Lebesgue integral and Cantor set 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.
 A: 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...
A: If $f\colon X \to \mathbb{R}$ where $X \subseteq \mathbb{R}^n$ , $V\subseteq X$ and $\mu(V)=0$ then $\int\limits_{V}f=0$.
Proof:
$$\int\limits_Vf=\int\limits_Vf^+-\int\limits_Vf^-$$
where $f^+=\max(f,0)$ and $f^-=\max(-f,0)$. According to Lebesgue integration scheme we have:
$$\int\limits_Vf^+=\lim\limits_{\Delta\to 0^{+}}\lim\limits_{n\to +\infty}\sum_{i=1}^n\Delta\mu(A_i)\tag{1}$$
where $A_i=\{x\in V:f^+(x)>i\Delta\}$.
Note that $A_i\subseteq V$ , so $0\leqslant \mu(A_i)\leqslant \mu(V)=0$ and If I am taking into account $(1)$, I'll otain $\int\limits_Vf^+=0$, this same about $f^-$ what completes the proof.
A: The fact that you're integrating a function over a set of measure $0$ and then adding to that the integral of a function whose values are all $0$ means the integral will be $0$.
So your reasoning is correct.
The hint doesn't seem to have any bearing.
A: (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 $|f|$.)  
