One-to-one correspondence between measures and linear functionals satisfying the MCP

Question

Let $(X,\Sigma)$ be a fixed measurable space. Let $M$ denote the set of $\Sigma$-measurable functions $f:X\to [-\infty,\infty]$, and let $M^+=\{f\in M:f\geq 0\}$.

A linear functional on $M^+$ is a function $T:M^+\to [0,\infty]$ which satisfies $T(\alpha f+\beta g)=\alpha T(f)+\beta T(g)$ for all $\alpha,\beta\in[0,\infty]$ and $f,g\in M^+$ (This is well-defined since $\alpha f+\beta g\in M^+$).

A functional $T:M^+\to [0,\infty]$ satisfies the monotone convergence property (MCP) if $T(\sup_n f_n)=\sup_n T (f_n)$ for any pointwise nondecreasing sequence $(f_n)\subset M^+$ (This is well-defined since $\sup_n f_n\in M^+$).

Am asked to answer the following questions:

1. Show that to each measure $\mu:\Sigma\to [0,\infty]$ correspond a unique linear functional $T:M^+\to [0,\infty]$ which satisfies the MCP and the property $\mu(A)=T(1_A)$ for all $A\in\Sigma$.

2. Show that each linear functional $T:M^+\to [0,\infty]$ satisfying the MCP induces a measure $\mu:\Sigma\to [0,\infty]$ via the assignment $\mu(A):=T(1_A)$ for $A\in\Sigma$.

3. Show that there is a one-to-one correspondence between measures on $(X,\Sigma)$ and linear functionals on $M^+$ satisfying the MCP.

4. For a given measure $\mu:\Sigma\to [0,\infty]$, show that the Lebesgue integral $\int d\mu:M^+\to [0,\infty]$ is the unique functional $T:M^+\to [0,\infty]$ satisfying the MCP, $T(f+g)=T(f)=T(g)$ for all $f,g \in M^+$, and $\mu(A):=T(1_A)$ for $A\in\Sigma$.

My attempt:

1. The Lebesgue integral $\int d\mu:M^+\to [0,\infty]$ already satisfies the required properties, so only uniqueness remains to be shown. Suppose $T_1,T_2$ both satisfies the required properties. Then $T_1,T_2$ agree if $f\in M^+$ is simple, say $f=\sum_{i=1}^n \alpha_i 1_{A_i}$ with $\alpha_i\in[0,\infty]$ and $A_i\in \Sigma$, because $$T_1\big(f)=\sum_{i=1}^n \alpha_i \mu(A_i)=T_2\big(f\big)$$ Now if $f\in M^+$ is arbitrary, then $f$ is the pointwise limit of a nondecreasing sequence of simple functions $(f_n)\subset M^+$ , and so from the MCP, $$T_1\big(f)=T_1(\sup_n f_n)=\sup_n T_1(f_n)=\sup_n T_2(f_n)=T_2(\sup_n f_n)=T_2(f)$$ It follows that $T_1=T_2$, which proves uniqueness.

2. We check the conditions for a measure. We have $\mu(\emptyset)=T(1_\emptyset)=T(01_\emptyset)=0T(1_\emptyset)=0$. If $(A_n)\subset \Sigma$ is pairwise disjoint then $\sum_{i=1}^n1_{A_i}\in M^+$ increases pointwise to $1_{\cup_{i=1}^\infty A_i}$, and so from the MCP $$\mu(\cup_{i=1}^\infty A_i)=T(1_{\cup_{i=1}^\infty A_i})=\sup_n T(\sum_{i=1}^n 1_{A_i})=\sup_n \sum_{i=1}^n \mu(A_i)=\sum_{i=1}^\infty \mu(A_i)$$ So $\mu$ is indeed a measure on $\Sigma$.

3. The assignment $\mu\mapsto \int d\mu$ is one-to-one by $1$ and onto by $2$. So it is a bijection from the set of measures on $(X,\Sigma)$ to the set of of linear functionals on $M^+$ satisfying the MCP.

4. The Lebesgue integral $\int d\mu:M^+\to [0,\infty]$ already satisfies the required properties, so only uniqueness remains to be shown. Suppose $T$ also satisfies the required properties. We only need to show that $T(\alpha f)=\alpha T(f)$ for all $\alpha\in[0,\infty]$, which will imply linearity of $T$, and then uniqueness will follow from $1$.We have $$T(0 f)=T(1_\emptyset)=\mu(\emptyset)=0=0T(f)$$ $$T(n f)=nT(f) \quad n\in \mathbb N, \quad \text{by induction using } T(f+g)=T(f)+T(g)$$ $$T(f)=T(n \frac{1}{n}f)=nT(\frac{1}{n}f)\implies\frac{1}{n}T(f)=T(\frac{1}{n}f) \quad n\in \mathbb N, $$ Therefore $T(rf)=rT(f)$ for any positive rational $r$. If $\alpha \in(0,\infty]$ choose an increasing sequence $(r_n)$ of positive rationals converging to $r$. Then $(r_nf)\subset M^+$ is a nondecreasing sequence converging pointwise to $\alpha f$ and so from the MCP $$T(\alpha f)=T(\sup_n r_nf)=\sup_nT(r_nf)=\sup_n r_n T(f)=\alpha T(f)$$

Is this correct? Am I missing something? Thanks a lot for your help.

Answer 1

Your work looks complete and thorough to me!