Exercise 1.4.49 (Characterisation of the unsigned integral)

$$ As always, from the additivity we can immediately deduce monotonicity, i.e., for $f,g\in U(X,B):f\le g$ we have

We demonstrate the existence and the uniqueness of $\mu$.

• (Existence)
  We argue that $\mu (E)=I(1_E)$ satisfies the desired criteria. First, we must check whether $\mu$ is a measure at all. Obviously, $\mu$ is a non-negative function.

  (i)

  item We have $\mu (\varnothing )=I(1_{\varnothing })=I(0)=I(0\cdot 0)=0\cdot I(0)=0$.

  (ii)

  Let $E_1,E_2,\dots$ be the sequence of disjoint measurable sets. We then have, by combining finite additivity and the stability of $I$ under limits, $$\mu \left(\bigcup _{n=1}^{\infty }{E}_n\right)=I\left(1_{\bigcup _{n=1}^{\infty }{E}_n}\right)=I\left(\sum _{n=1}^{\infty }{1}_{E_n}\right)=I\left(\underset{N\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}\sum _{n=1}^N{1}_{E_n}\right)=\underset{N\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}I\left(\sum _{n=1}^N{1}_{E_n}\right)=\underset{N\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}\sum _{n=1}^{N}I\left(1_{E_n}\right)=\sum _{n=1}^{\infty }\mu (E_n)$$

  Now we check whether $\mu$ satisfies the desired property. We have

  $${\int \; <!--\; nolimits\; -->}_{X}fd\mu =\sup <!--\; FUNCTION\; APPLICATION\; -->\left\{\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{X}hd\mu :\begin{array}{c}h\text{ is simple }\hfill \\ h\le f\hfill \end{array}\right\}.$$

  The integral of any simple $h$ w.r.t. $\mu$ can be broken down to $\mathrm{Simp}{\int }_{X}hd\mu ={\sum }_{i=1}^n{c}_i\mu (F_i)$ for some non-negative reals $c_1,\dots ,c_n\in [0,+\infty ]$ and sets $F_1,\dots ,F_n\subseteq X$ such that $h(F_i)=\{c_i\}$. We then have

  $$\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{X}hd\mu =\sum _{i=1}^n{c}_i\mu (F_i)=\sum _{i=1}^n{c}_{i}I(1_{F_i})=\sum _{i=1}^{n}I(c_i{1}_{F_i})=I\left(\sum _{i=1}^n{c}_i{1}_{F_i}\right)=I(h)$$

  Thus, the theorem is true for any simple function. Then, it is not hard to extend this property to any unsigned function using the Monotone Convergence property of $I$. From the supremum set in the definition of integral we pick a sequence of non-decreasing $0\le h_1\le h_2\le \dots$ simple functions ${(h_n)}_{n\in \mathbb{N}}$ such that $\int h_n\to \int \; <!--\; nolimits\; -->f$. On one hand, we have:

  $${\int \; <!--\; nolimits\; -->}_{X}fd\mu =\underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}{\int \; <!--\; nolimits\; -->}_X{h}_{n}d\mu =\underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}{h}_n={\int \; <!--\; nolimits\; -->}_{X}hd\mu$$

  on the

  $${\int \; <!--\; nolimits\; -->}_{X}fd\mu =\underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}{\int \; <!--\; nolimits\; -->}_X{h}_{n}d\mu =\underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}I\left(h_n\right)=I\left(\underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}{h}_n\right).$$

• (Uniqueness)
  Now lets assume there is another ${\mu }^{\prime }$ such that

  $$\forall <!--\; FUNCTION\; APPLICATION\; -->f\in U(X,B):I(f)={\int \; <!--\; nolimits\; -->}_{X}fd{\mu }^{\prime }.$$

  In particular, for all measurable $E\in B$ we have

  $$\mu (E)={\int \; <!--\; nolimits\; -->}_X{1}_{E}d\mu =I(1_E)={\int \; <!--\; nolimits\; -->}_X{1}_{E}d{\mu }^{\prime }={\mu }^{\prime }(E)$$

  by compatibility of measure and indicator integral.