1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Upper envelope proof (Royden)

  1. Apr 27, 2013 #1
    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. jcsd
  3. Apr 27, 2013 #2

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    A step function is defined piecewise on finitely many intervals - the indicator function of the rationals is NOT a step function
     
  4. Apr 27, 2013 #3
    Aha! Thank you!! My boneheaded self kept reading "step function" but thinking "simple function" :redface:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Upper envelope proof (Royden)
  1. Upper bound proof (Replies: 11)

Loading...