I was reading Lebesgue Integration of arbitrary functions from my teacher's note.
Definition : $f$ is a non-negative integrable function defined on a measurable set $E$, then we define the lebesgue integral of $f$ as $$\int_E f d\mu = \text{sup} \{\int_E s d\mu : s\le f \text{where s is a simple function}\}$$
And from here he straight went to some observations and one of the observations was :
If $f$ is a non-negative measurable function defined on $(E, \sigma, \mu)$ and $A$ is a measurable subset of $E$. Then
$$ \int_A f d\mu = \int_E \chi_A f d\mu $$
Now when I'm trying to prove it from definition I'm having a lot of difficulty. Just a reminder that the property of integral :
$f$ is a non-negative measurable function defined on $(E, \sigma, \mu)$ and $A, B$ are measurable disjoint subset of $E$. Then
$$ \int_{A \cup B} f d\mu = \int_A f d\mu + \int_B f d\mu$$
haven't been introduced by now. So I need a proof without the help of this result. Also I looked up at Royden and Rudin analysis books but I couldn't find any independent proof of these two properties.
Any help is appreciated.
Note : The function mentioned above is not necessarily bounded and the measure of the set $E$ is not necessarily finite.
Best Answer
The restriction operation : $$\Big\{s:E\to[0,+\infty)~\Big|~ s \text{ is a simple function and }s\leq \chi_Af\Big\}\\\hookrightarrow \Big\{s:A\to[0,+\infty)~\Big|~s\text{ is a simple function and }s\leq f\Big\}$$ is a bijection, and for a non-negative simple function being zero off $A$, we have : $$\int_E s\text d\mu= \int_As\text d\mu$$
Therefore, using the definition of the Lebesgue integral, we find : $$\int_A f \text d\mu= \int_E\chi_Af\text d\mu$$