# Rudin Remark 11.23

1. May 25, 2008

### ehrenfest

[SOLVED] Rudin Remark 11.23

1. The problem statement, all variables and given/known data
Prove: If f is measurable and bounded on E, and if $\mu(E) < + \infty$, then f is Lebesgue integrable on E.

2. Relevant 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.

3. 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?

Last edited: May 25, 2008
2. May 25, 2008

### morphism

Try to invoke the monotonicity of the integral.

3. May 26, 2008

### zhentil

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.

4. May 26, 2008

### zhentil

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