Originally Posted by

**zebra2147** I am working on the following exercise and could use some help.

Prove that if $\displaystyle \[

f(x)=

\begin{cases}

1, &\text{if $x > 0$}\\

-1, &\text{if $x<0$}\\

\end{cases}

\]$, then $\displaystyle f$ is integrable on $\displaystyle [-1,1]$ and $\displaystyle \int_{-1}^{1}f=0$.[Hint: Any lower sum is$\displaystyle \leq 0$ and some lower sum $\displaystyle =0$. Hence, the lower integral $\displaystyle \int_{-1}^{1}f=0$].

Here is what I have tried so far...

Let $\displaystyle s$ be a lower step function such that $\displaystyle s(x)\leq f(x)$ for all $\displaystyle x\in [-1,1]$. Then, let the lower sum for $\displaystyle f=\sum s=\sum_{i=1}^{n}A_{i}(x_{i}-x_{i-1})$. Then, we have that $\displaystyle A_{i}<f(x)$ for any lower step function. Also, $\displaystyle x_{i-1}<x<x_{i}$. Thus, for any $\displaystyle x<0$, $\displaystyle x_{i}-x_{i-1}>0$. So, $\displaystyle A_{i}(x_{i}-x_{i-1})<0$.

So for $\displaystyle \[

f(x)=

\begin{cases}

1, &\text{if $x > 0$}\\

-1, &\text{if $x<0$}\\

\end{cases}

\]$, we see that any lower sum $\displaystyle \leq 0$ and some lower sum is $\displaystyle =0$. Thus, $\displaystyle sup\{ \sum s|\text{s is a lower step function}\} =0$

=lower integral $\displaystyle \int_{-1}^{1}f=0$.

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

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

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.