Lebesgue integral on a null set is zero Sorry for this (maybe too) elementary question but if $N\subset\mathbb{R}^2$ is a null set for the Lebesgue measure, i.e. $\lambda(N)=0$, and $f$ is an Lebesgue integrable function why is then
$$
\int_N f\, d\lambda=0?
$$

Suppose $N$ is a set such that $\lambda(N) = 0$.  ($\lambda$ can be any measure, not just Lebesgue.)
Let's consider positive functions only because if $f$ is not positive, we know $f = f^{+} - f^{-}$ ($f$ can be decomposed into its positive and negative parts) and since $$\int \limits_{N} f \,d\lambda = \int\limits_{N} f^{+} \,d\lambda - \int \limits_{N} f^{-} \,d\lambda$$ if each of the terms on the right hand side is $0$ then so will the left hand side.
So assume without loss of generality that $f$ is positive.  How is the Lebesgue integral defined?  It is defined as:
$$\int\limits_{N} f \,d\lambda := \sup \left \{ \int\limits_{N} s \,d\lambda : 0 \leq s \leq f \text{ and } s \text{ is simple}   \right \}$$
Ok, so if we have any hope of calculating $\int\limits_{N} f \,d\lambda$, at this point it looks like we can do it if we can calculate $\int\limits_{N} s \,d\lambda$ for any simple function $s$ with $0 \leq s \leq f$.
Alright, let's work on that then.  Suppose $s$ is a simple function such that $0\leq s \leq f$.  We know $s(x) = \sum \limits_{i = 1}^{n} a_{i} \chi_{A_{i}}(x)$ because that's the general form of a simple function. ($\chi$ is an indicator function.)
Also, $$\int \limits_{N} s(x) \,d\lambda = \sum \limits_{i = 1}^{n} a_{i} \lambda(A_{i} \cap N).$$
But $A_{i} \cap N \subseteq N$ and $\lambda(N) = 0$, so $\lambda(A_{i} \cap N) = 0$.
Then our integral becomes  $$\int \limits_{N} s(x) \,d\lambda = \sum \limits_{i = 1}^{n} (a_{i} \cdot 0) = \sum \limits_{i =1}^{n} 0 = 0.$$
So, on the null set $N$ (null set just means set of measure $0$), we found that if $f$ is positive, then for any simple function $s$ with $0 \leq s \leq f$, the integral of $s$ over $N$ is $0$.  So $\sup \left \{ \int\limits_{N} s \,d\lambda : 0 \leq s \leq f \text{ and } s \text{ is simple}   \right \}$ is really just the supremum of a set of zeros, which makes it $0$.  Thus, $\int \limits_{N} f \,d\lambda = 0$.
