I'm asking for proof verification of the part b) of the following problem.
Please note that $\mathfrak{M}_L$ denote the Lebesgue sigma algebra.
For all proof, fix $\epsilon > 0$
My proof:
(I'm using $\phi$ to denote the function in the question, sorry for the confusion.)
Claim: $\lim_{x \to \infty} \phi(x) = \mu_L(E)$
For this proof, we will denote $\mathbb{N} = \{0,1,2,3..... \}$
Let $E_0 = E \cap (-\infty, 0]$, and $E_n = E \cap (n-1, n]$ for $n \geq 1$.
Note that
$\{E_n\}$ is a disjoint collection since $(-\infty, 0]$ and $(n-1,n]$ are disjoint pairwise.
$\cup_{n \in \mathbb{N}} E_n = E$ since $(-\infty, 0]$ and $(n-1,n]$ form a partiton of $\mathbb{R}$
Hence, we have that
$$\mu_L(E) = \mu_L(\cup_{n \in \mathbb{N}} E_n) = \sum_{n \in \mathbb{N}} \mu_L(E_n)$$
Hence, we can choose $M$ so that $\mu_L(E) - \epsilon < \sum_{i = 0}^M \mu_L(E_i) < \mu_L(E)$
Now note that for all $x > M$,
$$\mu_L(E) \geq \phi(x) = \mu_L(E \cap (-\infty, x]) \geq \mu_L(E \cap (-\infty, M]) = \mu_L(\cup_{i=0}^M E_i) > \mu_L(E) - \epsilon$$
Hence, $\lim_{x \to \infty} \phi(x) = \mu_L(E)$
Claim: $\lim_{x \to -\infty} \phi(x) = 0$
Proof:
Consider $E_0 = E \cap (0, \infty)$ and $E_n = E \cap (-n, -n+1]$. $\{E_n\}$ is a partition of $E$. Hence, $\mu_L(E) = \sum_{n \in \mathbb{N}} \mu_L(E_n)$.
We can choose $M$ so that $\mu_L(E) - \epsilon < \sum_{i = 0}^M \mu_L(E_i) < \mu_L(E)$. Then $\sum^\infty_{i = M+1} \mu_L(E_i) < \epsilon$.
Choose such $M$. Then for all $x < -M-1$,
$$\phi(x) = \mu_L(E\cap (-\infty, x]) \leq \mu_L(E \cap (-\infty, -M]) = \mu_L(E) - \mu_L(\cup_{i=0}^{M} E_i) = \mu_L(E) - \sum_{i = 0}^{M} \mu_L(E_i) < \epsilon$$
Hence, $\lim_{x \to -\infty} \phi(x) = 0$
Thanks for your help!
The proof for (a) seems obvious. Consider the fact that x1, x2 belong to R, and x1> x2. For the aggregate function, the value on the right must be greater than 0, and it is increasing. For (b), Because it is considered on R, it can be obtained by using the contradiction method when it goes to negative infinity. The case of positive infinity is very trivial. Refer to the measurement theory of Euclidean space. In fact, this example is much like An example of measures extended to an algebra //Cannot be made into latex due to conditions