Restricting σ-algebra

Question

$\mathfrak{G}$ is σ-algebra.

If E $\in \mathfrak G$ , prove that resticting μ on $\mathfrak T$ = { A $\cap$ E : A $\in \mathfrak G$ } , μ is still a positive measure. Hence, for measureable function f : X $\rightarrow \mathbb R^*$, $\int_E fdμ$ can be defined, then prove $\int_E fdμ = \int_X fχ_Edμ$.

I tried to solve this problem by several steps.

1. f is a simple function.
2. f is not simple but nonnegative function.
3. general function.

When f is a simple function, it was easy to show that the last equation holds. (Because $\int_X fχ_Edμ = \sum c_i μ(E \cap E_i) = \int_E fdμ$ ) I want to claim that when it comes to nonnegative function f, by just applying sup front of sum, equality still holds.

I'm not sure that when f is nonnegative function, the last equality holds. Is it okay to use $sup{\sum c_i μ(E\cap E_i)}$? Does this value equal to each of integral?

Answer 1

Let $f:X\to[0,\infty]$ be measurable. We can then find a non-decreasing sequence $\{f_j\}_{j\in\mathbb{N}}$ of simple functions $f_j:X\to[0,\infty)$ such that $f=\lim_{j\to\infty}f_j$ pointwise. It follows by the Monotone Convergence Theorem that

$$\int_E f~\mathrm{d}\mu=\lim_{j\to\infty}\int_Ef_j~\mathrm{d}\mu=\lim_{j\to\infty}\int_Xf_j\chi_E~\mathrm{d}\mu=\int_Xf\chi_E~\mathrm{d}\mu.$$

For a general measurable function $f:X\to\overline{\mathbb{R}}$ whose integral exists we have that

$$\int_E f~\mathrm{d}\mu=\int_Ef^+~\mathrm{d}\mu-\int_Ef^-~\mathrm{d}\mu=\int_Xf^+\chi_E~\mathrm{d}\mu-\int_Xf^-\chi_E~\mathrm{d}\mu=\int_X(f\chi_E)^+~\mathrm{d}\mu-\int_X(f\chi_E)^-~\mathrm{d}\mu=\int_X f\chi_E~\mathrm{d}\mu.$$