Let $f: [0,1] \to \mathbb{R}$
$f(x) = 0$ if $x\in C$
$f(x) = 1$ otherwise
Show that $f$ is Riemann Integrable and compute
$$\int_{0}^{1} f(x) dx$$
I called this the "reverse" characteristic function since normally the $1$ and $0$ have their places switched. When I did this problem, I thought the answer was the same as the normal characteristic function.
We want to show $U(f) = L(f) = \int_{0}^{1} f(x) dx$
Let $P$ be a partition
First, notice that $L(f,P) = 0$ because there are points in $[0,1]$ that are in the Cantor set.
This means $L(f) = 0$
Now, $U(f,P) = \frac{2^n}{3^n}$ since this is the length of the nth subinterval and the supremum is $1$ due to their being points in $[0,1]$ that are not in the Cantor set.
Taking the infimum on both sides,
$U(f) = inf (U(f,P)) = inf \frac{2^n}{3^n} = 0$ because as $n \to \infty$, this expression goes to $0$.
Therefore, the function is RI since $U(f) = L(f)$ and the value is $0$.
However, my classmates have told me that the answer is $1$. My professor neither confirmed nor denied this, but he did say that my answer was wrong. Considering that the normal characteristic function has a value of $0$ for this problem, it would make sense that this version has $1$ as an answer. However, I do not see how.
Can anyone explain this and show the steps?
Thank you!
Best Answer
If $g$ is the characteristic function of $C$ the $f(x)=1-g(x)$ for all $x$. Hence, $\int_0^{1} f(x)dx=\int_0^{1} 1dx-\int_0^{1} g(x)dx=1-0=1$,