Lebesgue Integration on a subset of a set X Let $(X, \mathcal{A}, μ)$ be a measure space. Let $f:X \rightarrow \mathbb{R}$ be a measurable function. Let $A \in \mathcal{A}$ with μ(A)=0. Then prove that
$$\int_A f dμ=0.$$
My try:
$$\int_A f dμ=\int_X (f \chi_A) dμ=0$$
How $\int_X (f \chi_A) dμ=0$
What are the next steps to reach at the conclusion?
Pls need help in this problem.
 A: Let $$g_n(x) = \begin{cases}|f(x)| & \text{if }|f(x)| \leq n \\ n & \text{otherwise} \end{cases}$$
Observe that for each fixed $x$, the sequence $g_n(x)$ is monotonically increasing and converges to $|f(x)|$. Therefore the monotone convergence theorem applies to go from the second line to the third line below:
$$\begin{aligned}
\left|\int_X f \chi_A \ d\mu\right| &\leq \int_X |f \chi_A|\ d\mu \\
&= \int_X \left(\lim_{n \to \infty}g_n \chi_A\ \right) d\mu \\
&= \lim_{n \to \infty} \int_X g_n \chi_A\ d\mu \\
&\leq \lim_{n \to \infty} \int_X n \chi_A\ d\mu \\
&= \lim_{n \to \infty} n \int_X \chi_A\ d\mu \\
&= \lim_{n \to \infty} n\mu(A) \\
&= \lim_{n \to \infty} 0 \\
&= 0
\end{aligned}$$
A: From Dr. Murthy's comments( I think this proof is simpler).
It's enough to prove the result for a non-negative function $f.$ Assume $f \geq 0.$ Since $f$ is measureable and non-negative, there exists a sequence of monotonically increasing simple function  $(s_n)$ such that $ s_n$ converges to $f.$ Then by the MCT,
$ \int_{A} f  =   
\lim_{n \to \infty} \int_{A} s_{n} = 0.$ The last equality is due to definition of integrals.
A: Let $f$ be a positive measurable function and $A\subseteq \mathbb R^n$ s.t. $m(A)=0$.
For any simple function $s\in\mathcal S_f=\{s:s\text{ simple and measurable such that } 0\le s\le f\}\implies$
$$0\le \sum_{i=1}^m c_i m(A\cap E_i)=0$$
where $\{c_1,\dots,c_m\}$ is the image of $s$,  $E_i=s^{-1}(c_i)$ and the sum is the definition of integral of a simple function on a measurable set $A$.
Remember that given $f:E\subset \mathbb R^n\to [0,+\infty]$, with $E\in \mathcal M(\mathbb R^n)$ you define
$$\int_E s(x)d\mu:=\int_{\mathbb R^n}s(x)\chi_E(x)=\sum_{i=1}^m c_i m(E_i\cap E)$$
$$\int_E f(x)d\mu:=\sup_{s\in\mathcal S_f}\int_E s(x)d\mu=\sup_{s\in\mathcal S_f}\int_{\mathbb R^n}s(x)\chi_E(x)d\mu.$$
