Let $(X, \mathcal{M}, \mu)$ be a measure space. Suppose $E \in \mathcal{M}$ and $f \in L^+$ where $L^+$ is a space of measurable functions from $X$ to $[0, \infty]$. $\int_E f$ is defined by $\int_X f\chi_E$ where $\chi_E$ is a characteristic function of $E$. Now every time we want to use some property that is true for integrals over whole space $X$, we have to do manipulations with $\chi_E$. For example, let $f_n$ be a sequence in $L^+$ such that $f_n(x) \nearrow f(x)$ for all $x \in E$. Suppose we know that monotone convergence theorem is true for integrals over $X$ and we want to prove
$$\int_E f = \lim_{n\to\infty} \int_E f_n.$$
Since $f_n\chi_E \nearrow f\chi_E$ we have
$$\int_E f = \int_X f\chi_E = \lim_{n\to\infty} \int_X f_n\chi_E = \lim_{n\to\infty} \int_E f_n.$$
I know it's easy but is there a way to avoid such manipulations? I want to know that equality above and many other statements about integrals are true because $E$ is a measure space in its own right (seems much more natural).
More precisely, for $E \in \mathcal{M}$ define
$$\mathcal{M}_E = \{ E \cap F \mid F \in \mathcal{M} \}.$$
It's easy to check that $\mathcal{M}_E$ is a $\sigma$-algebra. $(E, \mathcal{M}_E, \mu|_{\mathcal{M}_E})$ is a measure space and $\int_E f$ in space $(X, \mathcal{M}, \mu)$ is equal to $\int_E f|_E$ in space $(E, \mathcal{M}_E, \mu|_{\mathcal{M}_E})$ for every measurable $f$.
Is this a commonly accepted way of handling integrals over subset of measure space? If not, do I really have to use $\chi_E$ every time or there is some other way?
Best Answer
There are basically two interesting types of functions
$$f:X \to [0,\infty] , \mu-\text{ measurable} $$
and
$$f : X \to [- \infty, \infty] \text{ which is integrable, i.e. } \int f(x) \mu(dx) < \infty .$$
These functions are interesting, because you can define for both the integral $\int_X f(x) \mu(dx)$, but maybe the integral is infinity in the first case.
Many theorems are true for either non-negative measurable functions (e.g. monotonic convergence theorem) or integrable functions (dominant convergence theorem).
Now if you have proven the monotonic convergece theorem for a general measurable space, then you don't need to prove it for integrals restricted on a set, simply because $(E, \mathcal{M}_E, \mu|_{\mathcal{M}_E})$ is again a measure space, and thus the monotonic convergence theorem is also true here. As well as all other properties and theorems that are known for integrals on a measurable space.
So in your case,if you have non-negative functions $f_n(x) \nearrow f(x) $ for all $x\in E$ you immediately get
$$\int f(x) \mu|_{\mathcal{M}_E}(dx) = \lim_{n\to \infty} \int f_n(x) \mu|_{\mathcal{M}_E}(dx) $$
However, to be sure that the theorem really holds for all functions restricted to $E$ you need to understand the relation between integrable functions of $\mu_{\mathcal{M}_E}$ and functions $f$ restricted from $X$ to $E$. The relation is as follows:
Let $f:X \to [0,\infty]$ be $\mu$-measurable, or let $f:X \to [-\infty,\infty]$ be $\mu$ integrable. Define for some set $E \in \mathcal{M}$ the function $f':A \to [-\infty, \infty]$ with $f'(x) := f(x)$. Then, we have
$$ \int f'(x) \mu_{\mathcal{M}_E}(dx) = \int_A f(x) \mu(dx) $$