Upper function and lebesgue integrals

[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?

 

[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.

 

[tacman]

Your analysis is correct. The issue here is that while constant functions can be considered upper functions, they do not satisfy the definition of an increasing sequence of step functions. This is because a constant function is not strictly increasing, as it has the same value for all values of x. Therefore, it cannot be included in the set of step functions used in the definition of an upper function.

In theorem 10.6(b), the assumption that c must be non-negative is necessary because it ensures that the function cf is still an upper function. If c is negative, then cf is not an upper function and the theorem does not hold.

In theorem 10.14(a), the functions f and g are already assumed to be Lebesgue integrable, which means they are both upper functions. Therefore, the addition of a constant function does not change their properties as upper functions. This is why the theorem holds for all real values of a and b.

In summary, while constant functions can be considered upper functions, they do not fit the definition of an increasing sequence of step functions and therefore cannot be used in theorems that rely on this definition.