Lebesgue integral and Cantor set

Question

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.

Answer 1

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...

Answer 2

(same idea as in your answer and other answers, but using a form of the clue: not on integrals, but on null a.e property)

If $f$ null a.e. then the integral of $f$ exists and it is null. This can be shown for simple functions first. Then the simple functions dominated by $f$ must be null a.e. (this is maybe where the clue comes in), hence their integrals are null. Finally, the integral of $f$ must be null (supremum of null numbers).

(I assumed $f$ is non-negative in reasoning I presented, but it can be extended to any function by using the fact that both positive and negative parts of $f$ are dominated by $|f|$.)