Let f be a measurable function on (X, F, µ ) with a σ-finite measure µ . Assumet that integral from X of |f| dµ < infinity. Using Fubini’s theorem, prove that integral from X of |f| dµ = integral from 0 to infinity of µ ({x :|f(x)|>t })dt

To prove the equality:

\[
\int_X |f| \, d\mu = \int_0^\infty \mu(\{ x \in X : |f(x)| > t \}) \, dt,
\]

we will use **Fubini's theorem** and the concept of **distribution functions**.

### Step-by-step proof:

#### 1. Define the set \( A_t \):
For each \( t \geq 0 \), define the set \( A_t \subseteq X \) as follows:

\[
A_t = \{ x \in X : |f(x)| > t \}.
\]

This set captures all points \( x \) in \( X \) where the magnitude of \( f(x) \) exceeds \( t \).

#### 2. Express \( |f(x)| \) as an integral:
The function \( |f(x)| \) can be written using an integral representation involving the indicator function of the set \( A_t \):

\[
|f(x)| = \int_0^\infty \mathbb{1}_{\{ |f(x)| > t \}} \, dt,
\]

where \( \mathbb{1}_{\{ |f(x)| > t \}} \) is the indicator function, which is equal to 1 if \( |f(x)| > t \) and 0 otherwise.

This expression essentially states that the magnitude \( |f(x)| \) is equal to the "area" under the indicator function curve for \( t \geq 0 \).

#### 3. Apply Fubini’s theorem:
Now, we integrate \( |f(x)| \) over \( X \) with respect to \( \mu \):

\[
\int_X |f(x)| \, d\mu(x) = \int_X \left( \int_0^\infty \mathbb{1}_{\{ |f(x)| > t \}} \, dt \right) d\mu(x).
\]

By Fubini’s theorem (which allows us to swap the order of integration because \( f \) is measurable and the integral of \( |f| \) is finite), we can rewrite this as:

\[
\int_X |f(x)| \, d\mu(x) = \int_0^\infty \left( \int_X \mathbb{1}_{\{ |f(x)| > t \}} \, d\mu(x) \right) dt.
\]

#### 4. Recognize the inner integral:
The inner integral \( \int_X \mathbb{1}_{\{ |f(x)| > t \}} \, d\mu(x) \) is simply the measure of the set \( \{ x \in X : |f(x)| > t \} \), which we have defined as \( A_t \). Therefore:

\[
\int_X \mathbb{1}_{\{ |f(x)| > t \}} \, d\mu(x) = \mu(A_t) = \mu(\{ x \in X : |f(x)| > t \}).
\]

#### 5. Final form:
Substituting this into the previous expression, we get:

\[
\int_X |f(x)| \, d\mu(x) = \int_0^\infty \mu(\{ x \in X : |f(x)| > t \}) \, dt.
\]

This completes the proof.

### Conclusion:
We have shown that:

\[
\int_X |f(x)| \, d\mu(x) = \int_0^\infty \mu(\{ x \in X : |f(x)| > t \}) \, dt
\]

using Fubini's theorem and the integral representation of \( |f(x)| \).

---

Would you like further details or clarifications on any part of this proof?

### Further exploration questions:
1. What are the conditions required for applying Fubini's theorem?
2. Can this result be extended to complex-valued functions?
3. How does the concept of distribution functions relate to this proof?
4. What is the significance of the measure \( \mu \) being σ-finite in this context?
5. How does the measure of \( A_t \) change as \( t \to \infty \)?

### Tip:
Remember that Fubini’s theorem can be applied when either the integral is finite, or both integrals involved are non-negative (Tonelli’s theorem handles non-negative cases).

Let f be a measurable function on (X, F, µ ) with a σ-finite measure µ. Assume that the integral over X of |f| dµ < infinity. Using Fubini’s theorem, prove that the integral from X of |f| dµ equals the integral from 0 to infinity of µ({x :|f(x)| > t}) dt.

Explore a detailed proof using Fubini's Theorem to show that the integral over X of |f| dµ equals the integral from 0 to infinity of µ({x : |f(x)| > t}) dt. This problem involves key concepts in measure theory and Lebesgue integration, ideal for graduate-level students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation