# Upper envelope proof (Royden)

1. Apr 27, 2013

### mathmonkey

1. The problem statement, all variables and given/known data

Let $f$ be a bounded function on [a,b] and let $h$ be the upper envelope of $f$. Then $R \overline{\int}_a^b f = \int _a^b h.$ (if $\phi \geq f$ is a step function, then $\phi \geq h$ except at a finite number of points, and so $\int _a^b h \leq R \overline{\int}_a^b f$. But there is a sequence $\phi _n$ of step functions such that $\phi _n \rightarrow h$ so that $\int _a^b h = lim \int _a^b \phi _n \geq R \overline{\int}_a^b f$).

2. Relevant equations

Given $f$, the upper envelope $h$ is defined as

$h(y) = \inf _{\delta > 0} \sup_{|x-y|< \delta} f(x)$

3. The attempt at a solution

My question with the problem is with this statement:

"if $\phi \geq f$ is a step function, then $\phi \geq h$ except at a finite number of points, and so $\int _a^b h \leq R \overline{\int}_a^b f$."

I may be misunderstanding the definition of upper envelope, but suppose $f$ is defined as:

$f(x) = 0$ for $x \in \mathbb{R} - \mathbb{Q}$ and $f(x) = 1$ for $x \in \mathbb{Q}$. Then the upper envelope should be $h(x) = 1$ everywhere? Since every open set around each point in [a,b] will contain a rational number. On the other hand, $f$ itself is a step function, so if we set $\phi = f$, we have a step function such that $\phi \geq f$, but there are uncountably many points where $h \geq \phi$ which is in contradiction with the statement in the problem. Am I misunderstanding anything here?

Thanks for any help, it is greatly appreciated!

Last edited: Apr 27, 2013
2. Apr 27, 2013

### Office_Shredder

Staff Emeritus
A step function is defined piecewise on finitely many intervals - the indicator function of the rationals is NOT a step function

3. Apr 27, 2013

### mathmonkey

Aha! Thank you!! My boneheaded self kept reading "step function" but thinking "simple function"