple yet key operation, plays a crucial role in various mathematical analyses and proofs.
Expressed as \(|x|\), the absolute value of a number \(x\) is its distance from zero on the number line. This function has two main rules:
  • \(|x| = x\) if \(x \ge 0\)
  • \(|x| = -x\) if \(x < 0\)

By applying these rules, the absolute value simplifies expressions while maintaining positivity. When considering measurable functions, the absolute value is significant because it preserves the essential properties of a function while converting all values to non-negative results.

In the context of measurability, if a function \(f\) is measurable, then the function \(|f|\) is also measurable.
This is particularly useful when extending measurability to more complex or unfavorably valued functions, allowing us to still apply measure-theoretic techniques effectively.
Through this function, mathematicians can transform and analyze data while ensuring the structure and rules of measure theory are respected.

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Most popular questions from this chapter

Prove that a function \(f\) defined on a closed interval [a, \(b]\) is p-measurable if and only if, given any \(\varepsilon>0\), there is a continuous function \(\varphi\) on \([a, b]\) such that \(\mu\\{x: f(x) \neq \varphi(x)\\}<\mathrm{E}\) Hint. Use Egorov's theorem. Comment. This result, known as Luzin's theorem, shows that a measurable function "can be made continuous by altering it on a set of arbitrarily small measure."

Let \(\left\\{f_{n}\right\\}\) be a sequence of measurable functions converging almost everywhere to a functionf. Prove that \(\left\\{f_{n}\right\\}\) converges almost everywhere to a function \(g\) if and only if \(f\) and \(g\) are equivalent.

Prove that if a sequence \(\left\\{f_{n}\right\\}\) of functions converges to \(\mathbf{f}\) in measure, then it contains a subsequence \(\left\\{f_{n_{k}}\right\\}\) converging to \(\mathbf{f}\) almost everywhere. Hint. Let \(\left\\{\delta_{n}\right\\}\) be a sequence of positive numbers such that $$ \lim _{n \rightarrow \infty} \delta_{n}=0 $$ and let \(\left\\{\varepsilon_{n}\right\\}\) be a sequence of positive numbers such that $$ \sum_{n=1}^{\infty} \varepsilon_{n}<\infty $$ Let \\{n) be a sequence of positive integers such that \(\mathrm{n}\), \(>n_{k-1}\) and $$ \mu\left\\{x:\left|f_{n_{k}}(x)-f(x)\right|>\delta_{k}\right\\}<\varepsilon_{k} \quad(k=1,2, \ldots) $$ Moreover, let $$ R_{i}=\bigcup_{k=f}^{\infty}\left\\{x:\left|f_{n_{k}}(x)-f(x)\right|>\delta_{k}\right\\}, \quad Q=\bigcap_{i=1}^{\infty} R_{i} $$ Then \(\mu\left(R_{i}\right) \rightarrow \mu(Q)\) as \(\mathbf{i} \rightarrow \infty\), since \(R_{1} \supset R_{2} \supset \cdot \ldots\) On the other hand, $$ \mu\left(R_{i}\right)<\sum_{k=1}^{m} \varepsilon_{k} $$ and hence \(\mu\left(R_{i}\right) \rightarrow 0\), so that \(\mu(Q)=0\). Now show that \(\left\\{f_{n_{n}}\right\\}\) converges to \(f\) on \(E-Q .\)

Prove that the Dirichlet function $$ f(x)= \begin{cases}1 & \text { if } \mathrm{x} \text { is rational } \\ 0 & \text { if } x \text { is irrational }\end{cases} $$ is measurable on every interval \([a, b]\).

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