f is integrable

**zebra2147** · January 30th 2011, 10:44 AM

I am working on the following exercise and could use some help.
Prove that if $ f(x)= \begin{cases} 1, &\text{if $x > 0$}\\ -1, &\text{if $x<0$}\\ \end{cases} $ , then $f$ is integrable on $[-1,1]$ and $\int_{-1}^{1}f=0$ .[Hint: Any lower sum is $\leq 0$ and some lower sum $=0$ . Hence, the lower integral $\int_{-1}^{1}f=0$ ].

Here is what I have tried so far...
Let $s$ be a lower step function such that $s(x)\leq f(x)$ for all $x\in [-1,1]$ . Then, let the lower sum for $f=\sum s=\sum_{i=1}^{n}A_{i}(x_{i}-x_{i-1})$ . Then, we have that $A_{i}<f(x)$ for any lower step function. Also, $x_{i-1}<x<x_{i}$ . Thus, for any $x<0$ , $x_{i}-x_{i-1}>0$ . So, $A_{i}(x_{i}-x_{i-1})<0$ .
So for $ f(x)= \begin{cases} 1, &\text{if $x > 0$}\\ -1, &\text{if $x<0$}\\ \end{cases} $ , we see that any lower sum $\leq 0$ and some lower sum is $=0$ . Thus, $sup\{ \sum s|\text{s is a lower step function}\} =0$
=lower integral $\int_{-1}^{1}f=0$ .

We can do something similar for $x>0$ ...For $x>0$ we have that $A_{i}>0$ and $x_{i}-x_{i-1}>0$ . Therefore, $A_{i}(x_{i}-x_{i-1})>0$ .

Then conclude that $\int_{-1}^{1}f=0$ .

I'm not sure if I have said enough for the proof to be complete or if what I have is even good. Please help me out.

**Drexel28** · January 31st 2011, 12:30 PM

Originally Posted by zebra2147

I am working on the following exercise and could use some help.
Prove that if $ f(x)= \begin{cases} 1, &\text{if $x > 0$}\\ -1, &\text{if $x<0$}\\ \end{cases} $ , then $f$ is integrable on $[-1,1]$ and $\int_{-1}^{1}f=0$ .[Hint: Any lower sum is $\leq 0$ and some lower sum $=0$ . Hence, the lower integral $\int_{-1}^{1}f=0$ ].

Here is what I have tried so far...
Let $s$ be a lower step function such that $s(x)\leq f(x)$ for all $x\in [-1,1]$ . Then, let the lower sum for $f=\sum s=\sum_{i=1}^{n}A_{i}(x_{i}-x_{i-1})$ . Then, we have that $A_{i}<f(x)$ for any lower step function. Also, $x_{i-1}<x<x_{i}$ . Thus, for any $x<0$ , $x_{i}-x_{i-1}>0$ . So, $A_{i}(x_{i}-x_{i-1})<0$ .
So for $ f(x)= \begin{cases} 1, &\text{if $x > 0$}\\ -1, &\text{if $x<0$}\\ \end{cases} $ , we see that any lower sum $\leq 0$ and some lower sum is $=0$ . Thus, $sup\{ \sum s|\text{s is a lower step function}\} =0$
=lower integral $\int_{-1}^{1}f=0$ .

We can do something similar for $x>0$ ...For $x>0$ we have that $A_{i}>0$ and $x_{i}-x_{i-1}>0$ . Therefore, $A_{i}(x_{i}-x_{i-1})>0$ .

Then conclude that $\int_{-1}^{1}f=0$ .

I'm not sure if I have said enough for the proof to be complete or if what I have is even good. Please help me out.

Could you explain your proof a little more. Which step function are you talking about?

Meanwhile...

See here? I assume this should really be $f(x)=-1$ on $x\leqslant 0$ . You clearly have that $f$ is integrable on $[-1,0]$ since it's constant, and it's integrable on $[0,1]$ since it's a one-point perturbation of a continuous function, which in the post i linked you to I proved was integrable. Moreover, it's clear that $\displaystyle \int_{-1}^0f(x)\text{ }dx=-1$ and $\displaystyle \int_0^1 f(x)\text{ }dx=1$ since by that same post we get that $\displaystyle \int_0^1 f(x)\text{ }dx=\int_0^1 1\text{ }dx$ .

**zebra2147** · January 31st 2011, 01:31 PM

Well this is my first attempt at an integral proof and I learned that the way to prove that the integral exists is to prove that the lower integral is equal to the upper integral. So I guess my proof was attempting to prove that the lower integral =0 and then use a similar proof to show that the upper integral =0.
It gets a little messy but basically i think I wanted to show that for a lower step function $s$ we see that for $f=\sum s=\sum_{i=1}^{n}A_{i}(x_{i}-x_{i-1})$ , when $x<0, A_{i}<0$ and when $x>0,$ we see that the value for $A_{i}$ is equal to $|A_{i}|$ when $x<0$ . Therefore, $\int_{-1}^{1}f=0$ .

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