Lebesgue Integral on Measurespace I am reading "Real and Complex Analysis" from Rudin and could not really solve a presumably simple question I came up with. 
In the chapter where the L-Integral is constructed the measure space - $(X,\mathfrak{M},\mu)$ with set, $\sigma$-algebra and measure, respectively - is considered.
Now every integral theorem (Monotone Convergence, Fatou, Dominated Convergence)  is stated by integrating over the whole set $X%$. Since Rudin also proves the approximation theorem of measurable function using simple functions on all $X$, I am wondering in which sense the theorems can be transfered to problems, where I indeed have the measure space above, but only want to integrate over a subset $S\subset X$. 
I could use something like $f\chi_S$ which is again measurable if $f$ and the characteristic function $\chi$ are measurable, but if $f$ would be continuous on $S$ then I would destroy this property i.g. by extending to $X$. 
It would also of course be necessary that $S$ lies in $\mathfrak{M}$.  
 A: Yes, you are right, you can define integration of $f$ over a subset $S \subset X$ to be integrating $f$ times the indicator function of $S$:
\begin{align*}
\int_Sf d\mu := \int_X f \cdot \chi_S d\mu.
\end{align*}
You are right, $f$ may be continuous on $S$, but may not be continuous on the whole space $X$. There are no problems though, since continuity is a stronger condition than integrability. 
Also, you are right that $S$ needs to be measurable, i.e., $S \in \mathfrak{M}$.
A: You don't need a new concept to define:
$$\int_Sf\operatorname{d}\mu.$$
In fact, if $(X,\mathfrak{M},\mu)$ is a measure space, $S\in\mathfrak{M}$ and $f$ is measurable from $(X,\mathfrak{M})$ to $\mathbb{C}$, then the following hold:


*

*$\mathfrak{M}_S:=\{F\in\mathfrak{M}\ |\ F\subset S\}$ is a $\sigma$-algebra of subset of $S;$

*$\mu|_{\mathfrak{M}_S}$ is a measure on $\mathfrak{M}_S;$

*$f|_S$ is measurable from $(S,\mathfrak{M}_S)$ to $\mathbb{C}.$


So, if $f|_S\in L^1(S,\mathfrak{M}_S,\mu|_{\mathfrak{M}_S})$ you can define:
$$\int_Sf\operatorname{d}\mu:=\int_Sf|_S\operatorname{d}\mu|_{\mathfrak{M}_S}.$$
Then, it is a theorem that $f|_S\in L^1(S,\mathfrak{M}_S,\mu|_{\mathfrak{M}_S})$ if and only if $f \chi_S\in L^1(X,\mathfrak{M},\mu)$ and in this case it holds that:
$$\int_Sf\operatorname{d}\mu=\int_Xf \chi_S\operatorname{d}\mu.$$
You can prove this theorem via standard machine (for the standard machine technique, see e.g. David Williams - Probability with martingales, chapter 5).
