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

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

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