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mathmonkey
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Homework Statement
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##).
Homework Equations
Given ##f##, the upper envelope ##h## is defined as
##h(y) = \inf _{\delta > 0} \sup_{|x-y|< \delta} f(x) ##
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!
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