Integral over null set is zero I am sorry for this elementary question, but i could not figure out a rigorous proof for why the Lebesgue integral of any function over a null set is zero.
Thanks for helping!
 A: Hello this was asked over 7 years ago so I am sure that this is most definitely no longer relevant for you in particular but I thought I would answer this for people in the future who might was a slightly different answer. Infact that statement holds for any measure not just Lebesgue.
Firstly a little about the notation I will use:

$\mathbb{1}_A$ is the "indicator function" of the set $A$ you may have instead seen this as $\chi_A$ or $\mathcal{1}_A$ or something else. It simply takes the value $1$ on $A$ and $0$ otherwise.
This proof will require:
Theorem 1: 
If $f = g$ $ \mu$ almost everywhere then $\int f d\mu = \int g d\mu$
(A proof of this theorem can be found at the bottom of this answer but this is a fairly simple result in most measure theory courses.)
Proof of main result
We wish to show that $\int_A f d \mu = 0 $ whenever $\mu(A) = 0 $
Firstly $\int_A f d \mu = \int f \mathbb{1}_A d\mu $
Secondly the function $g := f \mathbb{1}_A $ is $0$ almost everywhere as by the definition of indicator functions it is non zero only on $A$ which again has measure zero from the question.
More formally let us define $ N := \{ x \in X | g(x) \not = 0 \} $ any $x \in N $ must be in $A$ because $g(x)$ for $x \not \in A = 0 $ hence $N \subseteq A$ Hence $\mu(N) = 0 $ (monotonicity of measure)
Hence $g = 0 $ almost everywhere and so by Theorem 1:  
$\int_A f d\mu = \int f \mathbb{1}_A d \mu = \int g d\mu = $(Theorem 1)$ = \int 0 d\mu = 0 $
A: Start with the definition. The Lebesgue integral of a simple function $s = \sum_{j=1}^n \alpha_j \ \chi_{A_j}$ is:
$$
\int_E s \,d\mu = \sum_{j=1}^n \alpha_i \ \mu(E \cap A_j)
$$
If $\mu(E) = 0$, then $\mu(E \cap A_j) = 0$ for all $j$. Thus $\int_E s \,d\mu = 0$.
The Lebesgue integral of a nonnegative function $f$ is the supremum of integrals of all simple functions $s$ such that $0 \le s \le f$. Since all of these integrals are $0$, the supremum is $0$ too.
Since every real function $f$ can be written as $f = f^+ - f^-$ where $f^+$ and $f^-$ are both nonnegative, we have $\int_E f \, d\mu = 0$ too. The general result follows from the fact that every complex function $f$ can be written as $f = u + i v$ where $u$ and $v$ are real.
A: Consider that $$\int_E f(x)\leq \sup|f(x)|\cdot m(E).$$ With the convention that $\infty\cdot 0=0$, we have $$\left|\int_E f(x)\right|\leq \sup|f(x)|\cdot m(E)=0.$$
