0

Theorem 1.5 Comparison Theorem on page 67 of Apostal's Calculus 1. The theorem states $s(x) < t(x)$ for all $x\in [a,b]$ then $\int_a^b s(x)dx < \int_a^b t(x)dx$. Note: $s$ and $t$ are step functions in this context.

I have trouble grasping why how this is true. This is my line of thought, what if the function $t(x) = \sqrt 2$ for $x\in[a,b]$, and $s(x)<t(x)=\sqrt2$ for $x\in[a,b]$, $t$ is a line and the area of the line is 0. $\int_a^b s(x)dx = \int_a^b t(x)dx - 0$. Which part am I misunderstanding, thanks to anyone who can help.

sewie
  • 1
  • Assuming these are Riemann integrals (and that $a < b$), essentially you need to show that $t-s$ is continuous at some point in $[a,b]$. Then the result follows trivially. See here and here for more. But this result is highly nontrivial, especially for chapter 1 of a calculus book. –  Sep 19 '18 at 17:04
  • You've omitted a very important detail. I checked my copy of Apostol. The assumption here is that $s$ and $t$ are step functions, not arbitrary integrable functions. That makes the proof much easier. I'm editing to add this detail. –  Sep 19 '18 at 17:09

2 Answers2

1

Given a fixed $s(x),$ where $s(x) < t(x)$ for all $x \in [a,b],$
consider $s_1(x) = s(x) + \dfrac{1}{2} [t(x) - s(x)].$

Further consider $s_{n+1}(x) = s_n(x) + \dfrac{1}{2} [t(x) - s_n(x)].$

Clearly, for any $n\in\mathbb{Z^+}, s_n(x) < t(x)$ for all $x \in [a,b].$
Let $u(x)$ denote the $lim_{n->\infty}s_n(x).$

Further, inspired by the OP's original line of thought, for all $x\in[a,b], u(x) = t(x).$
Therefore, $\int_a^b u(x) dx = \int_a^b t(x)dx.$

However, $s(x)$ itself is a fixed function.
Let v(x) denote $t(x) - s(x).\;$ Then, for all $x \in [a,b], v(x) > 0.$
Therefore, $\int_a^b t(x)dx - \int_a^b s(x)dx = \int_a^b \{t(x) - s(x)\}dx = \int_a^b v(x) dx > 0.$

user2661923
  • 35,619
  • 3
  • 17
  • 39
0

The integral $\int_a^b t(x)\,dx$ is not the area of the line formed by the graph of $y(t(x)$, but the area under the graph and the $x$-axis (assuming $t(x)>0$.) Thus, when $t(x)=\sqrt2$, we have $$ \int_a^b t(x)\,dx=\int_a^b \sqrt2\,dx=(b-a)\sqrt2>0. $$

Julián Aguirre
  • 76,354
  • if $\int_a^b t(x)dx$ is the area under $y=\sqrt2$ on $[a,b]$ and the $x$-axis, then wouldn't $\int_a^b s(x)dx = \int_a^b t(x)dx$, which contradicts the Theorem, which says if $s(x)<t(x)$ for every $x$ in the interval then the integral of $s(x)$ over the domain $[a,b]$ is less than that of $t(x)$. – sewie Sep 19 '18 at 16:14
  • My answer does not mention $s(x)$ at all. It just shows the error in your argument. – Julián Aguirre Sep 19 '18 at 16:19
  • Sorry, if I'm misunderstanding but I didn't mean that $\int_a^b t(x)dx$ to be the area of the line $y=\sqrt2$. I'm trying to say that on the interval $a<x<b$ $t(x)=\sqrt2$, so the integral is as you said $(b-a)\sqrt2$. My question has to do with defining $s(x)$ to be the union of $\int_a^b s_i(x)dx$ on the open interval, and for all $s_i(x)$ is a constant $ c_i < \sqrt2$ such that $c_1\leq c_2\leq \cdots \leq c_k < \sqrt2$. – sewie Sep 19 '18 at 16:37
  • I do not understand what you mean by union of $\int_a^bs_i(x),dx$. – Julián Aguirre Sep 19 '18 at 16:39
  • An example of what $\bigcup_{i=1}^{\infty} \int_a^b s_i(x)dx$ could be $\int_a^b 1 dx \cup \int_a^b 1.4 dx \cup \int_a^b 1.41 dx \cup \cdots \cup \int_a^b 1.414213562dx \cup \cdots$ Sorry I think my notation is messed up...not sure if it means what I want to say. – sewie Sep 19 '18 at 16:46
  • I certainly do not know what $\int_a^b f\cup \int_a^bg$ means. – Julián Aguirre Sep 19 '18 at 16:56