Upper function and lebesgue integrals

1. Aug 10, 2007

travis0868

I am comparing theorem 10.6(c) and 10.14(a) in Apostol's Mathematical Analysis.

My question is this:

Are constant functions considered upper functions? They certainly seem to fit the definition 10.4 for upper functions:

A real-valued function f defined on an interval I is called an upper function on I, and we write $f \in U(I)$, if there exists an increasing sequence of step functions ${s_n}$ such that:
$$a)\ s_n \nearrow\ f\ a.e.\ on\ I,$$
and
$$b)\ \lim_{x\rightarrow \infty} \int_\textrm{I} s_n \is\ finite.$$

There there's definition 10.12 for Lebesgue integrable functions:

We denote by L(I) the set of all functions f of the form f = u - v where $u\in U(I)$ and $v\in U(I)$. Each function f in L(I) is said to be Lebesgue-integrable on I and its integral is defined by the equation $\int_\textrm{I} f = \int_\textrm{I} u - \int_\textrm{I} v$

Here's thm 10.14(a):

$$Assume\ f \in L(I)\ and\ g \in L(I).\ Then\ we\ have\ (af + bg) \in L(I)\ for\ every\ real\ a\ and\ b\ \int_\textrm{I} (af + bg) = a \int_\textrm{I} f + b \int_\textrm{I} g.$$

Assuming that a constant function is an upper function, let v = 0. Choose some arbitrary upper function u. Let f = u - 0 and thus f is a member of L(I). Then $\int_\textrm{I} f = \int_\textrm{I} u$. By thm 10.14, $\int_\textrm{I} -f = - \int_\textrm{I} f$. Thus $\int_\textrm{I} -f = - \int_\textrm{I} u.$

But it is not always true that

integral -u = - integral u

when u is an upper function according to Thm 10.6(b):

$$Assume\ f \in U(I)\ and\ g \in U(I).\ Then:\ cf \in U(I)\ for\ ever\ constant\ c \geq 0\ and \int_\textrm{I} cf = c \int_\textrm{I} f$$

Notice that c must be >= 0. There's even a problem in the text that shows that thm 10.6(b) isn't always true if that assumption is violated.

My point is that if constant functions are allowed as upper functions, then thm 10.14(a) implies thm 10.6(b) should be true for all real c, not just non-negative c. Any thoughts on where my logic is wrong?

Last edited: Aug 10, 2007
2. Aug 10, 2007

travis0868

I see my mistake. A constant is an upper function, but:

if u $\in U(I)$ and f = u. Then f $\in L(I)$. But $\int -f = - \int u$. The negative must be on the outside of the integral sign for u.

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