# A question on the integral of a step function

• Feb 1st 2010, 02:27 PM
zzzhhh
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):
http://www.tuchuan.com/a/2010020206564827627.jpg

A step function on a general interval I and its integral over I are defined as follows (Page 253):
http://www.tuchuan.com/a/201002020701166419.jpg
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!
http://www.tuchuan.com/a/2010020207023017927.jpg
• Feb 7th 2010, 05:02 PM
zzzhhh
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}}
0 & x \in (b - \frac{1}{n},b) \\
s_n^ + (x) & {\rm{elsewhere}} \\
\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.