I need to evaluate the integral $\int_{[0,1]} f \; d\mu $ using Lebesgue integral
when $d\mu$ is Borel measurement and $f$ is given by:
$$ f(x) = \begin{cases}x &x\in C, \\
0&x\in[0,1]\setminus C ,
\end{cases}$$
$C$ is Cantor set.
I understand that $\mu(C)=0$, so doesn't it mean that-
$$\int_{[0,1]} f \; d\mu =\int_{[0,1]\backslash\ C} f \; d\mu+\int_C f \; d\mu = 0+\int_C x \; d\mu$$
and $\int_C x \; d\mu =0$ because $\mu(C)=0$ ?
I think I am misunderstanding something since I also get this "clue":
if $ m\leq f(x) \leq M $, then $ \int_Am\; d\mu\leq \int_Af\; d\mu \leq \int_AM\; d\mu $.
Thank you for your help.
Best Answer
Your answer and your argument are correct, the integral is $\int_{[0,1]}f(x)dx=0$. I don't get the clue either... Maybe they wanted an argument like this $\forall x \in [0,1]$ it holds that $f(x) <1$. Say that $f(x)=x <1$ only on a set of measure $0$, then run through your argument again... Although that seems quite pointless really...