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Math Help - A question on the integral of a step function

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    A question on the integral of a step function

    In Apostol's "Mathematical Analysis", a step function on closed interval [a,b] is defined as follows (Page 148):


    A step function on a general interval I and its integral over I are defined as follows (Page 253):

    But in the proof of Theorem 10.10 in page 259 (see figure below), the underlined sentence asserts the existence of \int _J s_n^+ over an arbitrary subinterval J. If s_n^+ assumes a constant value \ne 0 in some [a,b]\subseteq I, while J=(c,d) with c<a and a<d<b, then s_n^+ is not a step function on J because we can not find a closed interval in J such that s_n^+ is 0 outside of this closed interval. In turn, the integral \int_J s_n^+ is undefinable. How to handle this problem? Thanks!
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    Although s_n^+ is not a step function on J for the case stated above, it can be approximated by a sequence of step functions with right endpoints arbitrarily approching d. Based on this idea, We can construct a generating sequence so that f\in U(J) with the same integral value. Let I=(m,n), I_1=(m,c), I_2=[c,n).
    Generally speaking, if I=(a,b], I'=(a,b), f\in U(I) does not implies f\in U(I'). But if f\geq 0 a.e. on I, f\in U(I') does hold. Suppose \{s_n\} generates f on I, define s_n^+ as does the book, then \{s_n^+\} generates f on I. On I' let's define t_n(x)=\left\{ \begin{array}{*{20}{ll}}<br />
0 & x \in (b - \frac{1}{n},b) \\ <br />
s_n^ + (x) & {\rm{elsewhere}} \\ <br />
\end{array} \right.\ (when n is sufficiently large. Set t_n=0 for small n's). Then we can verify the following one by one: t_n is a step function on I', increasing (because so is s_n^+), t_n(x)=s_n^+(x)\to f(x) (ignore the possible first finite terms being 0), so t_n\nearrow f a.e. on I'. In addition, \int_{I'}t_n\leq\int_{I'}s_n^+ which is bounded above, so \lim\limits_{n \to \infty}\int_{I'}t_n exists. This proves f\in U(I') with \{t_n\} being the generating sequence.
    Next, it is clear that if f\in U(I) and f\in U(I'), \int_I f=\int_{I'} f because we can define the value arbitarily at b for each member of the generating sequence of f on I' to get a generating sequence of f on I, without changing the value of integral of the step functions.
    Let's return to the proof of the textbook. To ensure that s_n^+ is still a step function on J, the following condition must be met: if s_n^+ lies over J, that is, the (a,b) of the defining interval of s_n^+ intersects interior of J, then the endpoint of J lying in [a,b] must be closed. So we first assume the common endpoint of I_1 and I_2 (namely c) is closed for both I_1 and I_2, then the proof works correctly and the theorem holds. But if I_1 is open at c, it may not true that s_n^+ is still a step function on I_1. Now suppose I_1' is the same subinterval as I_1 except being closed at c (that is, I_1'=(m,c]), then we already have f\in U(I_1'). Since f\geq 0 by hypothesis, from the above analysis we have f\in U(I_1) too, and \int_{I_1}f=\int_{I_1'}f, hence we obtain \int_I f=\int_{I_1'}f+\int_{I_2}f=\int_{I_1}f+\int_{I_2}f, as required.
    Last edited by zzzhhh; February 7th 2010 at 06:37 PM.
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