# Solve Rudin Remark 11.23: Proving Lebesgue Integrability

• ehrenfest
In summary, Rudin defines the Lebesgue integral and shows that if f is measurable and bounded on E, and if the measure of E is finite, then f is Lebesgue integrable on E. This follows directly from the definition of the Lebesgue integral and the monotonicity and countable additivity properties of the measure. The definition of a simple function ensures that the sets E_i are disjoint, and if they are not, they can be split into new disjoint sets. Therefore, the proof follows from the definition and properties of the Lebesgue integral.

#### ehrenfest

[SOLVED] Rudin Remark 11.23

## Homework Statement

Prove: If f is measurable and bounded on E, and if $\mu(E) < + \infty$, then f is Lebesgue integrable on E.

## Homework Equations

Here is how Rudin defines the Lebesgue integral. Everyone probably already knows this but anyway:

Suppose
(51) $$s(x) = \sum_{i=1}^n c_i K_{E_i} (x) \mbox{ (for x \in X, c_i >0)}$$

(where K is the indicator function) is measurable, and suppose E is a measurable set. We define

(52) $$I_E (s) = \sum_{i=1}^n c_i \mu(E \cap E_i)$$

If f is measurab;e and nonnegative, we define the Lebesgue integral of f over the set E as

(53) $$\int_E f d\mu = \sup I_E (s)$$

where the sup is taken over all measurable simple functions s such that 0 \leq s \leq f.

Let f be measurable, and consider the two integrals

(55) $$\int_E f^+ d\mu \mbox{ and} \int_E f^- d\mu$$

where $f^+ = max(f,0)$ and $f^-= min(f,0)$. If at least one of the two integrals is finite, we define

(56) $$\int_E f d\mu = \int_E f^+ d\mu - \int_E f^- d\mu$$

If both integrals on the RHS are finite, we say the f is Lebesgue integrable.

## The Attempt at a Solution

Apparently this should follow directly from the definition, although I am having trouble figuring out why. Assume B is a bound for f. Then all of the c_i have to less then or equal to B (at least if E_i is nonempty). Then we someone need to get a bound for the I_E. So, we probably want to invoke the countable additivity of the set function mu but we then need to know that all of the $E_i \cap E$ are disjoint and we do not know that, right?

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Try to invoke the monotonicity of the integral.

The definition of a simple function is that the Ei are disjoint. If they're not, you can split them into new Ei that are disjoint. For instance, if
E1 = (0,1/2), c1 = 1
E2 = (1/4, 3/4), c2 = 2
E3 = (1/2, 1), c3 = 1.
you can easily separate the Ei and ci so that the new Ei are disjoint.

Also, the reason that this follows directly from the definition is that it is the definition. What are you trying to prove?

## 1. How do we define Lebesgue integrability?

Lebesgue integrability is a way to measure the size of a function on a given interval. It takes into account not just the magnitude of the function, but also its oscillations and variations.

## 2. What is the significance of Rudin Remark 11.23 in proving Lebesgue integrability?

Rudin Remark 11.23 states that if a function is bounded and has a finite number of discontinuities on a given interval, then it is Lebesgue integrable on that interval. This is important because it provides a necessary and sufficient condition for a function to be Lebesgue integrable.

## 3. How can we prove Lebesgue integrability using Rudin Remark 11.23?

To prove Lebesgue integrability using Rudin Remark 11.23, we need to show that the function in question is bounded and has a finite number of discontinuities on the given interval. This can be done by analyzing the behavior of the function and identifying any points where it is discontinuous.

## 4. Can Rudin Remark 11.23 be applied to all functions?

No, Rudin Remark 11.23 can only be applied to functions that are bounded and have a finite number of discontinuities on the given interval. If a function is unbounded or has an infinite number of discontinuities, then this remark cannot be used to prove its Lebesgue integrability.

## 5. Are there other methods for proving Lebesgue integrability?

Yes, there are other methods for proving Lebesgue integrability, such as the Riemann-Lebesgue theorem, the Lebesgue dominated convergence theorem, and the Lebesgue monotone convergence theorem. These methods may be more useful for functions that do not meet the criteria stated in Rudin Remark 11.23.