# How do you show that a monotone increasing real function on a closed interval [a, b]

• Mar 12th 2008, 07:54 PM
Kelles
How do you show that a monotone increasing real function on a closed interval [a, b]
Hello Everyone,
I have two questions on measure theory:
1)
How do you show that a monotone increasing real function on a closed interval [a, b] is Lebesgue measurable function.
2)
Suppose A is a subset of the closed interval [0,1] which is dense. (That is, the closure of A is all of [0,1].) suppose U is an open set with A⊆U. Must it be true that m(U) ≥ 1?

Thank you!
• Mar 12th 2008, 08:03 PM
ThePerfectHacker
Quote:

Originally Posted by Kelles
Hello Everyone,
I have two questions on measure theory:
1)
How do you show that a monotone increasing real function on a closed interval [a, b] is Lebesgue measurable function.

It happens to be integrable. First note $\displaystyle f$ is bounded since $\displaystyle f(a) \leq f \leq f(b)$. (Also assume that $\displaystyle f(a)\not = f(b)$ because otherwise [tex]f[/math[ would be a constant function, and there is nothing to prove).

Let $\displaystyle U(f,P)$ be the upper Darboux sum with respect to partition $\displaystyle P$ of $\displaystyle [a,b]$ and $\displaystyle L(f,P)$ be the lower Darboux sum with respect to partition $\displaystyle P$. To show this function is integrable we need to show for any $\displaystyle \epsilon >0$ we can find a partition $\displaystyle P$ such that $\displaystyle U(f,P)-L(f,P) < \epsilon$. Note, since $\displaystyle f$ is monotone it means $\displaystyle U(f,P) - L(f,P) = \sum_{k=1}^n [f(x_k)-f(x_{k-1})](x_k - x_{k-1})$. Let $\displaystyle \delta$ be $\displaystyle \min_{1\leq k \leq n} (x_k - x_{k-1})$. Then we see that $\displaystyle U(f,P)-L(f,P) \leq \sum_{k=1}^n [f(x_k)-f(x_{k-1})]\delta = [f(a)-f(b)]\delta$. Thus, if $\displaystyle \delta < \epsilon/[f(b)-f(a)]$ then $\displaystyle U(f,P)-L(f,P)<\epsilon$. This means given any $\displaystyle \epsilon > 0$ we can create equal subdivision of the interval until the interval length $\displaystyle \leq \delta$ and then we are have $\displaystyle U(f,P)-L(f,P)<\epsilon$. Thus $\displaystyle f$ is integrable.
• Mar 12th 2008, 09:58 PM
Kelles
Is integrability of a function means that the function Lebesgue measurable ?
If a function is integrable, does it mean that the function Lebesgue measurable ?

Thank you very much.
• Mar 13th 2008, 07:26 AM
ThePerfectHacker
Quote:

Originally Posted by Kelles
If a function is integrable, does it mean that the function Lebesgue measurable ?

I would assume so. I never studied measure theory, but from I understand Lebesque integrability is a generalization of Riemann integrability. Thus, if a function is Riemann integrable then it is Lebesque integrable.